What Are Advanced Integration Techniques in Mathematics?

[alyafey22]

Re: Integration lessons (continued)

4.7.2 Exercises

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Exercise 1

\(\displaystyle \int^\infty_0 \text{erfc}^2(x)\,dx = \frac{2-\sqrt{2}}{\sqrt{\pi}}
\)​

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Proof

Integration by parts

$$I = x\text{erfc}^2(x)\left. \right]^\infty_0 -2\int^\infty_0 x\text{erfc}'(x)\text{erfc}(x)\,dx $$

The first integral goes to 0

$$I = -2\int^\infty_0 x\,\text{erfc}'(x)\text{erfc}(x)\,dx$$

The derivative of the complementary error function

$$\text{erfc}'(x) = (1-\text{erf}(x))'= -\frac{2}{\sqrt{\pi}}e^{-x^2}$$

$$I= \frac{4}{\sqrt{\pi}}\int^\infty_0xe^{-x^2}\text{erfc}(x)\,dx$$

Integration by parts again we have

$$I=\frac{2}{\sqrt{\pi}}e^{-x^2}\text{erfc}(x)\left. \right]^\infty_0-\frac{4}{\pi}\int^\infty_0e^{-2t^2}\,dt $$

At infinity the integral goes to 0 and at 0 we have

$$\frac{2}{\sqrt{\pi}}e^{-x^2}\text{erfc}(x)_{x=0}=\frac{2}{\sqrt{\pi}}(1-\text{erf}(0))=\frac{2}{\sqrt{\pi}}$$

$$\int^\infty_0e^{-2t^2}\,dt = \frac{\sqrt{\pi}}{2\sqrt{2}}$$

Collecting the results together we have

$$I = \frac{2}{\sqrt{\pi}}-\frac{\sqrt{\pi}}{2\sqrt{2}} \times \frac{4}{\pi}=\frac{2-\sqrt{2}}{\sqrt{\pi}} $$Exercise 2

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\(\displaystyle \int^\infty_0 \sin(x^2) \text{erfc}(x) \,dx =\frac{\pi-2 \coth^{-1}\sqrt{2}}{4 \sqrt{2\pi}}\)​

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Proof

Using the substitution $ x=\sqrt{t}$

$$\frac{1}{2}\int^\infty_0 \sin(t) t^{-\frac{1}{2}}\text{erfc}(\sqrt{t}) \,dx $$

Consider the function

$$I(a) = \frac{1}{2}\int^\infty_0 \sin(t) t^{-\frac{1}{2}}\text{erfc}(a\sqrt{t}) \,dt$$

Differentiating with respect to $a$ we have

$$I'(a) = \frac{-1}{\sqrt{\pi}}\int^\infty_0 \sin(t) e^{-a^2t}\,dt=\frac{-1}{\sqrt{\pi}}\cdot \frac{1}{a^4+1}$$

Now integrating with respect to $a$

$$I(a) = \frac{-1}{\sqrt{\pi}}\int^a_0 \frac{dx}{x^4+1}+C$$

To evaluate the constant we take $a \to \infty$

$$I(\infty) = \frac{-1}{\sqrt{\pi}}\int^\infty_0\frac{dx}{x^4+1}+C$$

The function has an anti-derivative and the value is

$$\frac{-1}{\sqrt{\pi}} \int^\infty_0\frac{dx}{x^4+1}=-\frac{\sqrt{\pi}}{2\sqrt{2}}$$

and knowing that

$$\text{erfc}(\infty)=0 \,\,\, \implies \,\,\,\, C =\frac{\sqrt{\pi}}{2\sqrt{2}} $$

Finally we get

$$I(a) = \frac{-1}{\sqrt{\pi}}\int^a_0 \frac{dx}{x^4+1}+\frac{\sqrt{\pi}}{2\sqrt{2}}$$

Plugging $a=1$ we have our integral

$$I(1) = \int^\infty_0 \sin(x^2) \text{erfc}(x) \,dx=\frac{\sqrt{\pi}}{2\sqrt{2}}-\frac{1}{\sqrt{\pi}}\int^1_0 \frac{dx}{x^4+1}$$

Also knowing that

$$\frac{1}{\sqrt{\pi}}\int^1_0 \frac{dx}{x^4+1}=\frac{\pi+2 \coth^{-1}\sqrt{2}}{4 \sqrt{2\pi}}$$

$$ \int^\infty_0 \sin(x^2) \text{erfc}(x) \,dx=\frac{\pi-2 \coth^{-1}\sqrt{2}}{4 \sqrt{2\pi}}$$

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[HW]

Can you find closed forms for

\(\displaystyle \int^\infty_0 \text{erfc}^3(x)\, dx = ?\)

\(\displaystyle \int^\infty_0 \text{erfc}^4(x)\, dx = ? \)

What about

\(\displaystyle \int^\infty_0 \text{erfc}^n(x)\, dx = ? \)

 

[alyafey22]

Re: Integration lessons (continued)

4.Integration using special functions (continued)

4.8 Exponential integral function

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Definition

\(\displaystyle E(x)=\int^\infty_x\frac{e^{-t}}{t}\,dt\)​

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4.8.1 Exercises

Prove that

$$\lim_{x \to 0}\left[\log(x)+E(x) \right]=-\gamma$$

Integration by parts for $E(x) $

$$E(x)=e^{-t}\log(t) \left.\right]^\infty_x+\int^\infty_x\log(x)e^{-t}\,dt$$

$$E(x)=-e^{-x}\log(x)+\int^\infty_x\log(x)e^{-t}\,dt$$

$$\lim_{x \to 0}\left[\log(x)+E(x) \right]=\lim_{x \to 0}\left(\log(x)-e^{-x}\log(x)\right)+\int^\infty_0\log(x)e^{-t}\,dt$$

The first limit goes to 0

$$\lim_{x \to 0}\left[\log(x)+E(x) \right]=\int^\infty_0\log(x)e^{-t}\,dt=\psi(1)=-\gamma$$

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Prove that

$$\int^\infty_0 x^{p-1} E(ax)\,dx =\frac{\Gamma(p)}{pa^{p}} \,\,\, p>0$$Integrating by parts we have

$$\int^\infty_0 x^{p-1} E(ax)\,dx=\frac{1}{p}x^pE(ax) \left. \right]^\infty_0+\frac{1}{ap} \int^\infty_0 x^{p-1}e^{-ax}\,dx$$

The first limit goes to 0

$$I=\frac{1}{ap} \int^\infty_0 x^{p-1}e^{-ax}\,dx= \frac{1}{pa^p }\int^\infty_0 x^{p-1}e^{-x}\,dx=\frac{\Gamma(p)}{p \, a^p}$$

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Prove the more general case

$$\int^\infty_0 x^{p-1}e^{ax}E(ax) \, dx = \frac{\pi}{\sin(a\pi)}\cdot \frac{\Gamma(p)}{a^p}$$

Switch to the integral representation

$$\int^\infty_0 x^{p-1}e^{ax}\int^\infty_{ax}\frac{e^{-t}}{t}\,dt \, dx $$

Use the substitution $t=ax \,y$

$$\int^\infty_0\int^\infty_{1} x^{p-1}e^{ax}\frac{e^{-ax\,y}}{y}\,dy \, dx $$

By switching the two integrals

$$\int^\infty_1\frac{1}{y}\int^\infty_{0} x^{p-1}e^{-ax(y-1)} dx\,dy $$

By the Laplace identities

$$\frac{\Gamma(p)}{a^p}\int^\infty_1 \frac{1}{y(y-1)^p} \, dy$$

Now let $y=1/x$

$$\frac{\Gamma(p)}{a^p}\int^1_0 x^{p-1}(1-x)^{-p} \, dx$$

Using the the reflection formula for the Gamma function

$$\frac{\Gamma(p)}{a^p}\int^1_0 x^{p-1}(1-x)^{-p} \, dx = \frac{\pi}{\sin(a\pi)}\cdot \frac{\Gamma(p)}{a^p}$$

 

[alyafey22]

Re: Integration lessons (continued)

7.8.1 Exercises (continued)

Prove that

$$\int^\infty_0 e^{z}E^2(z) \, dz = \frac{\pi^2}{6}$$

Using the integral representation


$$E^2(z) = \int^\infty_1\int^\infty_1\frac{e^{-x z}e^{-y z}}{x y}\,dx \,d y$$

$$\int^\infty_0 e^{z}E^2(z) \, dz = \int^\infty_0 \int^\infty_1 \,\int^\infty_1\frac{e^{-z(x+y-1)}}{x y}\,dx \,d y \, dz$$

Switching the integration

$$ \int^\infty_1 \frac{1}{y} \int^\infty_1 \frac{1}{x}\,\int^\infty_0e^{-z(x+y-1)}\,dz\,dx \,d y $$

$$\int^\infty_1 \frac{1}{y} \int^\infty_1\frac{1}{x(x+y-1)}\,dx \, dy$$

The inner integral is an elmetnary integral

$$\int^\infty_1\frac{1}{x(x+y-1)}\,dx = -\frac{\log(y)}{1-y}$$

$$\int^\infty_1 \frac{\log(y)}{y(y-1)} \,dy$$

Now use the substitution $y=1/x$

$$-\int^1_0 \frac{\log(x)}{(1-x)} \,dx=-\int^1_0\frac{\log(1-x)}{x}\,dx = \text{Li}_2(1)=\frac{\pi^2}{6}$$

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Prove that

$$\int^\infty_0 z^{p-1} E^2(z)\,dz=\frac{2\Gamma(p)}{p^2}\, _2F_1(p,p;p+1;-1)$$

Consider the general case

$$F(p)=\int^\infty_0 z^{p-1} E^2(z)\,dz$$

Integrating by parts

$$F(p)=\frac{2}{p} \int^\infty_0 z^{p-1}e^{-z}E(z)\,dz$$

Write the integral representation

$$\frac{2}{p} \int^\infty_0 z^{p-1}e^{-z}\int^\infty_1 \frac{e^{-zt}}{t},dtdz$$

$$\frac{2}{p} \int^\infty_1\frac{1}{t}\int^\infty_0 z^{p-1}e^{-z(1+t)} \,dz\,dt$$

Take the inner integral

$$\frac{2\Gamma(p)}{p} \int^\infty_1\frac{dt}{t(1+t)^p}$$

Use the substitution $t=1/x$

$$\frac{2\Gamma(p)}{p} \int^1_0\frac{x^{p-1}}{(1+x)^p}\,dx$$

Using the Hypergeomtirc identity

$$\beta(c-b,b) \, _2F_1(a,b;c;z)=\int_0^1 \frac{x^{b-1}(1-x)^{c-b-1}}{(1-xz)^a}\, dx$$

put $c=p+1;b=p;a=p;z=-1$

$$\beta(1,p) \, _2F_1(p,p;p+1;-1)=\int_0^1 \frac{x^{p-1}}{(1+x)^p}\, dx$$

Hence the result

$$\int^\infty_0 z^{p-1} E^2(z)\,dz=\frac{2\Gamma(p)}{p^2}\, _2F_1(p,p;p+1;-1)$$

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[HW]

Find a general formula for

$$\int^\infty _0 x^n E^2(x) \, dx = ? \,\,\,\,\,\, n \in \mathbb{N}$$

 

[alyafey22]

Re: Integration lessons (continued)

4.Integration lessons (continued)

4.9.Complete elliptic integrals

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Complete elliptic of first kind

\(\displaystyle K(k)= \int^{\frac{\pi}{2}}_0\frac{d\theta}{\sqrt{1-k^2\sin^2 \theta }}=\int^1_0\frac{dx}{\sqrt{1-x^2}\sqrt{1-k^2x^2}}\)​

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Complete elliptic of second kind

\(\displaystyle E(k)= \int^{\frac{\pi}{2}}_0\sqrt{1-k^2\sin^2 \theta }\, d\theta =\int^1_0\frac{\sqrt{1-k^2x^2}}{\sqrt{1-x^2}}dx\)​

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Some special values

$$K(i) = \frac{1}{4\sqrt{2\pi}}\Gamma^2\left( \frac{1}{4}\right)$$

By definition we have

$$K(i)=\int^1_0\frac{dx}{\sqrt{1-x^2}\sqrt{1+x^2}}=\int^1_0 \frac{dx}{\sqrt{1-x^4}}$$

Let $x=\sqrt[4]{t}$ we have $dx=\frac{1}{4}t^{\frac{-3}{4}}\,dt$

$$K(i)=\frac{1}{4}\int^1_0t^{\frac{-3}{4}}(1-t)^{\frac{-1}{2}}\,dt $$

By beta function

$$K(i) = \frac{\Gamma\left( \frac{1}{4}\right)\Gamma\left( \frac{1}{2}\right)}{4\Gamma\left( \frac{3}{4}\right)}$$

By reflection formula

$$\Gamma\left( \frac{1}{4}\right)\Gamma\left( \frac{3}{4}\right)=\pi \csc\left( \frac{\pi}{4}\right)=\pi \sqrt{2}$$

$$K(i) = \frac{\Gamma^2\left( \frac{1}{4}\right)\Gamma\left( \frac{1}{2}\right)}{4\pi \sqrt{2}}=\frac{\Gamma^2\left( \frac{1}{4}\right)}{4\sqrt{2\pi}}$$

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$$ E(i)=\frac{\Gamma^2\left( \frac{1}{4}\right)}{4\sqrt{2\pi}}+\frac{\Gamma^2\left(\frac{3}{4} \right)}{\sqrt{2\pi}} $$

By definition we have

$$E(i)=\int^1_0\frac{\sqrt{1+x^2}}{\sqrt{1-x^2}}dx$$

Separating the two integrals

$$E(i)=\int^1_0\frac{1+x^2}{\sqrt{1-x^4}}dx=\int^1_0\frac{1}{\sqrt{1-x^4}}dx+\int^1_0\frac{x^2}{\sqrt{1-x^4}}dx$$

The first integral is $K(i)$ for the second integral use $x=\sqrt[4]{t}$

$$\frac{1}{4}\int^1_0t^{\frac{3}{4}-1}(1-t)^{-\frac{1}{2}}\,dt=\frac{\Gamma\left( \frac{3}{4}\right)\Gamma\left( \frac{1}{2}\right)}{4\Gamma\left( \frac{5}{4}\right)}=\frac{\Gamma^2\left(\frac{3}{4} \right)}{\sqrt{2\pi}}$$

Hence we have

$$E(i) = K(i) +\frac{\Gamma^2\left(\frac{3}{4} \right)}{\sqrt{2\pi}} = \frac{\Gamma^2\left( \frac{1}{4}\right)}{4\sqrt{2\pi}}+\frac{\Gamma^2\left(\frac{3}{4} \right)}{\sqrt{2\pi}}$$

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4.9.2.Some integrals involving elliptic integrals

$$\int^1_0 K(k) \, dk = 2G \,\,\,\,\, ;\,G=\text{Catalan's constant }$$

$$I=\int^1_0 \int^1_0\frac{1}{\sqrt{1-x^2}\sqrt{1-k^2x^2}} \,dx\, dk$$

Switching the two integrals

$$I=\int^1_0 \frac{1}{\sqrt{1-x^2}}\int^1_0\frac{1}{\sqrt{1-k^2x^2}} \,dk\, dx$$

$$I=\int^1_0 \frac{\arcsin x}{x\sqrt{1-x^2}} dx$$

Now let $\arcsin x = t$ hence we have $x=\sin t$

$$I=\int^{\frac{\pi}{2}}_0 \frac{t}{\sin \, t} dt$$

The previous integral is a representation of the constant

$$G=\frac{I}{2} \,\,\, \implies \,\,\, I=2G$$

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[HW]

Find the Values

$$E(0) \,\,\, , \,\,\,\, K \left( \frac{1}{\sqrt{2}}\right)$$

 

[alyafey22]

Re: Integration lessons (continued)

4.9.3.Properties of elliptic integrals

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\(\displaystyle \tag{1}K(\sqrt{k}) = \frac{1}{\sqrt{1-k}}K\left(\sqrt{\frac{k}{k-1} }\right)\,\,;\,k\notin [1,\infty)\)​

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Starting by the integral representation

$$K(k)=\int^1_0 \frac{dx}{\sqrt{1-x^2}\sqrt{1-k^2x^2}}$$

By using that $x=\sqrt{1-y^2}$

$$\int^1_0\frac{y\,dy}{\sqrt{1-y^2}\sqrt{1-(1-y^2)}\sqrt{1-k^2(1-y^2)}}$$

$$\int^1_0\frac{ dy}{\sqrt{1-y^2}\sqrt{1-k^2+k^2y^2}}=\int^1_0\frac{ dy}{\sqrt{1-k^2}\sqrt{1-y^2}\sqrt{1-\frac{k^2}{k^2-1}y^2}}$$

$$K(k)=\frac{1}{\sqrt{1-k^2}}K\left( \sqrt{\frac{k^2}{k^2-1}}\right)$$

which is equivalent to the property by taking the root

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\(\displaystyle \tag{2} E(\sqrt{k}) = \sqrt{1-k}\,E\left(\sqrt{\frac{k}{k-1}} \right)\,\,;\,k\notin [1,\infty)\)​

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Staring by the integral representation

$$E(\sqrt{k})=\int^1_0 \frac{\sqrt{1-kx^2}}{\sqrt{1-x^2}}\,dx$$

By using that $x=\sqrt{1-y^2}$

$$E(\sqrt{k})=\sqrt{1-k}\int^1_0 \frac{\sqrt{1-\frac{k}{k-1}y}}{\sqrt{1-y^2}}\,dx=\sqrt{1-k}\,E\left(\sqrt{\frac{k}{k-1} }\right)$$

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Using the above formulas we can have some values For the value $k=-1$

$$K(i) = \frac{1}{\sqrt{2}}K\left(\frac{1}{\sqrt{2}} \right)$$

$$K\left(\frac{1}{\sqrt{2}} \right) = \sqrt{2}K(i)= \frac{1}{4\sqrt{\pi}}\Gamma^2\left( \frac{1}{4}\right)$$

Similarly we have

$$E\left ( \frac{1}{\sqrt{2}}\right)= \frac{\Gamma^2\left( \frac{1}{4}\right)}{8\sqrt{\pi}}+\frac{\Gamma^2\left(\frac{3}{4} \right)}{2\sqrt{\pi}}$$

Hence using $k\to -k$

$$K(\sqrt{k}\,i) = \frac{1}{\sqrt{1+k}}K\left(\frac{\sqrt{k}}{\sqrt{1+k}}\right)\,\,$$

$$E(\sqrt{k}\,i) = \sqrt{1+k}\,E\left(\frac{\sqrt{k}}{\sqrt{1+k}}\right)\,\,$$

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[alyafey22]

Re: Integration lessons (continued)

4.9.4.Elliptic integrals as a Hypergeometric function

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Definition

\(\displaystyle K(k)=\frac{\pi}{2}{}_2F_1\left(\frac{1}{2},\frac{1}{2},1,k^2 \right)
\)​

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Using the integral representation of the hypergeometric function

$$ \beta(c-b,b) \, _2F_1(a,b,c,z)=\int_0^1 \frac{t^{b-1}(1-t)^{c-b-1}}{(1-tz)^a}\, dt$$

Now use the substitution $t=x^2$ and $z=k^2$

$$ \beta(c-b,b) \, _2F_1(a,b,c,k^2)=2\int_0^1 \frac{x^{2b-1}(1-x^2)^{c-b-1}}{(1-k^2x^2)^a}\, dx$$Put $a=\frac{1}{2}\,;$ $b=\frac{1}{2}$ and $c=1$$$ \int_0^1 \frac{dx}{\sqrt{1-x^2}\sqrt{1-k^2x^2}}\, =\frac{1}{2}\beta(1/2,1/2) \, _2F_1\left(\frac{1}{2},\frac{1}{2},1,k^2 \right)$$

By the beta function we have

$$\frac{1}{2}\beta(1/2,1/2)=\frac{1}{2}\Gamma^2\left(\frac{1}{2} \right)=\frac{\pi}{2}$$

Hence the result

$$ K(k) =\frac{\pi}{2}\, _2F_1\left(\frac{1}{2},\frac{1}{2},1,k^2 \right)$$

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By the same approach we have

\(\displaystyle E(k) =\frac{\pi}{2}\, _2F_1\left(\frac{1}{2},-\frac{1}{2},1,k^2 \right)\)​

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A relation using the Quadratic transformation Start by the Quadratic transformation

$$ _2F_1\left( a,b,2b,\frac{4z}{(1+z)^2} \right)=(1+z)^{2a}\,_2F_1\left(a,a-b+\frac{1}{2},b+\frac{1}{2},z^2 \right). $$

Hence we can deduce by putting $a=b=1/2$

$$ K\left( \frac{2\sqrt{k}}{1+k} \right)=(1+k)K(k) $$

Or we have

$$ K(k)=\frac{1}{k+1}K\left( \frac{2\sqrt{k}}{1+k} \right) $$

Hence we have for $k=\frac{1}{\sqrt{2}}$

$$K\left(2 \sqrt{-4+3 \sqrt{2}} \right)=\frac{1+\sqrt{2}}{\sqrt{2}}K\left(\frac{1}{\sqrt{2}} \right)= \frac{1+\sqrt{2}}{4\sqrt{2\pi}}\Gamma^2\left( \frac{1}{4}\right)$$

For the elliptic integral of second kind using the hypergeomtric representation with $a=\frac{-1}{2}$ and $b=\frac{1}{2}$

$$ _2F_1\left( -1/2,1/2,1,\frac{4z}{(1+z)^2} \right)=(1+z)^{-1}{}_2F_1\left(-1/2,-1/2,1,z^2 \right)$$

The later hypergeometric series can be written in terms of elliptic integrals using some general contiguity relations

$${}_2F_1\left(-1/2,-1/2,1,z^2 \right)=\frac{2}{\pi}\left( 2 E(k)+(k^2-1)K(k)\right)$$

So we have

$$2 E(k)+(k^2-1)K(k)=(k+1)E\left(\frac{2\sqrt{k}}{1+k} \right)$$

For $k=\frac{1}{\sqrt{2}}$

$$E\left(2 \sqrt{-4+3 \sqrt{2}} \right)=\frac{\sqrt{2}}{1+\sqrt{2}}\left[\frac{\Gamma^2\left( \frac{1}{4}\right)}{8\sqrt{\pi}}+\frac{\Gamma^2\left(\frac{3}{4} \right)}{\sqrt{\pi}} \right]$$

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Values for complex arguments Start by the following

$$K\left(\frac{2\sqrt{k}}{1+k} \right)=\int^1_0 \frac{1}{\sqrt{1-x^2}\sqrt{1-\frac{4k}{(1+k)^2}x^2}}\,dx$$

By some simplifications we have

$$E\left(\frac{2\sqrt{k}}{1+k} \right)=(1+k)\int^1_0 \frac{1}{\sqrt{1-x^2}\sqrt{(1+k)^2-4k \, x^2}}\,dx$$

Use $x=\sqrt{1-y^2}$

$$\int^1_0 \frac{1+k}{\sqrt{1-y^2}\sqrt{(1+k)^2-4k\,(1-y^2)}}\,dy=\frac{1+k}{1-k}\int^1_0 \frac{1}{\sqrt{1-y^2}\sqrt{1+\frac{4k}{(1-k)^2}y^2}}\,dy$$

Hence we have

$$K\left(\frac{2\sqrt{k}}{1+k} \right)=\frac{1+k}{1-k}K\left(\frac{2\sqrt{-k}}{1-k} \right)$$

Similarly we have

$$E\left(\frac{2\sqrt{k}}{1+k} \right)=\frac{1-k}{1+k}E\left(\frac{2\sqrt{-k}}{1-k} \right)$$

Using these formulas and the results we got earlier we have for $x=1/\sqrt{2}$
$$K\left(2 \sqrt{-4-3 \sqrt{2}} \right)=\frac{\pi\,\sqrt{\pi}}{4} \cdot \frac{2-\sqrt{2}}{\Gamma^2\left( \frac{3}{4}\right)}$$

$$E\left(2 \sqrt{-4-3 \sqrt{2}} \right)=\frac{(2+\sqrt{2})\left(\pi^2+4 \Gamma^4\left( \frac{3}{4}\right)\right)}{4\sqrt{\pi}\Gamma^2\left( \frac{3}{4}\right)}$$

 

[alyafey22]

Re: Integration lessons (continued)

4.9.5.Differentiation of elliptic integrals


Note : We should remove the variable $k$ and denote elliptic integrals $E$ and $K$ once there is no confusion. It is assumed that the variable is $k$ when we use these symbols.

Differentiation:

Interestingly the derivative of elliptic integrals can be written in terms of elliptic integralsDerivative of complete elliptic integral of second kind

$$\frac{d}{dk}E=\int^1_0 \frac{\frac{\partial\,}{\partial\,k}\sqrt{1-k^2 x^2}}{\sqrt{1-x^2}}\,dx$$

$$\frac{d}{dk}E=\int^1_0 \frac{-k\,x^2}{\sqrt{1-x^2}\sqrt{1-k^2\,x^2}}\,dx$$

Adding and subtracting 1 results in

$$\frac{1}{k}\int^1_0 \frac{\sqrt{1-k^2x^2}}{\sqrt{1-x^2}}\,dx-\frac{1}{k}\int^1_0 \frac{dx}{\sqrt{1-x^2}\sqrt{1-k^2\,x^2}}$$

Upon realizing the relation to elliptic integrals we conclude

\(\displaystyle \frac{d}{dk}E=\frac{E-K}{k}\)​

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For the complete elliptic integral of first kind we need more work

Start by the following

$$\frac{d}{dk}K=\int^1_0 \frac{1}{\sqrt{1-x^2}}\,\frac{\partial\,}{\partial\,k}\left[\frac{1}{\sqrt{1-k^2 x^2}}\right]dx$$

$$\frac{d}{dk}K=\int^1_0 \frac{kx^2}{\sqrt{1-x^2}\sqrt{1-k^2x^2}(1-k^2x^2)}\,dx$$

Adding and subtracting 1 we have

$$\frac{-1}{k}\int^1_0 \frac{1-kx^2-1}{\sqrt{1-x^2}\sqrt{1-k^2x^2}(1-k^2x^2)}\,dx=\frac{1}{k}\int^1_0 \frac{dx}{\sqrt{1-x^2}\sqrt{1-k^2x^2}(1-k^2x^2)}-\frac{K}{k}$$

Let us focus on the first integral

$$\int^1_0 \frac{1}{\sqrt{1-x^2}(1-\, k^2 \, x^2)^{\frac{3}{2}}}\,dx$$

Let $x=\sqrt{t}$ and we have $dx= \frac{1}{2\sqrt{t}}\,dt$

$$\frac{1}{2}\int^1_0 \frac{t^{-\frac{1}{2}}}{\sqrt{1-t}(1-\, k^2 \, t)^{\frac{3}{2}}}\,dx$$

Using the hypergeometric integral representation

$$\frac{1}{2}\int^1_0 \frac{t^{-\frac{1}{2}}}{\sqrt{1-t}(1-\, k^2 \, t)^{\frac{3}{2}}}=\frac{\pi}{2}{}_2F_1\left(\frac{3}{2},\frac{1}{2},1,k^2 \right)$$

Using the linear transformation

$$_2F_1\left(a,b,c,z \right)=(1-z)^{c-a-b}\,{}_2F_1\left(c-a,c-b,c,z \right)$$

We get by putting $k'=\sqrt{1-k^2}$

$$\frac{\pi}{2}{}_2F_1\left(\frac{3}{2},\frac{1}{2},1,k^2 \right)=\frac{1}{1-k^2}\frac{\pi}{2}{}_2F_1\left(-\frac{1}{2},\frac{1}{2},1,k^2 \right)=\frac{E}{k'^2}$$

So finally we get

\(\displaystyle \frac{d}{dk}K=\frac{1}{k}\left( \frac{E}{ k'^2}-K \right)\)​

Note : $k'$ is called the complementary modulus and it should be of interest for us in the next set of lectures.

 

[alyafey22]

Integration lessons continued ...

4.Integration using special functions (continued)


4.10. Euler sums

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General definition:

\(\displaystyle S_{p^{\,r},q} = \sum_{k\geq 1} \frac{(H_k^{(p)})^r}{k^q}\)​

Where we define the general harmonic number

\(\displaystyle H_k^{(p)} = \sum_{n=1}^k \frac{1}{n^p} \,\,\,;\,\, H^{(1)}_k \equiv H_k = \sum_{n=1}^k \frac{1}{n}\)​

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Euler sums were greatly studied by Euler, hence the name.

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Generating function:

\(\displaystyle \sum_{k\geq 1} H_k^{(p)}x^k = \frac{\mathrm{Li}_p(x)}{1-x}
\)​

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Proof

$$\sum_{k\geq 1} H_k^{(p)}x^k = \sum_{k\geq 1} \sum_{n=1}^k \frac{1}{n^p}x^k $$

By interchanging the two series we have

$$ \sum_{n\geq 1} \sum_{k\geq n} \frac{x^k}{n^p} =\sum_{n\geq 1}\frac{1}{n^p} \sum_{k\geq n}x^k $$

The inner sum is a geometric series


$$\frac{1}{1-x} \sum_{n\geq 1} \frac{x^n}{n^p} =\frac{\mathrm{Li}_p(x)}{1-x} $$

---

We can use this to generate some more functions by integrating. Hence assume $p=1$

$$\sum_{k\geq 1} H_k x^k = -\frac{\log(1-x)}{1-x} $$

Divide by $x$ and integrate to get

$$\sum_{k\geq 1} \frac{H_k}{k} x^k =\mathrm{Li}_2(x)+\frac{1}{2}\log^2(1-x)$$

Now divide by $x$ and integrate again

$$\sum_{k\geq 1} \frac{H_k}{k^2} x^k =\mathrm{Li}_3(x)+\frac{1}{2}\int^x_0
\frac{\log^2(1-t)}{t}\,dt$$

Now let us look at the integral

$$\int^x_0 \frac{\log^2(1-t)}{t}\, dt $$

Integrating by parts we get the following

$$\int^x_0 \frac{\log^2(1-t)}{t}\, dt = - \log(1-x) \text{Li}_2(x) -\int^x_0\frac{\text{Li}_2 (t)}{1-t} \, dt $$

Now we are left with the following integral

$$\int^x_0\frac{\text{Li}_2 (t)}{1-t} \, dt = \int^{1}_{1-x} \frac{\text{Li}_2 (1-t)}{t} \, dt $$

$$\int^{1}_{1-x} \frac{\frac{\pi^2}{6} -\text{Li}_2(t) - \log(1-t) \log(t)}{t} \, dt $$

$$ -\frac{\pi^2}{6}\log(1-x)- \int^{1}_{1-x} \frac{\text{Li}_2(t)}{t}\,dt -\int^{1}_{1-x}\frac{ \log(1-t) \log(t)}{t} \, dt $$

The first integral

• $$\int^{1}_{1-x} \frac{\text{Li}_2(t)}{t}\,dt =\text{Li}_3(1)-\text{Li}_3(1-x) $$

The second integral by parts we obtain

• $$ \int^{1}_{1-x}\frac{ \log(1-t) \log(t)}{t} =\text{Li}_3(1) +\log(1-x)\text{Li}_2(1-x)-\text{Li}_3(1-x) $$

Collecting the results together we obtain

$$\int^x_0\frac{\text{Li}_2 (t)}{1-t} \, dt =-\frac{\pi^2}{6}\log(1-x)-\text{Li}_2(1-x) \log(1-x)+2\, \text{Li}_3(1-x)- 2 \zeta(3) $$

Hence we solved the integral

$$\int^x_0 \frac{\log^2(1-t)}{t}\, dt = - \log(1-x) \text{Li}_2(x) +\frac{\pi^2}{6}\log(1-x)+\text{Li}_2(1-x) \log(1-x)-2\, \text{Li}_3(1-x)+ 2 \zeta(3) $$

So we have got our Harmonic sum

$$\sum_{k\geq 1} \frac{H_k}{k^2} \, x^{k} = \text{Li}_3(x)+\frac{1}{2} \left( - \log(1-x) \text{Li}_2(x) +\frac{\pi^2}{6}\log(1-x)+\text{Li}_2(1-x) \log(1-x)-2\, \text{Li}_3(1-x)+ 2 \zeta(3) \right) $$$$\sum_{k\geq 1} \frac{H_k}{k^2} \, x^{k} = \text{Li}_3(x)-\, \text{Li}_3(1-x)+\, \log(1-x) \text{Li}_2(1-x) +\frac{1}{2}\log(x) \log^2(1-x)+\zeta(3) $$The expression can be further simplified but I will leave it for the reader.

To be continued ...

 

[alyafey22]

4.Integration using special functions (continued)


4.10. Euler sums (continued)

---

General definition:

\(\displaystyle \sum_{n=1}^\infty \frac{H_n}{n^q}= \left(1+\frac{q}{2} \right)\zeta(q+1)-\frac{1}{2}\sum_{k=1}^{q-2}\zeta(k+1)\zeta(q-k)\,\,;q\geq 2\)​

This can be proved using complex analysis or using basic techniques of series manipulations as in this answer.

---

Where by definition we have for $q=2$

$$\sum_{n=1}^\infty \frac{H_n}{n^2}= 2\zeta(3)$$

Let us look at a way of solving that sum using integration

First we need that

$$H_n = \int^1_0 \frac{1-x^n}{1-x}\,dx$$

proof

$$\int^1_0 \frac{1-x^n}{1-x}\,dx =\sum_{k\geq 0} \int^1_0 x^k-x^{n+k}\,dx $$

$$\sum_{k\geq 0} \frac{1}{k+1}-\frac{1}{n+k+1} = H_n$$

Plugging the result in the sum we get

$$\int^1_0 \frac{1}{1-x}\sum_{n = 1}^\infty\frac{1-x^n }{n^2}\,dx = \int^1_0 \frac{\zeta(2)-\mathrm{Li}_2(x)}{1-x}dx$$

Now use the duplication formula for dilogarithm

$$\zeta(2)-\mathrm{Li}_2(x) = \mathrm{Li}_2(1-x)+\log(x)\log(1-x)$$

Hence we have

$$\int^1_0 \frac{\mathrm{Li}_2(1-x)+\log(x)\log(1-x)}{1-x}dx = \int^1_0 \frac{\mathrm{Li}_2(x)+\log(1-x)\log(x)}{x}dx$$

First integral

$$\int^1_0 \frac{\mathrm{Li}_2(x)}{x} = \mathrm{Li}_3(1) =\zeta(3)$$

Second integral

$$\int^1_0 \frac{\log(1-x)\log(x)}{x}dx =\int^1_0 \frac{\mathrm{Li}_2(x)}{x} = \zeta(3) $$

Finally

$$\sum_{n=1}^\infty \frac{H_n}{n^2} = \zeta(3)+\zeta(3) = 2\zeta(3)$$

---

Examples

$$\int^1_0\frac{\log^2(1-x)\log(x)}{x} = -\frac{\pi^4}{180}$$

Using the generating function

$$\sum_{k\geq 1}H_k x^{k-1} = -\frac{\log(1-x)}{x(1-x)}$$

By integrating both sides

$$\sum_{k\geq 1}\frac{H_k}{k} x^{k} =\mathrm{Li}_2(x) +\frac{1}{2}\log^2(1-x)$$

Or

$$\log^2(1-x) =2\sum_{k\geq 1}\frac{H_k}{k} x^{k} -2\mathrm{Li}_2(x)$$

plugging this in our integral we have

$$2\int^1_0\left(\sum_{k\geq 1}\frac{H_k}{k} x^{k} -\mathrm{Li}_2(x)\right)\frac{\log(x)}{x}\,dx$$

First integral

$$-2\int^1_0\mathrm{Li}_2(x)\frac{\log(x)}{x}\,dx =2\int^1_0\frac{\mathrm{Li}_3(x)}{x}\,dx = 2\zeta(4)$$

Second integral

$$2\sum_{k\geq 1}\frac{H_k}{k}\int^1_0x^{k-1}\log(x)\,dx$$

Using integration by parts twice and the formula presented first

$$-2\sum_{k\geq 1}\frac{H_k}{k^3} =-5\zeta(4)+\zeta^2(2)$$

Finally we get

$$\int^1_0\frac{\log^2(1-x)\log(x)}{x} =-5\zeta(4)+\zeta^2(2)+2\zeta(4) =\zeta^2(2)-3\zeta(4) $$

 

[alyafey22]

4.Integration using special functions (continued)

4.10. Euler sums (continued)

---

Show that

\(\displaystyle \int_{0}^{\infty} \, e^{-at} \sin(bt) \frac{\ln t}{t}\, dt =- \left( \frac{

\log(a^2+b^2) }{2} +\gamma \right) \arctan \left( \frac{b}{a} \right)\)​

---

Proof

We can start by the following integral

$$I(s) = \int_{0}^{\infty} t^{s-1} \, e^{-at} \, \sin(bt) dt $$

By using the the expansion of the sine function

$$ I(s)=\int_{0}^{\infty}t^{s-1} \, e^{-at}\sum_{n\geq 0} \frac{(-1)^n (bt)^

{2n+1}}{\Gamma(2n+2)} $$

By swapping the summation and integration

$$ I(s)= \sum_{n\geq 0} \frac{(-1)^n (b)^{2n+1}}{\Gamma(2n+2)} \int_{0}^

{\infty}t^{s+2n} \, e^{-at} dt= \frac{1}{a^s} \sum_{n\geq 0} \frac{(-1)^n

(b)^{2n+1} \Gamma (s+2n+1)}{\Gamma(2n+2) a^{2n+1}}$$

By differentiating and plugging $s=0$ we have
$$I'(0) = \sum_{n\geq 0} \frac{(-1)^n \psi_0(2n+1)}{2n+1} \left( \frac{b}

{a} \right)^{2n+1} -\log(a) \sum_{n\geq 0} \frac{(-1)^n }{2n+1} \left( \frac

{b}{a} \right)^{2n+1}$$

Now use that $\psi(n+1) +\gamma = H_n$

$$I'(0) = \sum_{n\geq 0} \frac{(-1)^n H_{2n}-\gamma}{2n+1} \left( \frac{b}

{a} \right)^{2n+1} -\log(a) \arctan \left( \frac{b}{a} \right) $$
$$I'(0) = \sum_{n\geq 0} \frac{(-1)^n H_{2n}}{2n+1} \left( \frac{b}{a}

\right)^{2n+1}-(\gamma +\log(a))\arctan \left( \frac{b}{a} \right) $$

Now we look at the harmonic sum

$$\begin{align} \sum_{k\geq 0}(-1)^k H_{2k} x^{2k}&= \sum_{k\geq 0}(-1)^k x^

{2k} \int^1_0 \frac{1-t^{2k}}{1-t} \, dt\\
&= \int^1_0 \frac{1}{1-t} \sum_{k\geq 0}(-1)^k x^{2k} \left(1-t^{2k}\right)

\, dt\\
&= \int^1_0 \frac{1}{1-t} \sum_{k\geq 0}(-1)^k \left(x^{2k}-(xt)^{2k}\right)

\, dt\\
&= \int^1_0 \frac{1}{1-t}\left(\frac{1}{1+x^2}-\frac{1}{1+t^2x^2}\right) \,

dt\\
&=\frac{1}{1+x^2} \int^1_0 \frac{1+t^2x^2-1-x^2}{(1-t)(1+t^2x^2)} \, dt\\
&=\frac{-x^2}{1+x^2} \int^1_0 \frac{(1-t^2)}{(1-t)(1+t^2x^2)} \, dt\\
&=\frac{-x^2}{1+x^2} \int^1_0 \frac{1+t}{(1+t^2x^2)} \, dt\\
&=\frac{-x^2}{1+x^2} \left( \int^1_0 \frac{1}{1+t^2x^2}+\frac{t}{1+t^2x^2} \,

dt \right)\\
&= \frac{-1}{2(1+x^2)} \left(2x \arctan (x) + \log(1+x^2) \right)
\end{align} $$

Using this we conclude by integrating

$$\sum_{k\geq 0}\frac{(-1)^k H_{2k}}{2k+1} x^{2k}=-\frac{1}{2} \log(1+x^2)

\arctan(x)$$

Hence the following

$$\sum_{k\geq 0}\frac{(-1)^k H_{2k}}{2k+1} \left(\frac{b}{a} \right)^{2k+1}=-

\frac{1}{2} \log \left( \frac{a^2+b^2}{a^2} \right) \arctan \left(\frac{b}{a}

\right)$$

Substituting that in our integral

$$\begin{align} \int_{0}^{\infty} \, e^{-at} \sin(bt) \frac{\ln t}{t}\, dt &=

-\left( \frac{1}{2} \log \left( \frac{a^2+b^2}{a^2} \right) + \gamma +\log(a)

\right) \arctan \left( \frac{b}{a} \right)\\
&=- \left( \frac{ \log(a^2+b^2) }{2} +\gamma \right) \arctan \left( \frac{b}

{a} \right)
\end{align}$$

---

Prove that

\(\displaystyle \begin{align}
\int^1_0 \frac{\mathrm{Li}_p(x)\,\, \mathrm{Li}_q(x)\, }{x}\, dx&= \sum_{n=1}^{p-1}(-1)^{n-1}\zeta(p-n+1)\zeta(q+n) -\frac{1}{2}\sum_{n=1}^{{p+q}-2}(-1)^{p-1}\zeta(n+1)\zeta({p+q}-n)\\ &+(-1)^{p-1}\left(1+\frac{{p+q}}{2} \right)\zeta({p+q}+1)\end{align}\)​

---

proof

We can see that

$$\int^1_0 \frac{\mathrm{Li}_p(x)\,\, \mathrm{Li}_q(x)\, }{x}\, dx = \sum_{k\geq 1}\sum_{n\geq 1}\frac{1}{k^{q}n^{p}(n+k)}$$

Let us first look at the following

$$\mathscr{C}(\alpha , k) =\sum_{n\geq 1}\frac{1}{n^{\alpha}(n+k)}\,\,\, ; \,\,\,\,\mathscr{C}(1, k)=\frac{H_k}{k} $$

We can find a general formula to find the sum

\begin{align}
\mathscr{C}(\alpha , k) &=\sum_{n\geq 1}\frac{1}{k\, n^{\alpha-1}}\left( \frac{1}{n}-\frac{1}{n+k}\right)\\ &= \frac{1}{k}\zeta(\alpha)-\frac{1}{k}\mathscr{C}(\alpha-1 , k)\\ &= \frac{1}{k}\zeta(\alpha)-\frac{1}{k^2}\zeta(\alpha-1)+\frac{1}{k^2}\mathscr{C}(\alpha-2 , k)\\ &= \sum_{n=1}^{\alpha-1}(-1)^{n-1}\frac{\zeta(\alpha-n+1)}{k^n}+(-1)^{\alpha-1}\frac{H_k}{k^\alpha}
\end{align}

Hence we have the general formula

$$\mathscr{C}(\alpha , k) = \sum_{n=1}^{\alpha-1}(-1)^{n-1}\frac{\zeta(\alpha-n+1)}{k^n}+(-1)^{\alpha-1}\frac{H_k}{k^\alpha}$$

Dividing by $k^{\beta}$ and summing w.r.t to $k$

$$\sum_{k\geq 1}\frac{\mathscr{C}(\alpha , k)}{k^{\beta}} = \sum_{n=1}^{\alpha-1}(-1)^{n-1}\zeta(\alpha-n+1)\zeta(\beta+n)+(-1)^{\alpha-1}\sum_{k\geq 1}\frac{H_k}{k^{\alpha+\beta}}$$

Now we use that

$$\sum_{n=1}^\infty \frac{H_n}{n^q}= \left(1+\frac{q}{2} \right)\zeta(q+1)-\frac{1}{2}\sum_{k=1}^{q-2}\zeta(k+1)\zeta(q-k)$$

Hence we have

$$\sum_{n=1}^\infty \frac{H_n}{n^{\alpha+\beta}}= \left(1+\frac{{\alpha+\beta}}{2} \right)\zeta({\alpha+\beta}+1)-\frac{1}{2}\sum_{k=1}^{{\alpha+\beta}-2}\zeta(k+1)\zeta({\alpha+\beta}-k)$$

And the generalization is the following formula

\begin{align}
\sum_{k\geq 1}\frac{\mathscr{C}(\alpha , k)}{k^{\beta}} &= \sum_{n=1}^{\alpha-1}(-1)^{n-1}\zeta(\alpha-n+1)\zeta(\beta+n) -\frac{1}{2}\sum_{n=1}^{{\alpha+\beta}-2}(-1)^{\alpha-1}\zeta(n+1)\zeta({\alpha+\beta}-n)\\ &+(-1)^{\alpha-1}\left(1+\frac{{\alpha+\beta}}{2} \right)\zeta({\alpha+\beta}+1)\end{align}

We conclude by putting that

\(\displaystyle \sum_{k\geq 1}\sum_{n\geq 1}\frac{1}{k^{q}n^{p}(n+k)} = \sum_{k\geq 1}\frac{\mathscr{C}(p, k)}{k^{q}}\)

 

[alyafey22]

4.Integration using special functions (continued)

4.10. Euler sums (continued)

We can relate the generalized harmonic number to the polygamma function

---

\(\displaystyle H^{(p)}_k = \zeta(p) +(-1)^{p-1}\frac{\psi_{p-1}(k+1)}{ (p-1)!} \)​

---

proof

$$H^{(p)}_k = \sum^k_{n=1}\frac{1}{n^p} =\zeta(p)-\sum^\infty_{n=k+1}\frac{1}{n^p}$$

Now let $n=i+k+1$

$$H^{(p)}_k = \sum^k_{n=1}\frac{1}{n^p} =\zeta(p)-\sum_{i\geq 0}\frac{1}{(i+k+1)^p}$$

We know that

$$(-1)^{p}\frac{\psi_{p-1}(k+1)}{ (p-1)!} =\sum_{i\geq 0} \frac{1}{(i+k+1)^{p}}\,\,p\geq 1$$

Hence we have

$$H^{(p)}_k = \zeta(p) +(-1)^{p-1}\frac{\psi_{p-1}(k+1)}{ (p-1)!} $$

We can use that to obtain a nice integral representation.

---

\(\displaystyle \sum_{k\geq 1}\frac{H^{(p)}_k}{k^q} = \zeta(p)\zeta(q) +(-1)^{p}\frac{1}{ (p-1)!}\int^1_0\frac{\mathrm{Li}_q(x)\log(x)^{p-1}}{1-x}\,dx\)​

---

Note that

$$\psi_0(a+1)= \int^1_0\frac{1-x^a}{1-x}\,dx$$

By differentiating with respect to $a$ , $p$ times we have

$$\psi_p(a+1) = \frac{\partial}{\partial a^p}\int^1_0\frac{1-x^a}{1-x}\,dx$$

$$\psi_p(a+1) = -\int^1_0\frac{x^a\log(x)^{p}}{1-x}\,dx$$

Let $a=k$

$$\psi_{p-1}(k+1) = -\int^1_0\frac{x^k\log(x)^{p-1}}{1-x}\,dx$$

Substituting that in our formula

$$H^{(p)}_k = \zeta(p) +(-1)^{p}\frac{1}{ (p-1)!}\int^1_0\frac{x^k\log(x)^{p-1}}{1-x}\,dx$$

Now divide by $k^q$ and sum with respect to $k$$$\sum_{k\geq 1}\frac{H^{(p)}_k}{k^q} = \zeta(p)\zeta(q) +(-1)^{p}\frac{1}{ (p-1)!}\int^1_0\frac{\mathrm{Li}_q(x)\log(x)^{p-1}}{1-x}\,dx$$

---

Symmetric formula

\(\displaystyle \sum_{k\geq 1} \frac{H^{(p)}_k}{k^q}+\sum_{k\geq 1} \frac{H^{(q)}_k}{k^p} =\zeta(p)\zeta(q)+\zeta(p+q)\)​

---

proof

Take the left side and swap the finite and infinite sums

$$\sum_{i\geq 1} \,\sum_{k\geq i}\frac{1}{i^p} \frac{1}{k^q}=\sum_{i\geq 1} \,\sum_{k\geq 1}\frac{1}{i^p} \frac{1}{k^q}-\sum_{i\geq 1}\frac{1}{i^p} \,\sum_{1\leq k \leq i-1} \frac{1}{k^q}$$

The second sum can be written as

$$\sum_{i\geq 1}\frac{1}{i^p} \,\sum_{1\leq k \leq i-1} \frac{1}{k^q} = \sum_{i\geq 1}\frac{1}{i^p} \,\left(\sum_{1\leq k \leq i}\frac{1}{k^q}-\frac{1}{i^p}\right)$$

By separating and changing the index we get

$$\sum_{k\geq 1}\frac{H^{(q)}_k}{k^p}-\zeta(p+q)$$

Hence we have

$$\sum_{k\geq 1} \frac{H^{(p)}_k}{k^q} =\zeta(p)\zeta(q)-\sum_{k\geq 1}\frac{H^{(q)}_k}{k^p}+\zeta(p+q)$$

$$\sum_{k\geq 1} \frac{H^{(p)}_k}{k^q}+\sum_{k\geq 1}\frac{H^{(q)}_k}{k^p} =\zeta(p)\zeta(q)+\zeta(p+q)$$for the special case $p=q=n$

$$\sum_{k\geq 1} \frac{H^{(n)}_k}{k^n} =\frac{\zeta^2(n)+\zeta(2n)}{2}$$

 

[alyafey22]

Examples

---

\(\displaystyle \sum_{k\geq 1}\frac{H_k^{(3)}}{k^2} = \frac{11\zeta(5)}{2}-2\zeta(2)\zeta(3)\)​

---

Using the symmetry formula

$$\sum_{k\geq 1}\frac{H_k^{(3)}}{k^2} = \zeta(2)\zeta(3)+\zeta(5)-\sum_{k\geq 1}\frac{H_k^{(2)}}{k^3}$$

Using the integral formula on the second sum

$$\sum_{k\geq 1}\frac{H_k^{(2)}}{k^3} =\zeta(2)\zeta(3)+ \int^1_0 \frac{\mathrm{Li}_3(x)\log(x)}{1-x}\,dx$$

Using integration by parts on the integral

$$\int^1_0 \frac{\mathrm{Li}_3(x)\log(x)}{1-x}\,dx =- \int^1_0\frac{\mathrm{Li}_2(x)\mathrm{Li}_2(1-x)}{x}\,dx$$

Let us think of solving

$$\int^1_0\frac{\mathrm{Li}_2(x)\mathrm{Li}_2(1-x)}{x}\,dx$$

Using the duplication formula

$$\mathrm{Li}_2(1-x) = \zeta(2)-\mathrm{Li}_2(x)-\log(x)\log(1-x)$$

$$ \int^1_0\frac{\mathrm{Li}_2(x)(\zeta(2)-\mathrm{Li}_2(x)-\log(x)\log(1-x))}{x}\,dx$$

The first integral

$$\zeta(2)\int^1_0 \frac{\mathrm{Li}_2(x)}{x}\,dx = \zeta(2)\zeta(3)$$

The third integral

$$\int^1_0\frac{\mathrm{Li}_2(x)\log(x)\log(1-x)}{x}\,dx = \frac{1}{2}\int^1_0\frac{\mathrm{Li}^2_2(x)}{x}\,dx$$

Finally we get

$$\int^1_0 \frac{\mathrm{Li}_3(x)\log(x)}{1-x}\,dx =\frac{3}{2}\int^1_0\frac{\mathrm{Li}^2_2(x)}{x}\,dx-\zeta(2)\zeta(3)$$

So

$$\sum_{k\geq 1}\frac{H_k^{(2)}}{k^3} = \frac{3}{2}\int^1_0\frac{\mathrm{Li}^2_2(x)}{x}\,dx$$

Hence we finally get that

$$ \sum_{k\geq 1}\frac{H_k^{(3)}}{k^2}=\zeta(2)\zeta(3)+\zeta(5)-\frac{3}{2}\int^1_0\frac{\mathrm{Li}^2_2(x)}{x}\,dx$$

Let us solve the integral

$$\int^1_0 \frac{\mathrm{Li}^2_2(x)}{x}\,dx$$

By series expansion

$$\sum_{k,n\geq 1}\frac{1}{(nk)^2}\int^1_0x^{n+k-1}\,dx =\sum_{k,n\geq 1}\frac{1}{(nk)^2(n+k)} $$

By some manipulations we get

$$\sum_{k\geq 1}\frac{1}{k^3}\sum_{n\geq 1}\frac{k}{n^2(n+k)}= \sum_{k\geq 1}\frac{1}{k^3}\sum_{n\geq 1}\frac{1}{n^2}-\sum_{k\geq 1}\frac{1}{k^3}\sum_{n\geq 1}\frac{1}{n(n+k)}$$

This can be simplified to conclude that

$$\int^1_0 \frac{\mathrm{Li}^2_2(x)}{x}\,dx = \zeta(2)\zeta(3)-\sum_{k\geq 1}\frac{H_k}{k^4}$$

Now using that

$$\sum_{k\geq 1}\frac{H_k}{k^4} = 3\zeta(5)-\zeta(2)\zeta(3)$$

Hence

$$\int^1_0 \frac{\mathrm{Li}^2_2(x)}{x}\,dx = 2\zeta(2)\zeta(3)-3\zeta(5)$$

Finally we get

$$ \sum_{k\geq 1}\frac{H_k^{(3)}}{k^2}=\zeta(2)\zeta(3)+\zeta(5)-\frac{3}{2}\left( 2\zeta(2)\zeta(3)-3\zeta(5)\right)=\frac{11\zeta(5)}{2}-2\zeta(2)\zeta(3)$$

 

[alyafey22]

Advanced Integration Techniques (back with new lessons)

4.Integration using special functions (continued)

4.11. Sine Integral function

---

We define the following

$$\mathrm{Si}(z) = \int^z_0 \frac{\sin(x) }{x}\, dx $$

A closely related function is the following

$$\mathrm{si}(z) = -\int^\infty_z \frac{\sin(x) }{x}\, dx $$

---

These function are related through the equation

$$\mathrm{Si}(z) = \mathrm{si}(z)+\frac{\pi}{2}$$

A closely related function is the $\mathrm{sinc}$ function

$$\mathrm{sinc}=\begin{cases}
1 & x=0 \\
\frac{\sin(x)}{x} & x \neq 0
\end{cases}$$

Using that we conclude that

$$\frac{d}{dx} \mathrm{Si}(x) = \mathrm{sinc}(x)$$

For the integration we conclude that

$$\int \mathrm{Si}(x)\,dx = \cos(x)+ x \,\mathrm{Si(x)}$$

Examples


Prove that

$$\int^\infty_0 \sin(x) \mathrm{si}(x) \, dx = -\frac{\pi}{4} $$

Using integration by parts we get

$$-\int^\infty_0 \frac{\sin(x)\cos(x)}{x} dx = -\frac{1}{2} \int^\infty_0 \frac{\sin(2x)}{x}\,dx $$

Let $2x = t $

$$-\frac{1}{2}\int^\infty_0 \frac{\sin(t)}{t} dx = -\frac{\pi}{4}$$

Prove that

$$\int^\infty_0 x^{\alpha -1}\,\mathrm{si}(x) \, dx =- \frac{\Gamma(\alpha)}{\alpha}\sin\left(\frac{\pi \alpha}{2}\right)$$

Using the integral representation

$$-\int^\infty_0 x^{\alpha -1}\,\int^\infty_x \frac{\sin(t)}{t}\,dt \, dx $$

Let $xy = t $

$$-\int^\infty_0 x^{\alpha -1}\,\int^\infty_1 \frac{\sin(xy)}{y}\,dy \, dx $$

Switching the integrals we get

$$-\int^\infty_1 \frac{1}{y} \int^\infty_0 x^{\alpha -1}\, \sin(xy) \, dx \,dy$$

Now let $xy = t $

$$-\int^\infty_1 \frac{1}{y^{\alpha+1}} \int^\infty_0 t^{\alpha -1}\, \sin(t) \, dx \,dy$$

The Mellin transform of the sine function is defined as

$$\mathcal{M}_s(\sin (x)) = \int^\infty_0 x^{s-1}\sin(x)\,dx = \Gamma (s) \sin\left( \frac{\pi s}{2} \right)$$

Hence we conclude that

$$ -\Gamma (\alpha) \sin\left( \frac{\pi \alpha}{2} \right)\int^\infty_1 \frac{1}{y^{\alpha+1}} = -\frac{\Gamma (\alpha)}{\alpha} \sin\left( \frac{\pi \alpha}{2} \right)$$

Prove that

$$\int^\infty_0 e^{-\alpha \, x}\,\mathrm{si}(x) \, dx =- \frac{\arctan (\alpha)}{\alpha}$$

Using the integral representation

$$-\int^\infty_0 e^{-\alpha \, x}\,\int^\infty_x \frac{\sin(t)}{t}\,dt \, dx $$

Let $xy = t $

$$-\int^\infty_0 e^{-\alpha \, x}\,\int^\infty_1 \frac{\sin(xy)}{y}\,dy \, dx $$

Switching the integrals

$$-\int^\infty_1 \frac{1}{y}\int^\infty_0 e^{-\alpha \, x}\, \sin(xy)\,dx \, dy $$

The inner integral is the laplace transform of the sine function

$$\mathcal{L}_s(\sin(at)) = \frac{a}{s^2+a^2}$$

Hence we conclude that

$$-\int^\infty_1 \frac{1}{y^2+\alpha^2} \,dy = -\frac{\arctan(\alpha)}{\alpha}$$

 

[alyafey22]

Re: Advanced Integration Techniques (back with new lessons)

4.Integration using special functions (continued)


4.11. Sine Integral function (continued)


Prove that

$$\int^\infty_0 \mathrm{si}(x) \log(x) \,dx =\gamma+1$$

We know that

$$\int^\infty_0 x^{\alpha -1}\,\mathrm{si}(x) \, dx =- \frac{\Gamma(\alpha)}{\alpha}\sin\left(\frac{\pi \alpha}{2}\right)$$

Differentiate with respect to $\alpha$

$$\int^\infty_0 x^{\alpha -1}\,\mathrm{si}(x) \,\log(x) dx = \frac{\Gamma(\alpha)}{\alpha^2}\sin\left(\frac{\pi \alpha}{2}\right)-\frac{\Gamma(\alpha)\psi(\alpha)}{\alpha}\sin\left(\frac{\pi \alpha}{2}\right)
+\frac{\pi}{2}\cos\left(\frac{\pi \alpha}{2}\right)$$

Let $\alpha \to 1$

$$\int^\infty_0 \mathrm{si}(x) \,\log(x) dx = 1- \psi(1) = 1-(-\gamma) = 1+\gamma$$

Find the integral

$$\int^\infty_0 \mathrm{si}(x) \, \sin(px) \, dx $$

Using integration by parts we get

$$-\left[\frac{\mathrm{si}(x) \cos(px)}{p} \right]^\infty_0+\frac{1}{p}\int^\infty_0 \frac{\sin(x)}{x} \, \cos(px) \, dx $$

Taking the limits

$$\lim_{x \to 0} \frac{\mathrm{si}(x) \cos(px)}{p} = \frac{\mathrm{si}(0)}{p} = -\frac{\pi}{2p}$$

$$\lim_{x \to \infty} \frac{\mathrm{si}(x) \cos(px)}{p} = 0$$

Hence we get

$$-\frac{\pi}{2p}+\frac{1}{p}\int^\infty_0 \frac{\sin(x)}{x} \, \cos(px) \, dx $$

The integral

$$\int^\infty_0 \frac{\sin(x)}{x} \, \cos(px) \, dx = \frac{1}{2}\int^\infty_0 \frac{\sin((p+1)x) -\sin((p-1)x)}{x}dx $$

Separate the integrals

$$I = \frac{1}{2}\int^\infty_0 \frac{\sin((p+1)x)}{x}dx -\frac{1}{2}\int^\infty_0 \frac{\sin((p-1)x)}{x}dx $$

If $p-1>0$ we get

$$I = \frac{\pi}{4}-\frac{\pi}{4} = 0$$

If $p-1<0$

$$I = \frac{1}{2}\int^\infty_0 \frac{\sin((p+1)x)}{x}dx +\frac{1}{2}\int^\infty_0 \frac{\sin((p-1)x)}{x}dx = \frac{\pi}{2} $$

If $p=1$ we have

$$I = \frac{1}{2}\int^\infty_0 \frac{\sin(2x)}{x}dx +0 = \frac{\pi}{4}$$

Finally we get

$$\int^\infty_0 \mathrm{si}(x)\sin(px) dx=
\begin{cases}
-\frac{\pi}{2p} & p > 1 \\
-\frac{\pi}{4p} & p < 1 \\
0 & p = 1
\end{cases}$$

 

[alyafey22]

Re: Advanced Integration Techniques (back with new lessons)

4.Integration using special functions (continued)


4.11. Sine Integral function (continued)


Prove that for $0<a<2$

$$\int^\infty_0 \mathrm{si}^2(x) \, \cos(ax) \, dx = \frac{\pi}{2a}\log(a+1) $$

Using integration by parts we get

$$\left[\frac{\mathrm{si}^2(x) \sin(ax)}{a} \right]^\infty_0-\frac{2}{a}\int^\infty_0 \frac{\mathrm{si}(x)\sin(x)}{x} \, \sin(ax) \, dx $$

Taking the limits

$$\lim_{x \to 0} \frac{\mathrm{si}^2(x) \sin(ax)}{a} = 0$$

$$\lim_{x \to \infty} \frac{\mathrm{si}^2(x) \sin(ax)}{a} = 0$$

Let the integral

$$I(a) = \int^\infty_0 \frac{\mathrm{si}(x)\sin(x)}{x} \, \sin(ax) \, dx$$

Differentiate with respect to $a$

$$I'(a) = \int^\infty_0 \mathrm{si}(x)\sin(x) \, \cos(ax) \, dx$$

Now use the product to sum trigonometric rules

$$I'(a) =\frac{1}{2} \int^\infty_0 \mathrm{si}(x)(\sin((a+1)x)-\sin((a-1)x)) \, dx$$

From the previous exercise we have

$$\int^\infty_0 \mathrm{si}(x)\sin((a+1)x) dx = \frac{-\pi}{4(a+1)}\,\,\,;\, a>0$$$$\int^\infty_0 \mathrm{si}(x)\sin((a+1)x) dx = 0\,\,\,;\, a<2$$

Hence we conclude that for $0<a<2$

$$I'(a) =-\frac{\pi}{4(a+1)}$$

Integrate with respect to $a$

$$I(a) =-\frac{\pi}{4}\log(a+1)+C$$

Let $a \to 0$

$$I(0) =0+C \,\,\, \to \,\,\, C = 0$$

Hence we have

$$\int^\infty_0 \frac{\mathrm{si}(x)\sin(x)}{x} \, \sin(ax) \, dx = -\frac{\pi}{4}\log(a+1)$$

Which implies that $$\int^\infty_0 \mathrm{si}^2(x) \, \cos(ax) \, dx = \frac{-2}{a}\left( -\frac{\pi}{4}\log(a+1)\right) = \frac{\pi}{2a}\log(a+1) $$

Find the integral, for $a \neq 1$

$$\int^\infty_0 \mathrm{si}(x) \cos(ax)\,dx $$

Use integration by parts to obtain

$$\frac{1}{a}\int^\infty_0\frac{\sin(x) \sin(ax)}{x}\,dx $$

Let the integral

$$I(t) = \int^\infty_0e^{-tx}\frac{\sin(x) \sin(ax)}{x}\,dx $$

Differentiate with respect to $t$

$$I'(t) = -\int^\infty_0e^{-tx}\sin(x) \sin(ax)\,dx $$

Use product to sum rules

$$I'(t) = \frac{1}{2}\int^\infty_0e^{-tx}(\cos((a+1)x)-\cos((a-1)x))\,dx $$

Now we can use the Laplace transform

$$I'(t) = \frac{1}{2}\left(\frac{t}{t^2+(a+1)^2}-\frac{t}{t^2+(a-1)^2} \right) $$

Integrate with respect to $t$

$$I(t) = -\frac{1}{4} \log\left(\frac{t^2+(a+1)^2}{t^2+(a-1)^2} \right) +C$$

After verifying the constant goes to 0, we have

$$\int^\infty_0e^{-tx}\frac{\sin(x) \sin(ax)}{x}\,dx = -\frac{1}{4} \log\left(\frac{t^2+(a+1)^2}{t^2+(a-1)^2} \right) $$

Let $t \to 0$

$$\int^\infty_0\frac{\sin(x) \sin(ax)}{x}\,dx = -\frac{1}{4} \log\left(\frac{a+1}{a-1} \right)^2 $$

We conclude that

$$\int^\infty_0 \mathrm{si}(x) \cos(ax)\,dx = -\frac{1}{4a} \log\left(\frac{a+1}{a-1} \right)^2$$

 

[alyafey22]

(AIT) Cosine Integral function

4.Integration using special functions (continued)


4.12. Cosine Integral function

---

Define

$$\mathrm{ci}(x) =- \int^\infty_x \frac{\cos(t)}{t}\,dt $$

A related function is the following

$$\mathrm{Cin}(x) = \int^x_0 \frac{1-\cos(t)}{t}\,dt$$

---

The derivative is

$$\frac{d}{dx}\mathrm{ci}(x) = \frac{\cos(x)}{x}$$

The integral

$$\int \mathrm{ci}(x) \,dx = x\mathrm{ci}(x)-\sin(x)$$

Prove the following

$$\mathrm{Cin}(x) = -\mathrm{ci}(x)+\log(x)+\gamma$$

Start by

$$\mathrm{Cin}(x) = \int^x_0 \frac{1-\cos(t)}{t}\,dt$$

Rewrite as

$$\mathrm{Cin}(x) =\int^\infty_0 \frac{1-\cos(t)}{t}\,dt- \int^\infty_x \frac{1-\cos(t)}{t}\,dt$$

Which simplifies to

$$\mathrm{Cin}(x) =\lim_{z \to \infty } \left[\int^z_0 \frac{1-\cos(t)}{t}\,dt- \log(z)\right] -\mathrm{ci}(x)+\log(x)$$

The limit goes to the Euler Maschorinit constant

$$\mathrm{Cin}(x) =\gamma -\mathrm{ci}(x)+\log(x)$$

Find the integral

$$\int^\infty_0 \mathrm{ci}(x) \, \cos(px) \, dx $$

Using integration by parts we get

$$\left[\frac{\mathrm{ci}(x) \sin(px)}{p} \right]^\infty_0-\frac{1}{p}\int^\infty_0 \frac{\cos(x)}{x} \, \sin(px) \, dx $$

Taking the limits

$$\lim_{x \to 0} \frac{\mathrm{ci}(x) \sin(px)}{p}=0$$

$$\lim_{x \to \infty}\frac{\mathrm{ci}(x) \sin(px)}{p} = 0$$

Hence we get

$$-\frac{1}{p}\int^\infty_0 \frac{\cos(x)}{x} \, \sin(px) \, dx$$

The integral

$$\int^\infty_0 \frac{\cos(x)}{x} \, \sin(px) \, dx = \frac{1}{2}\int^\infty_0 \frac{\sin((p-1)x) +\sin((p+1)x)}{x}dx $$

Separate the integrals

$$I = \frac{1}{2}\int^\infty_0 \frac{\sin((p+1)x)}{x}dx +\frac{1}{2}\int^\infty_0 \frac{\sin((p-1)x)}{x}dx $$

If $p-1>0$ we get

$$I = \frac{\pi}{4}+\frac{\pi}{4} = \frac{\pi}{2}$$

If $p-1<0$

$$I = \frac{\pi}{4}-\frac{\pi}{4} = 0 $$

If $p=1$ we have

$$I = \frac{1}{2}\int^\infty_0 \frac{\sin(2x)}{x}dx +0 = \frac{\pi}{4}$$

Finally we get

$$\int^\infty_0 \mathrm{ci}(x) \, \cos(px) \, dx =
\begin{cases}
-\frac{\pi}{2p} & p > 1 \\
-\frac{\pi}{4p} & p = 1 \\
0 & p < 1
\end{cases}$$

Find the integral for, $p>1$

$$\int^\infty_0 \mathrm{ci}(px) \mathrm{ci}(x)\,dx$$

Let

$$I(p) = \int^\infty_0 \mathrm{ci}(px) \mathrm{ci}(x)\,dx$$

Differentiate with respect to $p$

$$I'(p) = \frac{1}{p}\int^\infty_0 \mathrm{cos}(px) \mathrm{ci}(x)\,dx$$

If $p>1$ from the previous example we conclude that

$$I'(p) = \frac{1}{p}\left(\frac{-\pi}{2p}\right) = -\frac{\pi}{2p^2}$$

Integrate with respect to $p$

$$I(p) = \frac{\pi}{2p} + C$$

Take the limit $p \to \infty$, so $C = 0$.

 

[alyafey22]

Re: (AIT) Cosine Integral function

Prove that

$$\int^\infty_0 x^{\alpha-1}\mathrm{ci}(x)\,dx = -\frac{\Gamma(\alpha)}{\alpha}\cos\left( \frac{\alpha\pi}{2}\right)$$

Use the integral representation

$$-\int^\infty_0 x^{\alpha-1}\int^\infty_x \frac{\cos(t)}{t}\,dt\,dx$$

Let $t = yx$

$$-\int^\infty_0 x^{\alpha-1}\int^\infty_1 \frac{\cos(yx)}{y}\,dy\,dx$$

Switch the integrals

$$-\int^\infty_1\frac{1}{y}\int^\infty_0 x^{\alpha-1} \cos(yx)\,dx\,dy$$

Using the Mellin transform we get

$$-\Gamma(\alpha)\cos\left( \frac{\alpha\pi}{2}\right)\int^\infty_1\frac{1}{y^{1+\alpha}}\,dy = -\frac{\Gamma(\alpha)}{\alpha}\cos\left( \frac{\alpha\pi}{2}\right)$$

Prove that

$$\int^\infty_0 \mathrm{ci}(x) \log(x) \,dx = \frac{\pi}{2}$$

From the previous example we know

$$\int^\infty_0 x^{\alpha-1}\mathrm{ci}(x)\,dx = -\frac{\Gamma(\alpha)}{\alpha}\cos\left( \frac{\alpha\pi}{2}\right)$$

Differentiate with respect to $\alpha$

$$\int^\infty_0 x^{\alpha-1}\mathrm{ci}(x) \log(x)\,dx = \frac{\Gamma(\alpha)}{\alpha^2}\cos\left( \frac{\alpha\pi}{2}\right)-\frac{\Gamma(\alpha) \psi(\alpha)}{\alpha}\cos\left( \frac{\alpha\pi}{2}\right)+\frac{\pi}{2}\frac{\Gamma(\alpha)}{\alpha}\sin\left( \frac{\alpha\pi}{2}\right)$$

Take the limit $\alpha \to 1$

$$\int^\infty_0 \mathrm{ci}(x) \log(x)\,dx = 0-0+\frac{\pi}{2}\sin\left( \frac{\pi}{2}\right) = \frac{\pi}{2}$$

Prove that

$$\int^\infty_0 \mathrm{ci}(x) e^{-\alpha x}\,dx = -\frac{1}{\alpha}\log\sqrt{1+\alpha^2}$$

Use the integral representation

$$-\int^\infty_0 e^{-\alpha x}\int^\infty_1 \frac{\cos(yx)}{y}\,dy\,dx$$

Switch the integrals

$$-\int^\infty_1 \frac{1}{y}\int^\infty_0 e^{-\alpha x} \cos(yx)\,dx\,dy$$

Use the Laplace transformation

$$-\int^\infty_1 \frac{\alpha}{y(\alpha^2+y^2)}\,dy =- \frac{1}{2a}\log(1+\alpha^2) = -\frac{1}{\alpha}\log\sqrt{1+\alpha^2} $$

 

[alyafey22]

Re: (AIT) Cosine Integral function

Find the integral

$$\int^\infty_0 \mathrm{si}(qx) \mathrm{ci}(x) \, dx$$

Using the integral representation

$$\int^\infty_0 \mathrm{si}(qx) \int^\infty_1 \frac{\cos(yx)}{y}\,dy\, dx$$

Switch the integrals

$$\int^\infty_1 \frac{1}{y} \int^\infty_0 \mathrm{si}(qx) \cos(yx)\, dx\,dy$$

We also showed that

$$\frac{1}{2}\int^\infty_1 \frac{1}{y^2}\log\left( \frac{y+q}{y-q}\right)\,dy$$

We can prove that the anti-derivative

$$\left[\frac{\log(y)}{q}-\frac{1}{2q}\log(y^2-q^2)-\frac{1}{2y}\log\left( \frac{y+q}{y-q}\right)\right]^\infty_1$$

Which simplifies

$$\left[-\frac{1}{2q}\log\left(\frac{y^2-q^2}{y^2} \right)-\frac{1}{2y}\log\left( \frac{y+q}{y-q}\right)\right]^\infty_1$$

The limits

$$\lim_{y \to \infty }\frac{1}{2q}\log\left(\frac{y^2-q^2}{y^2} \right)+\frac{1}{2y}\log\left( \frac{y+q}{y-q}\right) = 0 $$

The limit $y \to 1$
$$\frac{1}{2q}\log\left(1-q^2 \right)+\frac{1}{2}\log\left( \frac{1+q}{1-q}\right)$$

Can be written as

$$\frac{1}{4q}\log\left(1-q^2 \right)^2+\frac{1}{4}\log\left( \frac{1+q}{1-q}\right)^2$$

Prove that

$$\int^\infty_0 \frac{\mathrm{ci}(\alpha x)}{x+\beta}\,dx = -\frac{1}{2}\left\{ \mathrm{si(\alpha \beta)^2+\mathrm{ci}(\alpha \beta)^2}\right\}$$

Let the following

$$I(\alpha) = \int^\infty_0 \frac{\mathrm{ci}(\alpha x)}{x+\beta}\,dx $$

Differentiate with respect to $\alpha$

$$I'(\alpha) = \frac{1}{\alpha}\int^\infty_0 \frac{\cos(\alpha x)}{x+\beta}\,dx $$

Let $x+\beta = t$

$$I'(\alpha) = \frac{1}{\alpha}\int^\infty_\beta \frac{\cos(\alpha (t-\beta)}{t}\,dt $$

Use sum to product rules

$$I'(\alpha) = \frac{1}{\alpha}\int^\infty_\beta \frac{\cos(\alpha t) \cos(\alpha \beta) + \sin(\alpha t) \sin(\alpha \beta)}{t}\,dt $$

Separate the integrals

$$I'(\alpha) = \frac{\cos(\alpha \beta)}{\alpha}\int^\infty_\beta \frac{\cos(\alpha t) }{t}\,dt+ \frac{\sin(\alpha \beta)}{\alpha}\int^\infty_\beta\frac{\sin(\alpha t) }{t}\,dt $$

This simplifies to

$$I'(\alpha) = -\frac{\cos(\alpha \beta)}{\alpha}\mathrm{ci}(\alpha \beta)-\frac{\sin(\alpha \beta)}{\alpha}\mathrm{si}(\alpha \beta) $$

Integrate with respect to $\alpha$$$I(\alpha) = -\frac{1}{2}\left\{ \mathrm{si(\alpha \beta)^2+\mathrm{ci}(\alpha \beta)^2}\right\}+ C$$

If $\alpha \to \infty $ we have $C =0$.

 

[alyafey22]

(AIT) Logarithm integral function

4.Integration using special functions (continued)

4.13. Logarithm Integral function

---

Define


$$\mathrm{li}(x) = \int^x_0 \frac{dt}{\log(t)}$$

---

Prove that

$$\int^1_0 \mathrm{li}(x)\,dx = -\log(2)$$

Let the following

$$I(a) = \int^1_0 \int^x_0 \frac{e^{-a\log(t)}dt}{\log(t)}\,dx $$

Differentiate with respect to $a$

$$I'(a) = - \int^1_0 \int^x_0 t^{-a}\,dt\,dx$$

$$I'(a) = \frac{1}{a-1} \int^1_0 x^{1-a}\,dx$$

Which reduces to

$$I'(a) = \frac{1}{(a-1)(2-a)} = \frac{1}{2-a}-\frac{1}{1-a}$$

Integrate with respect to $a$

$$I(a) = \log\left( \frac{1-a}{2-a}\right)+C$$

Take the limit $a \to \infty$ we get $C = 0$

$$\int^1_0 \int^x_0 \frac{e^{-a\log(t)}dt}{\log(t)}\,dx =\log\left( \frac{1-a}{2-a}\right)$$

Let $a \to 0$

$$\int^1_0 \mathrm{li}(x)\,dx =\log\left( \frac{1}{2}\right) = -\log(2)$$

Find the following

$$\int^1_0 x^{p-1}\mathrm{li}(x)\,dx$$

Let the following

$$I(a) = \int^1_0 x^{p-1}\int^x_0 \frac{e^{-a\log(t)}dt}{\log(t)}\,dx $$

Differentiate with respect to $a$

$$I'(a) = - \int^1_0 x^{p-1}\int^x_0 t^{-a}\,dt\,dx$$

$$I'(a) = \frac{1}{a-1} \int^1_0 x^{p-a}\,dx$$

Which reduces to

$$I'(a) = \frac{1}{(a-1)(p-a+1)} =\frac{1}{p} \left\{\frac{1}{p-a+1}-\frac{1}{1-a} \right\}$$

Integrate with respect to $a$

$$I(a) = \frac{1}{p}\log\left( \frac{1-a}{p-a+1}\right)+C$$

Take the limit $a \to \infty$ we get $C = 0$

$$\int^1_0 x^{p-1}\int^x_0 \frac{e^{-a\log(t)}dt}{\log(t)}\,dx =\frac{1}{p}\log\left( \frac{1-a}{p-a+1}\right)$$

Let $a \to 0$

$$\int^1_0 x^{p-1}\int^x_0 \frac{e^{-a\log(t)}dt}{\log(t)}\,dx =\frac{1}{p}\log\left( \frac{1}{p+1}\right) = -\frac{1}{p}\log\left( p+1\right)$$

Find the following


$$\int^1_0\mathrm{li}\left(\frac{1}{x}\right) \sin(a\log(x))\,dx$$

Let the following

$$I(b) = \int^1_0 \sin(a\log(x))\int^{\frac{1}{x}}_0 \frac{e^{-b\log(t)}dt}{\log(t)}\,dx $$

Differentiate with respect to $b$

$$I'(b) =-\int^1_0 \sin(a\log(x))\int^{\frac{1}{x}}_0 t^{-b}\,dt\,dx $$

$$I'(b) =\frac{1}{b-1}\int^1_0 x^{b-1}\sin(a\log(x))\,dx $$

Let $\log(x) = -t$

$$I'(b) =\frac{1}{1-b}\int^\infty_0 e^{-tb}\sin(at)\,dt $$

Now use the Laplace transform

$$I'(b) =\frac{1}{1-b}\frac{a}{a^2+b^2} $$

Integrate with respect to $b$

$$I(b) = \frac{a\log(a^2+b^2) -2a\log(b-1)+2\arctan(b/a)}{2a^2+2}+C$$

Let $b \to \infty $

$$0 = \frac{\pi}{2(a^2+1)}+C$$

Hence we have

$$\int^1_0 \sin(a\log(x))\int^{\frac{1}{x}}_0 \frac{e^{-b\log(t)}dt}{\log(t)}\,dx= \frac{a\log(a^2+b^2) -2a\log(b-1)+2\arctan(b/a)}{2a^2+2}-\frac{\pi}{2(a^2+1)}$$

Let $b \to 0$

$$\int^1_0 \sin(a\log(x))\mathrm{li}(x)\,dx= \frac{a\log(a^2)}{2a^2+2}-\frac{\pi}{2(a^2+1)} = \frac{1}{a^2+1}\left(a\log(a) -\frac{\pi}{2}\right)$$

 

[alyafey22]

Re: (AIT) Logarithm integral function

Find the following integral

$$\int^1_0\frac{\mathrm{li}(x)}{x} \log^{p-1}\left(\frac{1}{x}\right)\,dx $$

Let the following

$$I(a) = \int^1_0\frac{1}{x} \left[\int^x_0 \frac{e^{-a\log(t)}}{\log(t)}dt \right]\log^{p-1}\left(\frac{1}{x}\right)\,dx $$

Differentiate with respect to $a$$$I'(a) = -\int^1_0\frac{1}{x} \left[\int^x_0t^{-a}dt \right]\log^{p-1}\left(\frac{1}{x}\right)\,dx $$

$$I'(a) = \frac{1}{a-1}\int^1_0x^{-a}\log^{p-1}\left(\frac{1}{x}\right)\,dx $$

Let $-\log(x) = t$

$$I'(a) = \frac{-1}{1-a}\int^\infty_0 e^{-(1-a)t}t^{p-1}\,dx $$

$$I'(a) = \frac{-1}{1-a}\frac{\Gamma(p)}{(1-a)^p} = -\frac{\Gamma(p)}{(1-a)^{p+1}} $$

Integrate with respect to $a$

$$I(a) = -\frac{\Gamma(p)}{p(1-a)^p}$$

Let $a \to 0 $, Hence

$$\int^1_0\frac{\mathrm{li}(x)}{x} \log^{p-1}\left(\frac{1}{x}\right)\,dx = -\frac{\Gamma(p)}{p} $$

Prove that


$$\int^\infty_1 \mathrm{li}\left(\frac{1}{x}\right) \log^{p-1}(x)\,dx = -\frac{\pi}{\sin(\pi p)} \Gamma(p)$$

Let the following

$$I(a) = \int^\infty_1 \mathrm{li}\left(x^{-a}\right) \log^{p-1}(x)\,dx = -\frac{\pi}{\sin(\pi p)} \Gamma(p)$$

Differentiate with respect to $a$

$$\frac{d}{da}\mathrm{li}\left(x^{-a}\right)=\frac{d}{da} \int^{x^{-a}}_0\frac{dt}{\log(t)}= \frac{x^{-a}}{a}$$

Hence we have

$$I'(a) = \frac{1}{a}\int^\infty_1 x^{-a}\log^{p-1}(x)\,dx $$

Let $\log(x) = t$

$$I'(a) = \frac{1}{a}\int^\infty_0 e^{-(a-1)t}t^{p-1}\,dt $$

Using the Laplace transform

$$I'(a) =\Gamma(p) \frac{1}{a(a-1)^p} $$

Take the integral

$$\int^\infty_1 I'(a)da = \Gamma(p)\int^\infty_1 \frac{1}{a(a-1)^p} da $$

The left hand-side

$$I(\infty)-I(1) = \Gamma(p)\int^\infty_1 \frac{1}{a(a-1)^p} da $$

Now since $I(\infty) = 0$

$$I(1) = -\Gamma(p)\int^\infty_1 \frac{1}{a(a-1)^p} da $$

Which implies that

$$\int^\infty_1 \mathrm{li}\left(\frac{1}{x}\right) \log^{p-1}(x)\,dx = -\Gamma(p)\int^\infty_1 \frac{1}{a(a-1)^p} da$$

Now let $t = a-1$

$$ \int^\infty_0 \frac{t^{-p}}{t+1} dt$$

Using the beta integral $x+y = 1$ and $x-1 = -p$ which implies that $x = 1-p,y=p$

Hence we have

$$ \int^\infty_0 \frac{t^{-p}}{t+1} dt = \beta(p,1-p) = \Gamma(p)\Gamma(1-p) = \frac{\pi}{\sin(\pi p)}$$

Finally we get

$$\int^\infty_1 \mathrm{li}\left(\frac{1}{x}\right) \log^{p-1}(x)\,dx = -\frac{\pi}{\sin(\pi p)} \Gamma(p)$$

 

[alyafey22]

(AIT) Clausen functions

4.Integration using special functions (continued)

4.14. Clausen functions

---

$$\mathrm{cl}_m(\theta) = \begin{cases}
\sum_{k=1}^\infty \frac{\sin(k\theta)}{k^m} & m \text{ is even} \\
\sum_{k=1}^\infty \frac{\cos(k\theta)}{k^m} & m \text{ is odd}
\end{cases}$$

---

Duplication formula

$$\mathrm{cl}_m(2 \theta) = 2^{m-1}(\mathrm{cl}_m(\theta)-(-1)^m \mathrm{cl}_m(\pi-\theta))$$

Proof

If $m$ is even then

$$\mathrm{cli}_m(\pi-\theta) = \sum_{k=1}^\infty \frac{\sin(k\pi -k\theta)}{k^m} = -\sum_{k=1}^\infty (-1)^k \frac{\sin(k\theta)}{k^m} $$

$$\sum_{k=1}^\infty (-1)^k \frac{\sin(k\theta)}{k^m} + \sum_{k=1}^\infty \frac{\sin(k\theta)}{k^m} = \frac{1}{2^{m-1}}\sum_{k=1}^\infty \frac{\sin(2k\theta)}{k^m} $$

This implies that

$$\mathrm{cl}_m(2 \theta) = 2^{m-1}(\mathrm{cl}_m(\theta)- \mathrm{cl}_m(\pi-\theta))$$

If $m$ is odd then

$$\mathrm{cli}_m(\pi-\theta) = \sum_{k=1}^\infty \frac{\cos(k\pi -k\theta)}{k^m} = \sum_{k=1}^\infty (-1)^k \frac{\cos(k\theta)}{k^m} $$

$$\sum_{k=1}^\infty \frac{\cos(k\theta)}{k^m} + \sum_{k=1}^\infty (-1)^k \frac{\cos(k\theta)}{k^m} = \frac{1}{2^{m-1}}\sum_{k=1}^\infty \frac{\sin(2k\theta)}{k^m} $$

Which implies that

$$\mathrm{cl}_m(2 \theta) = 2^{m-1}(\mathrm{cl}_m(\theta)+ \mathrm{cl}_m(\pi-\theta))$$

Collecting the results we have

$$\mathrm{cl}_m(2 \theta) = 2^{m-1}(\mathrm{cl}_m(\theta)-(-1)^m \mathrm{cl}_m(\pi-\theta))$$

Find the integral


$$\int^\pi_0 \mathrm{cl}_{m}(\theta) d\theta \,\,\,\,\,\,\,\,\,\, m \text{ is even}$$

Using the series representation

$$\int^\pi_0 \sum_{k=1}^\infty \frac{\sin(k\theta)}{k^{m}} d\theta$$

Swap the integral and the series

$$\sum_{k=1}^\infty \frac{1}{k^{m}}\int^\pi_0 \sin(k\theta) d\theta$$

The integral

$$\int^\pi_0 \cos(k\theta) d\theta =-\left[ \frac{1}{k}\sin(k\theta) \right]^\pi_0=\frac{-(-1)^k+1}{k}$$

We get the summation

$$\sum_{k=1}^\infty \frac{-(-1)^k+1}{k^{m+1}} =\zeta(m+1)+\eta(m+1)$$

Now use that

$$\eta(s) = (1-2^{1-s})\zeta(s)$$

$$\sum_{k=1}^\infty \frac{-(-1)^k+1}{k^{m+1}} =\zeta(m+1)+(1-2^{-m})\zeta(m+1) = \zeta(m+1)(2-2^{-m})$$

 

[alyafey22]

Re: (AIT) Clausen functions

Find the integral for $m$ is even

$$\int^\infty_0 \mathrm{cl}_m(\theta) e^{-n\theta}\, d\theta$$

Using the series representation

$$\int^\infty_0 \sum_{k=1}^\infty \frac{\sin(k\theta)}{k^m} e^{-n\theta}\, d\theta$$

Swap the integral and the series

$$\sum_{k=1}^\infty \frac{1}{k^m}\int^\infty_0 \sin(k\theta) e^{-n\theta}\, d\theta$$

Using the Laplace transform we have

$$\sum_{k=1}^\infty \frac{1}{k^{m-1}(k^2+n^2)}$$

Add and subtract $k^2$ and divide by $n^2$

$$\frac{1}{n^2}\sum_{k=1}^\infty \frac{k^2+n^2-k^2}{k^{m-1}(k^2+n^2)}$$

Distribute the numerator

$$\frac{1}{n^2}\zeta(m-1)-\frac{1}{n^2}\sum_{k=1}^\infty\frac{1}{k^{m-3}(k^2+n^2)}$$

Continue this approach to conclude that

$$\sum_{i= 1}^j(-1)^{i-1}\frac{1}{n^{2i}}\zeta(m-(2i-1))+\frac{(-1)^j}{n^{2j}}\sum_{k=1}^\infty\frac{1}{k^{m-(2j+1)}(k^2+n^2)}$$

Let $m-2j-1 = 1$ which implies that $j = m/2-1$

$$\sum_{i= 1}^{m/2-1}(-1)^{i-1}\frac{1}{n^{2i}}\zeta(m-(2i-1))+\frac{(-1)^{m/2-1}}{n^{m-2}}\sum_{k=1}^\infty\frac{1}{k(k^2+n^2)}$$

Now let us look at the sum

$$\sum_{k=1}^\infty\frac{1}{k(k^2+n^2)} = \sum_{k=1}^\infty\frac{1}{2ink}\left\{ \frac{1}{k-in}-\frac{1}{k+in}\right\}$$

Which can be written as

$$\sum_{k=1}^\infty\frac{1}{k(k^2+n^2)} = \frac{1}{2n^2}\sum_{k=1}^\infty\frac{1}{k}\left\{ \frac{in}{k+in}+\frac{-in}{k-in}\right\}$$

According to the digamma function

$$\sum_{k=1}^\infty\frac{1}{k(k^2+n^2)} = \frac{1}{2n^2}\left\{ \gamma+\psi(1+in)+\psi(1-in)+\gamma\right\}$$

which simplifies to

$$\sum_{k=1}^\infty\frac{1}{k(k^2+n^2)} = \frac{\psi(1-in)+\psi(1+in)+2\gamma}{2n^2}$$

Now we we can verify $\psi(1-in) = \overline{\psi(1+in)}$ which suggests that

$$\psi(1+in) +\psi(1-in) = 2\Re\left\{\psi(1+in) \right\} $$

Hence we have the sum

$$\sum_{k=1}^\infty\frac{1}{k(k^2+n^2)} = \frac{2\Re\left\{\psi(1+in) \right\}+2\gamma}{2n^2}=\frac{\Re\left\{\psi(1+in) \right\}+\gamma)}{n^2}$$

This concludes to

$$\sum_{j= 1}^{m/2-1}(-1)^{j-1}\frac{\zeta(m-(2i-1))}{n^{2j}}+(-1)^{m/2-1}\frac{\Re\left\{\psi(1+in) \right\}+\gamma}{n^m}$$

Find the following integral

$$\int^\infty_0 \mathrm{cl}_m(\theta) \theta^{n-1}\, d\theta$$

If $m$ is even

$$\int^\infty_0 \sum_{k=1}^\infty \frac{\sin(k\theta)}{k^m}\theta^{n-1}\, d\theta$$

Using the series representation

$$\sum_{k=1}^\infty \frac{1}{k^m}\int^\infty_0 \sin(k\theta) \theta^{n-1}d\theta$$

Let $\phi = k\theta $

$$\sum_{k=1}^\infty \frac{1}{k^{m+n}}\int^\infty_0 \sin(\phi) \phi^{n-1}d\phi$$

Now using the mellin transform

$$\zeta(m+n) \sin\left( \frac{\pi}{2}n\right)\Gamma(n)$$

If $m$ is odd
$$\zeta(m+n) \cos\left( \frac{\pi}{2}n\right)\Gamma(n)$$

 

[alyafey22]

Re: (AIT) Clausen functions

Clausen Integral function


$$\mathrm{cl}_2(\theta) = -\int^\theta_0 \log\left[2\sin\left(\frac{\phi}{2}\right)\right]d\phi$$

Start by the following

$$\mathrm{Li}_2(e^{i\theta}) = \sum_{k=1}^\infty \frac{e^{ik\theta}}{k^2}= \sum_{k=1}^\infty \frac{\cos(k\theta)}{k^2}+i\sum_{k=1}^\infty \frac{\sin(k\theta)}{k^2} $$

By the integral definition of the dilogarithm

$$\mathrm{Li}_2(e^{i\theta}) -\zeta(2)= -\int^{e^{i\theta}}_1 \frac{\log(1-x)}{x}\,dx$$

Let $x = e^{i\phi}$

$$\mathrm{Li}_2(e^{i\theta}) -\zeta(2)= -i\int^{\theta}_0 \log(1-e^{i\phi})d \phi$$

Let us look at the following

$$1-e^{i\phi} = 1-\cos(\phi)-i\sin(\phi) = 2\sin^2(\phi/2)-2i\sin(\phi/2)\cos(\phi/2)$$

Which simplifies to

$$1-e^{i\phi} =2\sin(\phi/2)\left[\sin(\phi/2)-i\cos(\phi/2)\right] = 2\sin(\phi/2) e^{-(i/2) (\pi-\phi)}$$

Hence our integral

$$\mathrm{Li}_2(e^{i\theta}) -\zeta(2)= -i\int^{\theta}_0 \log \left[2\sin(\phi/2) e^{-(i/2) (\pi-\phi)}\right]d \phi$$

Use the complex integral properties

$$\mathrm{Li}_2(e^{i\theta}) -\zeta(2)= -i\int^{\theta}_0 \log \left[2\sin(\phi/2)\right]d\phi +\frac{1}{4}(\pi-\theta)^2-\frac{1}{4}\pi^2$$

By equating the imaginary parts we have our result.

We can see the special value
$$\mathrm{cl}_2\left(\frac{\pi}{2}\right) = \sum_{k=1}^\infty \frac{\sin(k\pi/2)}{k^2}= \sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)^2} = G$$

Where $G$ is the Catalan's constant.

Prove the following


$$\mathrm{cl}_2(2\theta) =2\mathrm{cl}_2(\theta)-\mathrm{cl}_2(\pi -\theta) $$

Use the integral representation

$$\mathrm{cl}_2(2\theta) = -\int^{2\theta}_0 \log\left[2\sin\left(\frac{t}{2}\right)\right]dt$$

Let $t= 2\phi$

$$-2\int^{\theta}_0 \log\left[2\sin\left(\phi\right)\right]d\phi$$

Use double angle identity

$$-2\int^{\theta}_0 \log\left[4\sin\left(\frac{\phi}{2}\right) \cos\left(\frac{\phi}{2}\right)\right]d\phi$$

Separate the logarithms

$$-2\int^{\theta}_0 \log\left[2\sin\left(\frac{\phi}{2}\right) \right]d\phi-2\int^{\theta}_0 \log\left[2\cos\left(\frac{\phi}{2}\right) \right]d\phi$$

We can verify that

$$\mathrm{cl}_2(\pi -\theta ) = \int^{\theta}_0 \log\left[2\cos\left(\frac{\phi}{2}\right) \right]d\phi$$

Hence

$$\mathrm{cl}_2(2\theta) =2\mathrm{cl}_2(\theta)-\mathrm{cl}_2(\pi -\theta)$$

Using that

$$\mathrm{cl}_2(3\pi) =2\mathrm{cl}_2\left(\frac{3\pi}{2}\right)-2\mathrm{cl}_2\left(-\frac{\pi}{2}\right) $$

Since $\mathrm{cl}_2(3\pi) = 0$
$$\mathrm{cl}_2\left(\frac{3\pi}{2}\right)=\mathrm{cl}_2\left(-\frac{\pi}{2}\right) = - \mathrm{cl}_2\left(\frac{\pi}{2}\right)= -G $$

Prove that

$$\int^{2\pi}_0 \mathrm{cl}_2(x)^2dx = \frac{\pi^5}{90}$$

Using the series representation

$$\sum_{k=1}^\infty \sum_{n=1}^\infty \frac{1}{(nk)^2}\int^{2\pi}_0 \sin(kx) \sin(nx) dx$$

Consider the integral

$$\int_0^{2\pi} \sin(kx)\sin(nx)\,dx = \frac{1}{2}\int_0^{2\pi} \cos((k-n)x)-\cos((k+n)x)\,dx $$

Consider two cases

If $n = k $ then

$$\frac{1}{2}\int_0^{2\pi} 1-\cos(2nx)\,dx = \pi $$

If $n \neq k$

$$\frac{1}{2}\int_0^{2\pi} \cos((k-n)x)-\cos((k+n)x)\,dx =\frac{1}{2}\left[\frac{\sin((k-n)x)}{k-n}-\frac{\sin((k+n)x)}{k+n} \right]^{2\pi}_0 =0 $$

Hence we have

$$\int_0^{2\pi} \sin(kx)\sin(nx)\,dx=\begin{cases} 0 & n \neq k\\ \pi & n=k\end{cases}$$

We can write the series as

$$\sum_{n=1}^\infty \sum_{k=1}^\infty \frac{1}{(nk)^2} = \sum_{n\neq k}^\infty \frac{1}{(nk)^2}+ \sum_{n=1}^\infty \frac{1}{n^4} $$

Now since the integral $n\neq k $ goes to zero the result reduces to

$$\pi \sum_{n=1}^\infty \frac{1}{n^4} =\pi \zeta(4) = \frac{\pi^5}{90} $$

 

[alyafey22]

(AIT) Barnes G function

4.Integration using special functions (continued)


4.15. Barnes G function

---

Define the following

$$ G(z+1)=(2\pi)^{z/2}\exp\left(-\frac{z+z^2(1+\gamma)}{2}\right)\prod_{n=1}^{\infty}\left\{\left(1+\frac{z}{n} \right)^n\exp\left(\frac{z^2}{2n}-z\right) \right\}$$

---

Difference formula

$$G(z+1) = \Gamma(z) G(z)$$

From the series representation we have

$$ \frac{G(z+1)}{G(z)} = \sqrt{2 \pi} \exp \left(-z - \gamma z + \frac{\gamma}{2} \right) \prod_{k=1}^{\infty} \left(\frac{k+z}{k+z-1} \right)^{k} \exp \left( \frac{2z-1-2k}{2k} \right). $$

This can be written as

$$\frac{G(z+1)}{G(z)} = z\sqrt{2 \pi} \exp \left(-z + \frac{\gamma}{2} \right) \prod_{k=1}^{\infty} \left(\frac{k+z}{k+z-1} \right)^{k} \exp \left( -\frac{1+2k}{2k} \right)\left(1+\frac{z}{k}\right)\\ \frac{e^{- \gamma z }}{z} \prod_{k=1}^{\infty} \left(1+\frac{z}{k} \right)^{-1} e^{\left( \frac{z}{k} \right)}$$

Which simplifies to

$$\frac{G(z+1)}{G(z)} = z\Gamma(z)\sqrt{2 \pi} \exp \left(-z + \frac{\gamma}{2} \right) \prod_{k=1}^{\infty} \left(\frac{k+z}{k+z-1} \right)^{k} \exp \left( -\frac{1+2k}{2k} \right)\left(1+\frac{z}{k}\right)$$

It suffices to prove that

$$ z\sqrt{2 \pi} \exp \left(-z + \frac{\gamma}{2} \right) \prod_{k=1}^{\infty} \left(\frac{k+z}{k+z-1} \right)^{k} \exp \left( -\frac{1+2k}{2k} \right)\left(1+\frac{z}{k}\right) = 1$$

or

$$ \prod_{k=1}^{\infty} \left(\frac{k+z}{k+z-1} \right)^{k} \exp \left( -\frac{1+2k}{2k} \right)\left(1+\frac{z}{k}\right) = \frac{\exp \left(z -\frac{\gamma}{2} \right)}{z\sqrt{2 \pi}}$$Start by

$$\lim_{N \to \infty }\prod_{k=1}^{N} \left(\frac{k+z}{k+z-1} \right)^{k} \exp \left( -\frac{1+2k}{2k} \right)\left(1+\frac{z}{k}\right) $$

Notice

\begin{align}
\prod_{k=1}^{N} \left(\frac{k+z}{k+z-1} \right)^{k} \left(1+\frac{z}{k}\right)&= \frac{\prod_{k=1}^{N} (k+z)^k \prod_{k=1}^{N}\left(1+\frac{z}{k}\right)}{\prod_{k=1}^{N}(k+z-1)^k} \\[1em]
&= \frac{\prod_{k=1}^{N} (k+z)^k \prod_{k=1}^{N} (k+z)}{zN!\prod_{k=1}^{N-1}(k+z)^{k+1}} \\[1em]
&= \frac{(N+z)^{N+1}\prod_{k=1}^{N-1} (k+z)^k\prod_{k=1}^{N-1} (k+z)}{zN!\prod_{k=1}^{N-1}(k+z)^{k+1}}\\[1em]
&= \frac{(N+z)^{N+1}\prod_{k=1}^{N-1} (k+z)^{k+1}}{zN!\prod_{k=1}^{N-1}(k+z)^{k+1}}\\[1em]
&= \frac{(N+z)^{N+1}}{zN!}
\end{align}

The second product

$$\prod_{k=1}^{N} \exp \left( -\frac{1+2k}{2k} \right) =\exp \left( -\sum_{k=1}^N\frac{1+2k}{2k} \right) = e^{-\frac{1}{2}H_N-N} $$

Hence we have the following
$$e^{-\frac{1}{2}H_N-N}\times \frac{(N+z)^{N+1}}{z N! } $$

According to Stirling formula we have

$$e^{-\frac{1}{2}H_N-N}\frac{(N+z)^{N+1}}{z N! } \sim e^{-\frac{1}{2}H_N-N}\frac{(N+z)^{N+1}}{z(N/e)^N}\times \frac{1}{\sqrt{2\pi N}} $$

By some simplifications we have

$$\frac{e^{-\frac{1}{2}(H_N-\log N)}}{z}\left(1+\frac{z}{N}\right) \times \left(1+ \frac{z}{N}\right)^N\times \frac{1}{\sqrt{2\pi}} \sim \frac{\exp\left(-\frac{\gamma}{2} +z\right)}{z\sqrt{2\pi}}$$

Where we used that

$$\lim_{n \to \infty }H_n -\log(n)= \gamma$$

Reflection formula


$$\log \left\{\frac{G(1-z)}{G(1+z)} \right\} = z \log\left(\frac{\sin(\pi z)}{\pi}\right) + \frac{\mathrm{cl}_2(2\pi z)}{2\pi}$$

Start by the series expansion

$$\frac{G(1-z)}{G(1+z)} = \frac{(2\pi)^{-z/2}\exp\left(\frac{z-z^2(1+\gamma)}{2}\right)\prod_{n=1}^{\infty}\left\{\left(1-\frac{z}{n} \right)^n\exp\left(\frac{z^2}{2n}+z\right) \right\}}{(2\pi)^{z/2}\exp\left(-\frac{z+z^2(1+\gamma)}{2}\right)\prod_{n=1}^{\infty}\left\{\left(1+\frac{z}{n} \right)^n\exp\left(\frac{z^2}{2n}-z\right) \right\}}$$

This simplifies to

$$\frac{G(1-z)}{G(1+z)} = (2\pi)^{-z}e^z\prod_{n=1}^{\infty}\frac{(n-z)^n}{(n+z)^n} e^{2z}$$

Take the log of both sides

$$\log \left\{\frac{G(1-z)}{G(1+z)} \right\} = -z \log(2\pi)+z + \log\left \{ \prod_{n=1}^{\infty}\frac{(n-z)^n}{(n+z)^n} e^{2z} \right\}$$
Let the following

$$f(z) = \log\left\{\prod_{n=1}^{\infty}\frac{(n-z)^n}{(n+z)^n} e^{2z} \right\} = \sum_{n=1}^\infty n\log(n-z)-n\log(n+z)+2z$$

Differentiate with respect to $z$

$$f'(z) =\sum_{n=1}^{\infty}\frac{-n}{n-z}-\frac{n}{n+z}+2 = \sum_{n=1}^{\infty}\frac{-n(n+z)-n(n-z)+2(n^2-z^2)}{n^2-z^2} $$

Hence we have

$$\sum_{n=1}^{\infty}\frac{-2z^2}{n^2-z^2}$$

Now we can use the following

$$z\pi \cot \pi z = 1+\sum_{n=1}^{\infty}\frac{-2z^2}{n^2-z^2}
$$

Hence we conclude that

$$f'(z) = z\pi \cot \pi z-1$$

Integrate with respect to $z$

$$f(z) = \int^z_0 x\pi \cot(\pi x) \, dx -z$$

Hence we have

$$\log \left\{\frac{G(1-z)}{G(1+z)} \right\} = -z \log(2\pi)+\int^z_0 z\pi \cot(\pi x) \, dx$$

Now use integration by parts for the integral
\begin{align}
\int^z_0 x\pi \cot(\pi x) \, dx
&=z\log(\sin\pi z)-\int^z_0 \log(\sin\pi x) dx\\
&=z\log(2\sin\pi z)-\int^z_0 \log(2\sin\pi x) dx\\
&=z\log(2\sin\pi z)-\frac{1}{2\pi }\int^{2\pi z}_0 \log\left(2\sin\frac{x}{2}\right) dx\\
&=z\log(2\sin\pi z)+\frac{\mathrm{cl}_2(2\pi z)}{2\pi}\\
\end{align}

That implies

$$\log \left\{\frac{G(1-z)}{G(1+z)} \right\} = z\log(2\sin\pi z)-z \log(2\pi)+ \frac{\mathrm{cl}_2(2\pi z)}{2\pi}$$

which implies our result.

 

[alyafey22]

Re: (AIT) Barnes G function

Values at positive integers

$$G(n) = \prod^{n-1}_{k=1} \Gamma(k)$$

It can be proved by induction. For $G(1) = 1$, suppose

$$G(n) = \prod^{n-1}_{k=1} \Gamma(k)$$

We want to show

$$G(n+1) = \prod^{n}_{k=1} \Gamma(k)$$

By the difference formula

$$G(n+1) = \Gamma(n)G(n) = \Gamma(n) \prod^{n-1}_{k=1} \Gamma(k) =\prod^{n}_{k=1} \Gamma(k) $$

Loggamma integral

$$\int^z_0 \log \Gamma(x) dx =\frac{z}{2}\log(2\pi)+\frac{z(z-1)}{2}+z\log \Gamma(z)-\log G(z+1)$$

Take the log to the series representation

$$ \log G(z+1)=\frac{z}{2}\log(2\pi)-\frac{z+z^2(1+\gamma)}{2}+\sum_{n=1}^{\infty}n \log\left(1+\frac{z}{n} \right)+\frac{z^2}{2n}-z$$

Let the following

$$f(z) = \sum_{n=1}^{\infty}n \log\left(1+\frac{z}{n} \right)+\frac{z^2}{2n}-z$$

Differentiate with respect to $z$

$$f(z) = \sum_{n=1}^{\infty}\frac{n}{z+n}+\frac{z}{n}-1 =\sum_{n=1}^{\infty}\frac{z^2}{n(n+z)} $$

Now use the following

$$\psi(z) = -\gamma -\frac{1}{z}+\sum_{n=1}^\infty \frac{z}{n(n+z)}$$

which implies that

$$\sum_{n=1}^\infty \frac{z^2}{n(n+z)} = z\psi(z)+\gamma z+1$$

Hence we have

$$f(z) = z\psi(z)+\gamma z+1$$

Integrate with respect to $z$

$$f(z) = \int^z_0 x\psi(x) dx +\frac{\gamma z^2}{2}+z$$

which implies that

$$f(z) = z\log \Gamma(z) -\int^z_0 \log \Gamma(x) dx +\frac{\gamma z^2}{2}+z$$

Hence we have

$$ \log G(z+1)=\frac{z}{2}\log(2\pi)-\frac{z+z^2(1+\gamma)}{2}+z\log \Gamma(z) -\int^z_0 \log \Gamma(x) dx +\frac{\gamma z^2}{2}+z$$

By some rearrangements we have

$$\int^z_0 \log \Gamma(x) dx =\frac{z}{2}\log(2\pi)+\frac{z(z-1)}{2}+z\log \Gamma(z)-\log G(z+1)$$

 

[alyafey22]

Re: (AIT) Barnes G function

Relation to Hyperfactorial function


Prove for $n$ is a positive integer

$$G(n+1) = \frac{(N!)^n}{H(n)}$$

Where $H(n)$ is the hyperfactorial function

$$H(n) = \prod^n_{k=1}k^k$$

proof

We can prove it by induction for $n=0$ we have, $G(1)=1$,

suppose that

$$G(n) = \frac{\Gamma(n)^{n-1}}{H(n-1)}$$

we want to find

$$G(n+1) = \Gamma(n)G(n) = \Gamma(n) \frac{\Gamma(n)^{n-1}}{H(n-1)}$$

Notice that

$$H(n-1) = \prod^{n-1}k^k = \frac{\prod^{n}k^k}{n^n} = \frac{H(n)}{n^n }$$

We deduce that

$$G(n+1) = \Gamma(n)G(n) = \frac{\Gamma(n)^{n}\times n^n}{H(n)} = \frac{(N!)^n}{H(n)}$$

A related constant


We define the Glaisher-Kinkelin constant as

$$A = \lim_{n \to \infty}\frac{H(n)}{n^{n^2/2+n/2+1/12}e^{-n^2/4}}$$

Prove that

$$\lim_{n \to \infty}\frac{G(n+1)}{(2\pi)^{n/2}n^{n^2/2-1/12} e^{-3n^2/4}} = \frac{e^{1/12}}{A}$$

proof

From the previous result we have

$$\lim_{n \to \infty}\frac{(N!)^n}{H(n)(2\pi)^{n/2}n^{n^2/2-1/12} e^{-3n^2/4}}$$

Now use the Stirling approximation

$$(N!)^n \sim (2\pi)^{n/2}n^{n^2+n/2}e^{-n^2+1/12}$$

Hence we deduce that

$$\lim_{n \to \infty}\frac{(2\pi)^{n/2}n^{n^2+n/2}e^{-n^2+1/12}}{H(n)}\times\frac{1}{(2\pi)^{n/2}n^{n^2/2-1/12} e^{-3n^2/4}}$$

By simplifications we have

$$e^{1/12}\lim_{n \to \infty}\frac{n^{n^2/2+n/2+1/12}e^{-n^2/4}}{H(n)} = \frac{e^{1/12}}{A}$$

Exercise

$$\zeta'(2) = \frac{\pi^2}{6}\left(\log(2\pi)+\gamma-12\log A \right)$$

We already proved that

$$\log \left[\frac{G(n+1)}{(2\pi)^{n/2}n^{n^2/2-1/12} e^{-3n^2/4}} \right] \sim \frac{1}{12}-\log A$$

Let the following

$$f(n) = \log \left[\frac{G(n+1)}{(2\pi)^{n/2}n^{n^2/2-1/12} e^{-3n^2/4}} \right]$$

Use the series representation of the Barnes functions

$$f(n) = \log \left[ \frac{(2\pi)^{n/2}\exp\left(-\frac{n+n^2(1+\gamma)}{2}\right)\prod_{k=1}^{\infty}\left\{\left(1+\frac{n}{k} \right)^k\exp\left(\frac{n^2}{2k}-n\right) \right\}}{(2\pi)^{n/2}n^{n^2/2-1/12} e^{-3n^2/4}} \right]$$

Which reduces to

$$f(n) = -\frac{n+n^2(1+\gamma)}{2}+\sum^\infty_{k=1}\left\{k \log \left(1+\frac{n}{k} \right)+\frac{n^2}{2k}-n \right\}\\ -\left( \frac{n^2}{2}-\frac{1}{12}\right)\log(n)+\frac{3n^2}{4}$$

Differentiate with respect to $n$

$$f'(n) = -\frac{1}{2}-n-\gamma n+n\psi(n)+\gamma n+1-n\log(n)-\frac{n}{2}+\frac{1}{12n}+\frac{3n}{2}$$

Note that we already showed that

$$\frac{d}{dn}\sum^\infty_{k=1}\left\{k \log \left(1+\frac{n}{k} \right)+\frac{n^2}{2k}-n\right\}= n\psi(n)+\gamma n+1$$

By simplifications we have

$$f'(n) = n\psi(n)-n\log(n)+\frac{1}{12n}+\frac{1}{2}$$

Now use that

$$\psi(n) = \log(n)-\frac{1}{2n}-2\int^\infty_0 \frac{z dz}{(n^2+z^2)(e^{2\pi z}-1)}dz$$

Hence we deduce that

$$f'(n) =-2\int^\infty_0 \frac{nz dz}{(n^2+z^2)(e^{2\pi z}-1)}dz+\frac{1}{12n}$$

Integrate with respect to $n$

$$f(n) =-\int^\infty_0 \frac{z\log(n^2+z^2)}{(e^{2\pi z}-1)}dz+\frac{1}{12}\log(n)+C$$

Take the limit $n \to 0$

$$C = \lim_{n \to 0}f(n)-\frac{1}{12}\log(n)+\int^\infty_0 \frac{z\log(z^2)}{(e^{2\pi z}-1)}dz $$

Hence we have the limit

$$\lim_{n \to 0}\frac{G(n+1)}{(2\pi)^{n/2}n^{n^2/2-1/12} e^{-3n^2/4}}-\frac{1}{12}\log(n) = \lim_{n \to 0}\log \frac{G(n+1)}{(2\pi)^{n/2}n^{n^2/2} e^{-3n^2/4}} = 0$$

Hence we see that

$$C = 2\int^\infty_0 \frac{z\log(z)}{(e^{2\pi z}-1)}dz $$

Finally we have

$$f(n) =-\int^\infty_0 \frac{z\log(n^2+z^2)}{(e^{2\pi z}-1)}dz+\frac{1}{12}\log(n)+ 2\int^\infty_0 \frac{z\log(z)}{(e^{2\pi z}-1)}dz$$

$$f(n) =-\int^\infty_0 \frac{z\log\left(1+\frac{z^2}{n^2} \right)}{(e^{2\pi z}-1)}dz-\log(n^2)\int^\infty_0 \frac{z}{(e^{2\pi z}-1)}dz+\frac{1}{12}\log(n)\\+ 2\int^\infty_0 \frac{z\log(z)}{(e^{2\pi z}-1)}dz$$

Also we have

$$\int^\infty_0 \frac{z}{(e^{2\pi z}-1)}dz = \frac{1}{24}$$

That simplifies to

$$f(n)=-\int^\infty_0 \frac{z\log\left(1+\frac{z^2}{n^2} \right)}{(e^{2\pi z}-1)}dz+2\int^\infty_0 \frac{z\log(z)}{(e^{2\pi z}-1)}dz$$

Take the limit $n \to \infty $

$$2\int^\infty_0 \frac{z\log(z)}{(e^{2\pi z}-1)}dz = \frac{1}{12}-\log A$$

Now use that

$$
\begin{align}
2\int^\infty_0 \frac{z\log(z)}{(e^{2\pi z}-1)}dz
&= 2\int^\infty_0 \frac{z\log(z)}{e^{2\pi z}}\times \frac{1}{1-e^{-2\pi z}}dz\\
&= 2\sum_{n=0}^\infty \int^\infty_0 e^{-2\pi z(n+1)}z\log(z)\,dz\\
&= \sum_{n=1}^\infty \frac{\psi(2) − \log(2\pi)+\log(n)}{2\pi^2 n^2}\\
&=\frac{(\psi(2)-\log(2\pi))\zeta(2)+\zeta'(2)}{2\pi^2}\\
\end{align}
$$

Hence we conclude that

$$\zeta'(2) =(\log(2\pi)-\psi(2))\zeta(2) + 2\pi^2\left(\frac{1}{12}-\log A \right) = \zeta(2) (\log(2\pi)+\gamma-12 \log A)$$

 

[alyafey22]

Re: (AIT) Barnes G function

Relation to hurwitz zeta function


Prove that

$$\log G(z+1) - z \log \Gamma(z) = \zeta'(-1) - \zeta'(-1,z) $$

Start by the following

$$\zeta(s,z) = \frac{z^{-s}}{2}+\frac{z^{1-s}}{s-1}+2\int^\infty_0 \frac{\sin(s\arctan(x/z))}{(z^2+x^2)^{s/2}(e^{2\pi x}-1)}\,dx$$

Hence we have

$$\zeta'(-1,z) = -\frac{z\log(z)}{2}+\frac{z^2\log(z)}{2}-\frac{z^2}{4}+\int^\infty_0 \frac{x\log(x^2+z^2)+2z\arctan(x/z)}{(e^{2\pi x}-1)}\,dx$$

Now use that

$$\psi(z) = \log(z)-\frac{1}{2z}-2\int^\infty_0 \frac{x }{(z^2+x^2)(e^{2\pi x}-1)}dx$$

Which implies that

$$\int^\infty_0 \frac{2zx }{(z^2+x^2)(e^{2\pi x}-1)}dx=z\log(z)-\frac{1}{2}-z\psi(z)$$

By taking the integral

$$\int^\infty_0 \frac{x\log(x^2+z^2) -x\log(x^2)}{(e^{2\pi x}-1)}dx=\int^z_0x\log(x)\,dx-\int^z_0x\psi(x)\,dx-\frac{z}{2}$$

Which simplifies to

$$\int^\infty_0 \frac{x\log(x^2+z^2) }{(e^{2\pi x}-1)}dx=\zeta'(-1)-\frac{z^2}{4}+\frac{1}{2} z^2 \log(z)-z\log\Gamma(z)+\int^z_0\log\Gamma(x)\,dx-\frac{z}{2}$$

Also we have

$$2\int^\infty_0 \frac{x}{(x^2+z^2)(e^{2\pi x}-1)}dx=\log(z)-\frac{1}{2z}-\psi(z)$$

By integration we have

$$2\int^\infty_0 \frac{\arctan(x/z)}{(e^{2\pi x}-1)}dx=z+\frac{\log(z)}{2}-z \log(z)+\log\Gamma(z)+C$$

Let $z \to 1$ to evaluate the constant

$$2\int^\infty_0 \frac{\arctan(x/z)}{(e^{2\pi x}-1)}dx=z+\frac{\log(z)}{2}-z \log(z)+\log\Gamma(z)-\frac{1}{2}\log(2\pi)$$

Multiply by $z$

$$2\int^\infty_0 \frac{z\arctan(x/z)}{(e^{2\pi x}-1)}dx=z^2+\frac{z\log(z)}{2}-z^2 \log(z)+z\log\Gamma(z)-\frac{z}{2}\log(2\pi)$$

Substitute both integrals our formula

$$\zeta'(-1,z) =-\frac{z\log(z)}{2}+\frac{z^2\log(z)}{2}-\frac{z^2}{4}+z^2+\frac{z\log(z)}{2}-z^2 \log(z)+z\log\Gamma(z)\\-\frac{z}{2}\log(2\pi)+\zeta'(-1)-\frac{z^2}{4}+\frac{1}{2} z^2 \log(z)-z\log\Gamma(z)+\int^z_0\log\Gamma(x)\,dx-\frac{z}{2}$$

Which reduces to

$$\int_{0}^{z} \log \Gamma(x) \, \mathrm dx = \frac{z}{2} \log(2 \pi) + \frac{z(1-z)}{2} - \zeta^{'}(-1) + \zeta^{'}(-1,z)$$

We also showed that

$$\int_{0}^{z} \log \Gamma(x) \, \mathrm dx = \frac{z}{2} \log (2 \pi) + \frac{z(1-z)}{2} + z \log \Gamma(z) - \log G(z+1)$$

By equating the equations we get our result.

Prove that


$$\zeta'(-1) = \frac{1}{12}-\log A$$

Start by

$$\zeta(s) = \lim_{m \to \infty} \left( \sum_{k=1}^{m} k^{-s} - \frac{m^{1-s}}{1-s} - \frac{m^{-s}}{2} + \frac{sm^{-s-1}}{12} \right) \ , \ \text{Re}(s) >-3.$$

Differentiate with respect to $s$

$$\zeta'(s) = \lim_{m \to \infty} \left(- \sum_{k=1}^{m} k^{-s}\log(k) - \frac{m^{1-s}}{(1-s)^2} +\frac{m^{1-s}}{1-s}\log(m)+\frac{m^{-s}}{2}\log(m)\\ + \frac{m^{-s-1}}{12}-\frac{m^{-s-1}}{12}\log(m)\right) $$

Now let $s \to -1$

$$\zeta'(-1) = \lim_{m \to \infty} \left(- \sum_{k=1}^{m} k\log(k) - \frac{m^{2}}{4} +\frac{m^{2}}{2}\log(m)+\frac{m}{2}\log(m)\\ + \frac{1}{12}-\frac{1}{12}\log(m)\right) $$

Take the exponential of both sides

$$e^{\zeta'(-1)}= e^{1/12}\lim_{m \to \infty} \frac{m^{m^2/2+m/2-1/12} e^{-m^2/4}}{ e^{\sum_{k=1}^{m} k\log(k)}} = e^{1/12}\lim_{m \to \infty} \frac{m^{m^2/2+m/2-1/12} e^{-m^2/4}}{ H(m+1)} = \frac{ e^{1/12}}{A} $$

We conclude that

$$\zeta'(-1) = \frac{1}{12}-\log A$$

Prove that

$$G\left(\frac{1}{2}\right) = 2^{1/24} \pi^{-1/4}e^{1/8}A^{-3/2}$$

We know that

$$\log G(z)+ \log \Gamma(z)- z \log \Gamma(z) = \zeta'(-1) - \zeta'(-1,z) $$

Note that

$$\zeta \left(s, \frac{1}{2} \right) = (2^s-1)\zeta(s)$$

Which implies that

$$\zeta' \left(-1, \frac{1}{2} \right) = \frac{\log(2)}{2}\zeta(-1)-\frac{1}{2}\zeta'(-1)$$

Hence we have

$$\log G\left(\frac{1}{2} \right)+ \frac{1}{2}\log \Gamma\left(\frac{1}{2} \right)= \frac{3}{2}\zeta'(-1)-\frac{\log(2)}{2}\zeta(-1) $$

Using that we have

$$G\left(\frac{1}{2}\right) = 2^{1/24}\pi^{-1/4}e^{\frac{3}{2}\zeta'(-1)}$$

Note that

$$\zeta(-1) = -\frac{1}{12}$$

This can be proved by the functional equation of the zeta function.