Exploring Intriguing Integrals

[polygamma]

Here's an eclectic bunch.1) $ \displaystyle \int_{0}^{\infty} \frac{x^{2}}{(1+x^{5})(1+x^{6})} \ dx $2) $ \displaystyle \int_{0}^{\infty} \frac{1}{(1+x^{\varphi})^{\varphi}} \ dx $ where $\varphi$ is the golden ratio3) $ \displaystyle \int_{0}^{\infty} \sin \left(x^{2} + \frac{1}{x^{2}} \right) \ dx $4) $ \displaystyle \int_{0}^{\infty} e^{-x} \text{erf}(\sqrt{x}) \ dx $ where $\text{erf}(x)$ is the error function5) $\displaystyle \int_{0}^{2 \pi} \cos (\cos x) \cosh (\sin x) \ dx $

 

[Sherlock1]

1) $ \displaystyle \int_{0}^{\infty} \frac{x^{2}}{(1+x^{5})(1+x^{6})} \ dx $

Sub $x = \tan(t)$, then $\displaystyle I = \int_{0}^{\pi/2}\frac{\tan^2{t}~\sec^2{t}}{\left(1+\tan^{6}{t} \right)\left(1+\tan^5{t}\right)}\;{dt}.$ Sub $t \mapsto \frac{\pi}{2}-t$, then $ \displaystyle I = \int_{0}^{\pi/2}\frac{\tan^2{t}~\cot^2{t}}{\left(1+\cot^{6}{t} \right)\left(1+\cot^5{t}\right)}\;{dt}.$ Then by adding $\displaystyle 2I = \int_{0}^{\pi/2}\frac{2\sin^2{2t}}{3\cos{4t}+5}\;{dt} = \int_{0}^{\pi/2}\frac{1}{1+4\cot^2{2t}}\;{dt}$. Sub $t = \tan^{-1}{\ell}$ then $ \displaystyle I = \frac{1}{2}\int_{0}^{\infty}\frac{\ell^2}{\ell^6+1}\;{d\ell}$. Sub $y = \ell^3$ then $ \displaystyle I = \frac{1}{6}\int_{0}^{\infty}\frac{1}{y^2+1}\;{dy} = \frac{\pi}{12}.$

 

[sbhatnagar]

Nice Problems!

2) $ \displaystyle \int_{0}^{\infty} \frac{1}{(1+x^{\varphi})^{\varphi}} \ dx $ where $\varphi$ is the golden ratio

Consider \( I(a)=\displaystyle \int_{0}^{\infty} \frac{1}{(1+x^{a})^{a}} \ dx \)

\( I(a)=\displaystyle \int_{0}^{\infty} \left( \frac{1}{1+x^{a}} \right)^a dx \)

By the substitution \(\displaystyle u=\frac{1}{1+x^a} \) we obtain:

\( \displaystyle I(a)=\frac{1}{a}\int_{0}^{1}u^a \left( \frac{1}{u} \right)^2 \left( \frac{1}{u}-1\right)^{\frac{1}{a}-a}du=\frac{1}{a}\int_{0}^{1}u^{a-\frac{1}{a}-1}(1-u)^{\frac{1}{a}-1} du=\frac{\Gamma{\left( a-\frac{1}{a}\right)}\Gamma{\left( \frac{1}{a}\right)}}{a\Gamma{\left(a \right)}}\)

Therefore \( \displaystyle I(\varphi)= \frac{\Gamma{\left( \varphi-\frac{1}{\varphi}\right)}\Gamma{\left( \frac{1}{\varphi}\right)}}{\varphi\Gamma{\left( \varphi \right)}} \)

It is known that \( \displaystyle \frac{1}{\varphi}=\varphi-1 \).

So finally we get \( \displaystyle I(\varphi)=\frac{\Gamma(1)\Gamma(\varphi-1)}{\varphi \Gamma(\varphi)}= \dfrac{\Gamma(\varphi-1)}{\varphi \Gamma(\varphi)}=\boxed{1}\)

 

[Sherlock1]

So finally we get \( \displaystyle I(\varphi)=\frac{\Gamma(1)\Gamma(\varphi-1)}{\varphi \Gamma(\varphi)}= \boxed{\dfrac{\Gamma(\varphi-1)}{\varphi \Gamma(\varphi)}}\)

Which is 1. (Rofl)

I think there should be a clever way of doing this without employing the gamma function!

 

[polygamma]

Another approach for the first one is to let $\displaystyle x = \frac{1}{u}$.

My approach to the second one was the same as sbhatnagar except I immediately used $ \displaystyle \int_{0}^{\infty} \frac{1}{(1+x^{a})^{b}} \ dx = \frac{1}{a} B \left (\frac{1}{a},b- \frac{1}{a} \right) $

 

[sbhatnagar]

Problem 5

Let \(\displaystyle I= \int_{0}^{2 \pi} \cos (\cos x) \cosh (\sin x) \ dx \)

Note that:

\( \displaystyle \cos (\cos x) \cosh (\sin x) = \left(\frac{e^{i\cos{x}}+e^{-i\cos{x}}}{2}\right) \left( \frac{e^{\sin{x}}+e^{-\sin{x}}}{2}\right) = \cdots=\frac{\cos(e^{ix})+\cos{(e^{-ix}})}{2} \)\( \displaystyle I=\int_{0}^{2\pi} \frac{\cos(e^{ix})+\cos{(e^{-ix}})}{2} dx=\frac{1}{2} \int_{0}^{2\pi}\cos(e^{ix})+\cos{(e^{-ix}}) \ dx \)

Let \( \displaystyle I_1=\int_{0}^{2\pi} \cos(e^{ix})dx \) and \( \displaystyle I_2=\int_{0}^{2\pi} \cos(e^{-ix})dx \)

\( \displaystyle \begin{align*} I_1 &=\int_{0}^{2\pi} \cos(e^{ix})dx \\ &= \int_{0}^{2\pi} \Big( 1-\frac{e^{2ix}}{2!}+\frac{e^{4ix}}{4!}-\cdots \Big) dx \\ &= 2\pi-0 \\ &= 2\pi\end{align*} \)

\( \displaystyle \begin{align*} I_2 &=\int_{0}^{2\pi} \cos(e^{-ix})dx \\ &= \int_{0}^{2\pi} \Big( 1-\frac{e^{-2ix}}{2!}+\frac{e^{-4ix}}{4!}-\cdots \Big) dx \\ &= 2\pi-0 \\ &= 2\pi\end{align*} \)

Therefore \( \displaystyle I=\frac{I_1+I_2}{2}=\frac{4\pi}{2}= \boxed{2\pi} \)

 

[polygamma]

Again my solution is similar to sbhatnagar's.

$\displaystyle \cos (\cos x) \cosh (\sin x) = \text{Re} \ \cos (e^{ix}) $

So $\displaystyle \int_{0}^{2 \pi} \cos (\cos x) \cosh (\sin x) \ dx = \text{Re} \int_{0}^{2 \pi} \cos (e^{ix}) \ dx $

$ \displaystyle \text{Re} \int\limits_{|z|=1} \frac{\cos (z)}{iz} \ dz = \text{Re} \ 2 \pi \ \text{Res} \ \left[\frac{\cos (z)}{z}, 0 \right] = 2 \pi (1) = 2 \pi$

 

[Mathwebster]

Here's mine for the first

Let $\displaystyle I = \int_0^{\infty} \dfrac{x^2}{(1+x^5)(1+x^6)}dx$

Then under a change of variable $x \to 1/x$ gives

$\displaystyle I = \int_0^{\infty} \dfrac{x^7}{(1+x^5)(1+x^6)}dx$

Then adding gives

$\displaystyle 2I = \int_0^{\infty} \dfrac{x^2 +x^7}{(1+x^5)(1+x^6)}dx =\int_0^{\infty} \dfrac{x^2}{1+x^6}dx = \dfrac{\pi}{6} $

Solving for I gives $\pi/12$ and stated earlier.

 

[Mathwebster]

My solution to (3)

Let $\displaystyle I= \int_0^{\infty} \sin \left(x^2+\dfrac{1}{x^2}\right)dx$

Let $x \to \dfrac{1}{x}$ gives

$\displaystyle I= \int_0^{\infty} \dfrac{1}{x^2}\sin \left(x^2+\dfrac{1}{x^2}\right)dx$

Adding gives

$\displaystyle 2 I= \int_0^{\infty}\left( 1 + \dfrac{1}{x^2}\right) \sin \left(x^2+\dfrac{1}{x^2}\right)dx$

Now split the integral up into the following two

$\displaystyle 2 I= \int_0^{1}\left( 1 + \dfrac{1}{x^2}\right) \sin \left(x^2+\dfrac{1}{x^2}\right)dx + \int_1^{\infty}\left( 1 + \dfrac{1}{x^2}\right) \sin \left(x^2+\dfrac{1}{x^2}\right)dx$

Under the change of variable $x \to \dfrac{1}{x}$ the second integral becomes the first so we have

$\displaystyle I= \int_0^{1}\left( 1 + \dfrac{1}{x^2}\right) \sin \left(x^2+\dfrac{1}{x^2}\right)dx $

Now let $u =\dfrac{1}{x} - x$ so

$\displaystyle I = \int_0^{\infty} \sin (u^2 + 2)du = \sin 2 \int_0^{\infty}\cos u^2 \;du + \cos 2 \int_0^{\infty} \sin u^2 =\sqrt{\dfrac{\pi}{2}}\left( \sin 2 + \cos 2\right)$.

 

[Mathwebster]

And (4), if we re-write the integral

$\displaystyle \int_0^{\infty} e^{-x} \text{erf} \sqrt{x} dx = \frac{2}{\sqrt{\pi}}\int_0^{\infty} \int_0^{\sqrt{x}} e^{-x} e^{-y^2} dy dx$

Reversing the order of integration gives

$\displaystyle \frac{2}{\sqrt{\pi}}\int_0^{\infty} \int_{y^2}^{\infty} e^{-x} e^{-y^2} dx dy = \frac{2}{\sqrt{\pi}}\int_0^{\infty} e^{-2y^2} dy= \frac{2}{\sqrt{\pi}} \frac{\sqrt{2\pi}}{4} = \frac{1}{\sqrt{2}}$

 

[polygamma]

Another solution to the third integral.

$ \displaystyle \int_{0}^{\infty} \sin \left( x^{2}+ \frac{1}{x^{2}} \right) \ dx = \int_{0}^{\infty} \sin \Bigg( \Big( x - \frac{1}{x} \Big)^{2} +2 \Bigg) \ dx $

$ \displaystyle = \sin(2) \int_{0}^{\infty} \cos \Bigg( \Big(x-\frac{1}{x} \Big)^{2} \Bigg) \ dx + \cos(2) \int_{0}^{\infty} \sin \Bigg( \Big(x-\frac{1}{x} \Big)^{2} \Bigg) \ dx$$ \displaystyle \int_{0}^{\infty} f \Bigg( \Big(ax - \frac{b}{x} )\Big)^2 \Bigg) \ dx = \frac{1}{b} \int_{0}^{\infty} f (u^{2}) \ du \ \ a,b>0$so $ \displaystyle \int_{0}^{\infty} \sin \left( x^{2}+ \frac{1}{x^{2}} \right) \ dx= \sin(2) \int_{0}^{\infty} \cos (x^{2}) \ dx + \cos(2) \int_{0}^{\infty} \sin (x^{2}) \ dx$

$ = \displaystyle \frac{1}{2} \sqrt{\frac{\pi}{2}} \big( \sin(2)+\cos(2)\big) $

 

[Markov2]

Now split the integral up into the following two

$\displaystyle 2 I= \int_0^{1}\left( 1 + \dfrac{1}{x^2}\right) \sin \left(x^2+\dfrac{1}{x^2}\right)dx + \int_1^{\infty}\left( 1 + \dfrac{1}{x^2}\right) \sin \left(x^2+\dfrac{1}{x^2}\right)dx$

Note that in both cases you could've applied that ${x^2} + \dfrac{1}{{{x^2}}} = {\left( {x - \dfrac{1}{x}} \right)^2} + 2.$ So it's a bit shorter.

Didn't see previous post, but that is.

 

[sbhatnagar]

@Random Variable: These are very entertaining problems. Can I ask you where are these from? :)

 

[polygamma]

The first two and the last one I saw posted on the Art of Problem Solving forums a long time ago. I don't remember where I originally saw the third one. And the fourth one is on most Laplace transform tables. I have a notebook of hundreds.

 

[oasi]

Note that in both cases you could've applied that ${x^2} + \dfrac{1}{{{x^2}}} = {\left( {x - \dfrac{1}{x}} \right)^2} + 2.$ So it's a bit shorter.

Didn't see previous post, but that is.

i can't get it

 

[sbhatnagar]

i can't get it

Why not? Do you know how to multiply?

$$\begin{align*} \left( x-\frac{1}{x}\right)^2+2 &=x^2+\frac{1}{x^2}-2+2 \\ &= x^2+\frac{1}{x^2}\end{align*}$$