Where is my mistake $\int_{0}^{1}\frac{\log(1-x)\log(1-x^2)}{x}dx$

Question

Edit

As commented bellow by @Donald Splutterwit and @ Elliot Yu, it seems that my computation is numerically correct and the post is wrong! I also added a corollary from this computation.

I saw the following statement here

$$\boxed{\int_{0}^{1}\frac{\log(1-x)\log(1-x^2)}{x}dx=\frac{7}{4}\zeta(3)}$$

And I wanted to proof it. My approach was the following

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

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

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

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

let $1-x=t$

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

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

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

Now use the fact that

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

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

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

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

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The second integral is a little trickier

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

Observe the following

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

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

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

dividing both sides by $x$ and integrating from $0$ to $1$

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

Now we have to evaluate the three integrals on the RHS and we are done. I´ll state only the value of each integral since their proof is easy to find on this forum.

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

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

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

Putting all together we get

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

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Now summing the results of $I_{1}$ and $I_{2}$ we get the final result which differs form the above statement!

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

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

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

Answer 1

With $I_1=\int_0^1\frac{\ln^2(1-x)}x dx = 2\zeta(3)$, the second integral may be evaluated, perhaps more simply, as follows

\begin{align} I_{2}= & \int_{0}^{1}\frac{\ln(1-x)\ln(1+x)}{x}dx\\ =& \ \frac14 \int_{0}^{1}\underset{t=x^2}{ \frac{\ln^2(1-x^2)}{x} }dx\ - \frac14 \int_{0}^{1} \underset{t= \frac{1-x}{1+x}}{ \frac{\ln^2\frac{1-x}{1+x}} {x}}dx\\ =& \ \frac18 \int_0^1\frac{\ln^2(1-t)}t dt - \frac12 \int_{0}^{1}\underset{1-t\to t}{ \frac{\ln^2t}{1-t}} dt +\frac12 \int_{0}^{1} \underset{1-t^2\to t}{\frac{t\ln^2t}{1-t^2} }dt\\ = & \ \left( \frac18 - \frac12 +\frac1{16}\right) \int_0^1\frac{\ln^2(1-t)}t dt =-\frac5{16} \cdot 2\zeta(3) = -\frac58 \zeta(3) \end{align} As a result $${\int_{0}^{1}\frac{\log(1-x)\log(1-x^2)}{x}dx =I_1+I_2=\frac{11}{8}\zeta(3)}$$