An easy looking integral, not so easy...

Easy looking integral, not so easy…

I saw this post from @infseriesbot and wanted to proof at least the first one. After a hardwork to compute this easy looking integral, I found out that the result stated by @infseriesbot is incorrect! Exctly, this time @infseriesbot is wrong. Checking numerically confirms the computation here.

$\boxed{\int_{0}^{1}\frac{\log(1-x)\log(1-x^2)}{x}dx=\frac{11}{8}\zeta(3)}$

Proof

$I=\int_{0}^{1}\frac{\log(1-x)\log(1-x^2)}{x}dx=\int_{0}^{1}\frac{\log(1-x)\log[(1-x)(1+x)]}{x}dx$

$=\int_{0}^{1}\frac{\log(1-x)\log(1-x)}{x}dx+\int_{0}^{1}\frac{\log(1-x)\log(1+x)}{x}dx$

$=\underbrace{\int_{0}^{1}\frac{\log^2(1-x)}{x}dx}_{I_{1}}+\underbrace{\int_{0}^{1}\frac{\log(1-x)\log(1+x)}{x}dx}_{I_{2}}$

$I_{1}=\int_{0}^{1}\frac{\log^2(1-x)}{x}dx$

let 1-x=t

$I_{1}=\int_{0}^{1}\frac{\ln^2(t)}{1-t} dt$

$=\int_{0}^{1}\ln^2(t)\sum_{k=0}^{\infty}t^kdt$

$=\sum_{k=0}^{\infty}\int_{0}^{1}t^k\ln^2(t)dt$

Now use the fact that

$\boxed{\int_{0}^{1}x^m ln^n(x)dx=\frac{(-1)^{n}n! }{(m+1)^{n+1}}}$

for $n=2 \,\,\text{and} \,\, m=k$

$I_{1}=\sum_{k=0}^{\infty}\frac{(-1)^{2}2! }{(k+1)^{3}}$

$\boxed{\int_{0}^{1}\frac{\log^2(1-x)}{x}dx=2\zeta(3)}$

The second integral is a little trickier

$I_{2}=\int_{0}^{1}\frac{\log(1-x)\log(1+x)}{x}dx$

Observe the following

$\Left(\log(1-x)+\log(1+x) \Right)^2=\log^2(1-x)+2\log(1-x)\log(1+x)+\log^2(1+x)$

$\log(1-x)\log(1+x)=\frac{\Left(\log[(1-x)(1+x)] \Right)^2-\log^2(1-x)-\log^2(1+x)}{2}$

$\log(1-x)\log(1+x)=\frac{\log^2(1-x^2)-\log^2(1-x)-\log^2(1+x)}{2}$

dividing both sides by and integrating from to

$\int_{0}^{1}\frac{\log(1-x)\log(1+x)}{x}dx=\frac{1}{2}\int_{0}^{1}\frac{\log^2(1-x^2)}{x}dx-\frac{1}{2}\int_{0}^{1}\frac{\log^2(1-x)}{x}dx-\frac{1}{2}\int_{0}^{1}\frac{\log^2(1+x)}{x}dx$

Now we have to evaluate the three integrals on the RHS and we are done. I´ll state the value of each of the integrals and show the proof in the end.

$\int_{0}^{1}\frac{\log^2(1-x^2)}{x}dx=\zeta(3)$

The next integral is the same as $I_{1}$ so no need to proof it again.

$\int_{0}^{1}\frac{\log^2(1-x)}{x}dx=2\zeta(3)$

$\int_{0}^{1}\frac{\log^2(1+x)}{x}dx=\frac{1}{4}\zeta(3)$

Putting all together we get

$I_{2}=\int_{0}^{1}\frac{\log(1-x)\log(1+x)}{x}dx=\frac{1}{2}\Left\{\zeta(3)-2\zeta(3)-\frac{1}{4}\zeta(3) \Right\}$

$\boxed{\int_{0}^{1}\frac{\log(1-x)\log(1+x)}{x}dx=-\frac{5}{8}\zeta(3)}$

Now summing the results of $I_{1}$ and $I_{2}$ we get the final result!

$\int_{0}^{1}\frac{\log(1-x)\log(1-x^2)}{x}dx=2\zeta(3)-\frac{5}{8}\zeta(3)=\frac{11}{8}\zeta(3)$

Checking numerically we have $\zeta(3)=1.2020569 \cdots$ and $\frac{11}{8} \cdot 1.2020569\approx 1.6528282417625$ , which agrees with WolframAlpha!

A Corollary

We just computed the integral

$\int_{0}^{1}\frac{\log(1-x)\log(1+x)}{x}dx=-\frac{5}{8}\zeta(3)$

On the other hand we can show that this integral equals

$\sum_{n=1}^{\infty}\frac{(-1)^n H_{n}}{n^2}$ and conclude that

$\sum_{n=1}^{\infty}\frac{(-1)^{n-1} H_{n}}{n^2}=\frac{5}{8}\zeta(3)$

Proof:

Expanding $\log(1+x)$ in Taylor series we get

$I=\int_{0}^{1}\frac{\log(1-x)\log(1+x)}{x}dx=\sum_{n=1}^{\infty}\frac{(-1)^{n-1} }{n}\int_{0}^{1}x^{n-1}\log(1-x)dx$

Integrating by parts we get:

$I=\sum_{n=1}^{\infty}\frac{(-1)^{n-1} }{n}\bigg\{\frac{\log(1-x)(x^n-1)}{n}\Big|_{0}^{1}+\frac{1}{n}\int_{0}^{1}\frac{x^n-1}{1-x}dx \bigg\}$

$I=\sum_{n=1}^{\infty}\frac{(-1)^{n-1} }{n}\bigg\{-\frac{1}{n}\int_{0}^{1}\frac{1-x^n}{1-x}dx \bigg\}$

$I=\sum_{n=1}^{\infty}\frac{(-1)^{n-1} }{n}\bigg\{-\frac{1}{n}H_{n} \bigg\}$

$\boxed{\int_{0}^{1}\frac{\log(1-x)\log(1+x)}{x}dx=-\sum_{n=1}^{\infty}\frac{(-1)^{n-1}H_{n} }{n^2}}$

and therefore

$\boxed{\sum_{n=1}^{\infty}\frac{(-1)^{n-1} H_{n}}{n^2}=\frac{5}{8}\zeta(3)}$

Proof that $\int_{0}^{1}\frac{\log^2(1-x^2)}{x}dx=\zeta(3)$

Let $1-x^2=t \Rightarrow \,\, dx=-\frac{dt}{2x}$

$\int_{0}^{1}\frac{\log^2(1-x^2)}{x}dx=\frac{1}{2}\int_{1}^{0}\frac{\log^2(t)}{x}\frac{(-dt)}{x}$

$=\frac{1}{2}\int_{0}^{1}\frac{\log^2(t)}{x^2}dt=\frac{1}{2}\int_{0}^{1}\frac{\log^2(t)}{1-t}dt$

The last integral we have already computed, it´s $I_{1}$ ! and it´s equal to $2\zeta(3)$ , therefore

$\int_{0}^{1}\frac{\log^2(1-x^2)}{x}dx=\frac{1}{2}2\zeta(3)$

and

$\boxed{\int_{0}^{1}\frac{\log^2(1-x^2)}{x}dx=\zeta(3)}$

Now let´s proof the harder one!

$I=\int_{0}^{1}\frac{\log^2(1+x)}{x}dx$

Lets start with the substitution $1+x=\frac{1}{y} \, \Rightarrow \, dx=-\frac{dy}{y^2}$

$I=\int_{1/2}^{1}\frac{\log^2\Big( \frac{1}{y}\Big)}{\frac{1}{y}-1}\frac{dy}{y^2}=\int_{1/2}^{1}\frac{\log^2(y)}{y(1-y)}dy$

$=\int_{1/2}^{1}\frac{\log^2(y)}{y}dy+\int_{1/2}^{1}\frac{\log^2(y)}{(1-y)}dy$

$=\frac{1}{3}\log^3(y)\Big|_{1/2}^{1}+\int_{1/2}^{1}\frac{\log^2(y)}{(1-y)}dy$

$=\frac{\log^3(2)}{3}+\int_{1/2}^{1}\frac{\log^2(y)}{(1-y)}dy$

$=\frac{\log^3(2)}{3}+\int_{0}^{1}\frac{\log^2(y)}{(1-y)}dy-\int_{0}^{1/2}\frac{\log^2(y)}{(1-y)}dy$

$=\frac{\log^3(2)}{3}+2\zeta(3)-\int_{0}^{1/2}\frac{\log^2(y)}{(1-y)}dy$

Let $t=\frac{y}{1-y}$ in the last integral. $y=\frac{t}{1+t} \,\, \Rightarrow \,\, dy=\frac{dt}{(1+t)^2}$

$I=\frac{\log^3(2)}{3}+2\zeta(3)-\int_{0}^{1}\frac{\log^2\Big(\frac{t}{1+t} \Big)}{1-\frac{t}{1+t}}\frac{dt}{(1+t)^2}$

$=\frac{\log^3(2)}{3}+2\zeta(3)-\int_{0}^{1}\frac{\log^2\Big(\frac{t}{1+t} \Big)}{1+t}dt$

$=\frac{\log^3(2)}{3}+2\zeta(3)-\int_{0}^{1}\frac{\Big(\log(t)-\log(1+t) \Big)^2}{1+t}dt$

$I=\frac{\log^3(2)}{3}+2\zeta(3)-\underbrace{\int_{0}^{1}\frac{\log^2(t)}{1+t}dt}_{J_{1}}-\underbrace{\int_{0}^{1}\frac{\log^2(1+t)}{1+t}dt}_{J_{2}}+2\underbrace{\int_{0}^{1}\frac{\log(t)\log(1+t)}{1+t}dt}_{J_{3}}$ (1)

We now evaluate each of these three integrals separately

$J_{1}=\int_{0}^{1}\frac{\log^2(t)}{1+t}dt$

$=\int_{0}^{1}\log^2(t)\sum_{k=0}^{\infty}(-1)^kt^kdt$

$=\sum_{k=0}^{\infty}(-1)^k\int_{0}^{1}\log^2(t)t^kdt$

$\sum_{k=0}^{\infty}(-1)^k\frac{2}{(k+1)^3}=2\eta(3)$

using the formula that $\eta(s)=(1-2^{1-s})\zeta(s)$

We get

$\boxed{J_{1}=\int_{0}^{1}\frac{\log^2(t)}{1+t}dt=\frac{3}{2}\zeta(3)}$

$J_{2}=\int_{0}^{1}\frac{\log^2(1+t)}{1+t}dt$

To compute $J_{2}$ I will first evaluate the indefinite version and then plug the limits.

$\int\frac{\log^2(1+t)}{1+t}dt \qquad \text{let 1+t=u}$

$\int\frac{\log^2(u)}{u}du$

Now integrate by parts with $u=\log^2(u) \, \Rightarrow du=2\log(u)\frac{du}{u}$ and $v=\log(u)$

$\int\frac{\log^2(u)}{u}du=\log^3(u)-2\int\frac{\log^2(u)}{u}du$

$\int\frac{\log^2(u)}{u}du=\frac{\log^3(u)}{3}$

$\int\frac{\log^2(1+t)}{1+t}dt=\frac{\log^3(1+t)}{3}$

$\boxed{J_{2}=\int_{0}^{1}\frac{\log^2(1+t)}{1+t}dt=\frac{\log^3(2)}{3}}$

$J_{3}=\int_{0}^{1}\frac{\log(t)\log(1+t)}{1+t}dt$

To evaluate $J_{3}$ we again consider Integration by parts.

Let $u=log(t) \, \Rightarrow du=\frac{dt}{t}$

And $dv=\int\frac{\log(1+t)}{1+t}dt \, \Rightarrow v=\frac{\log^2(1+t)}{2}$ . To find we used the same process used to calculate $J_{2}$

$\int_{0}^{1}\frac{\log(t)\log(1+t)}{1+t}dt=\underbrace{\frac{\log^2(1+t)}{2}\log(t)}_{=0} \Big|_{0}^{1}-\frac{1}{2}\int_{0}^{1}\frac{\log^2(1+t)}{t}dt$