An interesting logarithm integral

[alyafey22]

Here is an integral that I seen in another forum

\(\displaystyle \int^1_0 \frac{\log(x)\log(x+1)}{1-x} \approx -0.5 \)

That was a value return by W|A . This thread will be dedicated to try to solve the integral . Any ideas will always be welcomed .

The integral might require some dilogarithm identities . The first step might be integration by parts .

 

[topsquark]

Here is an integral that I seen in another forum

\(\displaystyle \int^1_0 \frac{\log(x)\log(x+1)}{1-x} \approx -0.5 \)

That was a value return by W|A . This thread will be dedicated to try to solve the integral . Any ideas will always be welcomed .

The integral might require some dilogarithm identities . The first step might be integration by parts .

Looking at the dilogarithms I am lead to ask if the "log" operator is base 10 or e?

-Dan

 

[alyafey22]

It is a based e logarithm . In complex analysis we always assume \(\displaystyle \log\) to be the natural logarithm , it looks better than \(\displaystyle \ln \,\, \) (Cool).

 

[alyafey22]

Ok , I tried integration by parts

\(\displaystyle \int^1_0 \frac{\log(x) \log(1-x)}{1+x}\, dx +\int^1_0 \frac{\log(1+x) \log(1-x)}{x}\, dx\)

The first integral

I might start by the following

\(\displaystyle \int^1_0 \frac{x^{a-1}(1-x)^{b-1}}{1+x} = \int^1_0 \left( \sum_{n \geq 0 }(-x)^n \right) x^{a-1}(1-x)^{b-1} \, dx \)\(\displaystyle \sum_{n \geq 0 }(-1)^n \left( \int^1_0 x^{n+a-1}(1-x)^{b-1} \, dx \right) \)\(\displaystyle \int^1_0 \frac{x^{a-1}(1-x)^{b-1}}{1+x} =\Gamma(b) \sum_{n \geq 0 }(-1)^n \frac{\Gamma(n+a)}{\Gamma(n+a+b)} = \phi (a,b)\)

\(\displaystyle \int^1_0 \frac{x^{a-1} (1-x)^{b-1} \log(x) \log(1-x) }{1+x} = \frac{\partial }{\partial a} \left( \frac{\partial}{\partial b} \phi (a,b) \right)\)

\(\displaystyle \int^1_0 \frac{\log(x) \log(1-x) }{1+x} = \frac{\partial \phi (1,1) }{\partial a \, \partial b} \)

Our best bet is to find the sum

\(\displaystyle \sum_{n \geq 0 }(-1)^n \frac{\Gamma(n+a)}{\Gamma(n+a+b)}\)

 

[alyafey22]

\(\displaystyle I= \int^1_0 \frac{\log(x)\log(x+1)}{1-x}\, dx \)

I have got a possibly better idea

let \(\displaystyle x = 1-t \)

\(\displaystyle I=\int^1_0 \frac{\log(1-t)\log(2-t)}{t}\, dt = \int^1_0 \frac{\log(1-t)\left ( \log(1-\frac{t}{2}) + \log(2) \right)}{t}\, dt \)

\(\displaystyle I=\int^1_0 \frac{\log(1-t)\log(1-\frac{t}{2})}{t} \, dt +\log(2)\int^1_0 \frac{\log(1-t)}{t}\, dt \)

\(\displaystyle I=\int^1_0 \frac{\log(1-t)\log(1-\frac{t}{2})}{t} \, dt -\log(2) \text{Li}_2(1) = \int^1_0 \frac{\log(1-t)\log(1-\frac{t}{2})}{t}- \frac{\pi^2}{6} \log(2)\)

The next step might be to evaluate

\(\displaystyle \int^1_0 \frac{\log(1-t)\log(1-\frac{t}{2})}{t} \, dt \)

 

[chisigma]

Here is an integral that I seen in another forum

\(\displaystyle \int^1_0 \frac{\log(x)\log(x+1)}{1-x} \approx -0.5 \)

That was a value return by W|A . This thread will be dedicated to try to solve the integral . Any ideas will always be welcomed .

The integral might require some dilogarithm identities . The first step might be integration by parts .

Integrals of this type are analized in...

http://www.mathhelpboards.com/f49/integrals-natural-logarithm-5286/#post24247

... and in particular for an analytic function...

$$f(x) = \sum_{k=0}^{\infty} a_{k}\ x^{k}\ (1)$$... the following formula holds...

$$\int_{0}^{1} f(x)\ \ln^{n} x\ dx = \sum_{k=0}^{\infty} a_{k}\ \int_{0}^{1} x^{k}\ \ln^{n} x\ dx = (-1)^{n} n!\ \sum_{k=0}^{\infty} \frac{a_{k}}{(k+1)^{n+1}}\ (2) $$

In Your case is $f(x)= \frac{\ln (1+x)}{1-x}$ and is...

$$\frac{\ln (1+x)}{1-x} = x + \frac{x^{2}}{2} + \frac{5\ x^{3}}{6} + \frac{7\ x^{4}}{12} + \frac{47\ x^{5}}{60} + \frac{37\ x^{6}}{60} + \frac{319\ x^{7}}{420} + \frac{533\ x^{8}}{540} + ...\ (3)$$

... so that is...

$$\int_{0}^{1} \frac{\ln (1+x)}{1-x}\ \ln x\ dx = - \frac{1}{4} - \frac{1}{18} - \frac{5}{96} - \frac{7}{300} - \frac{47}{2160} - \frac{37}{2940} - \frac{319}{26680} - \frac{533}{43740} - ...\ (4)$$

Kind regards

$\chi$ $\sigma$

 

[alyafey22]

In the previous post we got the following integral

\(\displaystyle \int^1_0 \frac{\log(1-t)\log(1-\frac{t}{2})}{t} \, dt \)

After integration by parts we get

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

Adding the results together we get \(\displaystyle \int^1_0 \frac{\log(x) \log(1+x)}{1-x}\, dx =\int^1_0 \frac{\text{Li}_2(x)}{x-2}\, dx \)

 

[alyafey22]

\(\displaystyle \int^1_0 \frac{\text{Li}_2(x)}{x-2}\, dx \)

\(\displaystyle \frac{-1}{2} \int^1_0 \frac{\text{Li}_2(x)}{1-\frac{x}{2}}\, dx \)

\(\displaystyle \frac{-1}{2}\sum_{k\geq 0} \int^1_0 x^k \text{Li}_2(x)\, dx \)

\(\displaystyle \frac{-1}{2}\sum_{k\geq 0} \sum_{n\geq 1}\frac{1}{2^k n^2} \int^1_0 x^{k+n} dx \)

\(\displaystyle \frac{-1}{2}\sum_{k\geq 0} \sum_{n\geq 1}\frac{1}{2^k}\frac{1}{n^2(n+k+1)} \)

\(\displaystyle \frac{-1}{2}\sum_{k\geq 0}\frac{1}{2^k} \sum_{n\geq 1}\frac{1}{n^2(k+1)}+\frac{(k+1)}{n(k+1)^2(n+k+1)} \)

\(\displaystyle \frac{-1}{2}\sum_{k\geq 0}\frac{1}{2^k} \sum_{n\geq 1}\frac{1}{n^2(k+1)}\,\,+\,\,\frac{1}{2}\sum_{k\geq 0}\frac{1}{2^k} \sum_{n\geq 1}\frac{(k+1)}{n(k+1)^2(n+k+1)} \)

\(\displaystyle \frac{-\zeta(2)}{2}\sum_{k\geq 0}\frac{1}{2^k(k+1)}\,\, +\,\, \frac{1}{2}\sum_{k\geq 0}\frac{1}{2^k(k+1)^2 } \sum_{n\geq 1}\frac{(k+1)}{n(n+k+1)} \)

\(\displaystyle -\zeta(2) \log(2) +\sum_{k\geq 0}\frac{\psi(k+2) + \gamma}{2^{k+1}(k+1)^2 } \)

\(\displaystyle -\zeta(2) \log(2) +\sum_{k\geq 0}\frac{H_{k+1}}{2^{k+1}(k+1)^2 } \)

\(\displaystyle -\zeta(2) \log(2) +\sum_{k\geq 1}\frac{H_{k}}{2^{k}k^2 } \)

Last step is solve the Euler sum

\(\displaystyle \sum_{k\geq 1}\frac{H_{k}}{2^{k}k^2 } \)

 

[alyafey22]

In the previous post I used the following relation

\(\displaystyle \psi(n) = H_{n-1}-\gamma\)

Proof

By definition we have

\(\displaystyle \psi(n) = -\gamma +\sum_{k\geq 0}\frac{1}{k+1}-\frac{1}{k+n} \)

Since $n$ is an integer we can do the following

\(\displaystyle \sum_{k\geq 0}\frac{1}{k+n} = \sum_{k \geq n}\frac{1}{k} \)

\(\displaystyle \psi(n) = -\gamma +\sum_{k\geq 1}\frac{1}{k}- \sum_{k \geq n}\frac{1}{k} = -\gamma + \sum_{k=1}^{n-1}\frac{1}{k} \)

Now since harmonic numbers are defined as follows

\(\displaystyle \sum_{k=1}^{n}\frac{1}{k} = H_n\)

The proof is complete \(\displaystyle \square \).

 

[alyafey22]

An attempt to solve the Euler sum

\(\displaystyle \sum_{k\geq 1} \frac{H_k}{2^k k^2}\)

Rather we will try to generalize a little bit

\(\displaystyle \sum_{k\geq 1} \frac{H_k}{ k^2} x^k \)

Start by the following

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

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

Integrating both sides we have

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

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

Integrating again we obtain

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

A further attempt will be in the next post .

 

[alyafey22]

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

Integrating by parts we get the following

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

Now we are left with the following integral

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

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

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

The first integral

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

The second integral by parts we obtain

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

Collecting the results together we obtain \(\displaystyle \int^x_0\frac{\text{Li}_2 (t)}{1-t} \, dt =-\frac{\pi^2}{6}\log(1-x)-\text{Li}_2(1-x) \log(1-x)+2\, \text{Li}_3(1-x)- 2 \zeta(3) \)Hence we solved the integral \(\displaystyle \int^x_0 \frac{\log^2(1-t)}{t}\, dt = - \log(1-x) \text{Li}_2(x) +\frac{\pi^2}{6}\log(1-x)+\text{Li}_2(1-x) \log(1-x)-2\, \text{Li}_3(1-x)+ 2 \zeta(3) \)

So we have got our Harmonic sum \(\displaystyle \sum_{k\geq 1} \frac{H_k}{k^2} \, x^{k} = \text{Li}_3(x)+\frac{1}{2} \left( - \log(1-x) \text{Li}_2(x) +\frac{\pi^2}{6}\log(1-x)+\text{Li}_2(1-x) \log(1-x)-2\, \text{Li}_3(1-x)+ 2 \zeta(3) \right) \)\(\displaystyle \sum_{k\geq 1} \frac{H_k}{k^2} \, x^{k} = \text{Li}_3(x)-\, \text{Li}_3(1-x)+\, \log(1-x) \text{Li}_2(1-x) +\frac{1}{2}\log(x) \log^2(1-x)+\zeta(3) \)The expression can be further simplified but I will leave it like this
So we finally we got our solution (Whew)

 

[Prove It]

It is a based e logarithm . In complex analysis we always assume \(\displaystyle \log\) to be the natural logarithm , it looks better than \(\displaystyle \ln \,\, \) (Cool).

That's debatable. I think it looks better as ln, and it is clearly less time consuming to write ln :P

 

[alyafey22]

We already proved that

\(\displaystyle \int^1_0 \frac{\log(x) \log(1+x)}{1-x} \, dx = -\zeta(2) \log(2) +\sum_{k\geq 1}\frac{H_{k}}{2^{k}k^2 }\)

We found that

\(\displaystyle \sum_{k\geq 1} \frac{H_k}{k^2} \, x^{k} = \text{Li}_3(x)-\, \text{Li}_3(1-x)+\, \log(1-x) \text{Li}_2(1-x) +\frac{1}{2}\log(x) \log^2(1-x)+\zeta(3)\)

putting \(\displaystyle x =1/2 \) we get

\(\displaystyle \sum_{k\geq 1} \frac{H_k}{2^k k^2} = -\log(2) \text{Li}_2\left( \frac{1}{2}\right) -\frac{1}{2}\log^3(2)+\zeta(3)\)

\(\displaystyle \sum_{k\geq 1} \frac{H_k}{2^k k^2} = -\log(2) \left( \frac{\pi^2}{12}- \frac{1}{2}\log^2(2) \right) -\frac{1}{2}\log^3(2)+\zeta(3)\)

\(\displaystyle \sum_{k\geq 1} \frac{H_k}{2^k k^2} = -\frac{\pi^2}{12} \log(2)+\zeta(3)\)

\(\displaystyle \int^1_0 \frac{\log(x) \log(1+x)}{1-x} \, dx = -\frac{\pi^2}{6}\log(2) -\frac{\pi^2}{12} \log(2)+\zeta(3)\)

Hence we found the result was seeking for

\(\displaystyle \int^1_0 \frac{\log(x) \log(1+x)}{1-x} \, dx = -\frac{\pi^2}{4}\log(2)+\zeta(3) \)​