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While doing a little bit research on the "alternating Basel Problem" I have come across this related post which states the equality

$$\int_0^1 \frac{\ln(1+x)}x\mathrm dx=-\frac12\int_0^1 \frac{\ln x}{1-x}\mathrm dx\tag1$$

Using the Dilogarithm one can show that "alternating Basler Problem" is a direct consequence of this equation and yields to

$$\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^2}=\frac{\pi^2}{12}$$

Therefore I have no doubts to trust the author of the the cited post. However, I tried to verify the equality by myself and failed. For this purpose I enforced the subsitution $x\mapsto1+x$ within the integral on the right

$$\begin{align} -\frac12\int_0^1 \frac{\ln x}{1-x}\mathrm dx=-\frac12\int_{(0-1)}^{(1-1)} \frac{\ln(1+x)}{1-(1+x)}\mathrm dx=-\frac12\int_{-1}^{0} \frac{\ln(1+x)}x\mathrm dx \end{align}$$

But from hereon I am not sure how to proceed. Clearly now I have to show that

$$\begin{align} -\frac12\int_{-1}^0\frac{\ln(1+x)}x\mathrm dx&=\int_0^1 \frac{\ln(1+x)}x\mathrm dx\\ \frac12\int_0^1\frac{\ln(1-x)}x\mathrm dx&=\int_0^1 \frac{\ln(1+x)}x\mathrm dx\\ 0&=\int_0^1 \frac1x\left(\ln(1+x)-\frac12\ln(1-x)\right)\mathrm dx \end{align}$$

It seems like I have made a mistake somewhere inbetween since WolframAlpha does not agree with my reasoning. Additionally I have no idea how to proceed. To be honest I am quite confused right now.

First of all where exactly did I went wrong? Furthermore could someone provide a complete proof for the given equality? Please tell me when this question has been asked before.

Thanks in advance!

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  • $\begingroup$ As for where you messed up, it was when you went from $$-\frac{1}{2}\int_{-1}^0 \frac{\ln(1+x)}{x}dx$$ to $$\frac{1}{2}\int_0^1 \frac{\ln(1-x)}{x}dx$$ This is because you did a substitution, but you ALSO flipped the bounds of integration, so there should still be a negative sign. $\endgroup$ – Isaac Browne Oct 4 '18 at 20:49
  • $\begingroup$ @IsaacBrowne I guessed so but should it not be $x=-u\Rightarrow dx=-du$ and furthermore $$-\frac12\int_{-(-1)}^{-(0)}\frac{\ln(1-u)}{-u}(-du)=\frac12\int_{1}^{0}\frac{\ln(1-u)}{u}(du)$$ and therefore I can change the bounds of integrations nevertheless? $\endgroup$ – mrtaurho Oct 4 '18 at 20:57
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    $\begingroup$ Whoops, I guess you're not mistaken. But now I think I actually found the negative error in your initial substitution, since the denominator goes from $1-(1+y)$ to $y$ as it is now. $\endgroup$ – Isaac Browne Oct 4 '18 at 21:15
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    $\begingroup$ Your substitution was fine, but watch the denominator, it goes from $1-(1+y)=-y$ to $y\neq -y$ without any multiple of $-1$ being added. $\endgroup$ – Isaac Browne Oct 4 '18 at 21:22
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    $\begingroup$ Oh god dammit. I finally got what you have to tried to explain me for about an hour... I am so sorry for missing your point in such a stupid way. However thank you on the one hand for poiting it out and on the other hand for your notable patience! $\endgroup$ – mrtaurho Oct 4 '18 at 23:21
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HINT:

Note that we have

$$\begin{align} \frac12\int_0^1 \frac{\log(x)}{1-x}\,dx&\overbrace{=}^{x\mapsto x^2}\int_0^1 \frac{x\log(x^2)}{1-x^2}\,dx\\\\ &=\int_0^1 \log(x)\left(\frac{1}{1-x}-\frac{1}{1+x}\right)\,dx \end{align}$$

Can you finish now?

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  • $\begingroup$ Oh, that's nice! $\endgroup$ – Migos Oct 4 '18 at 20:49
  • $\begingroup$ @Aubyn Thank you! I hope it was useful too. $\endgroup$ – Mark Viola Oct 4 '18 at 20:50
  • $\begingroup$ @MarkViola To be honest I am still totally confused. Could you maybe provide a complete evaluation? $\endgroup$ – mrtaurho Oct 4 '18 at 21:06
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    $\begingroup$ @mrtaurho Integrate by parts the integral $\int_0^1 \frac{\log(x)}{x+1}\,dx$ with $u=\log(x)$ and $v=\log(x+1)$. What is the result? Now use this along with the HINT provided herein. Do you have it now? $\endgroup$ – Mark Viola Oct 5 '18 at 2:38
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    $\begingroup$ You're welcome. My pleasure. $\endgroup$ – Mark Viola Oct 5 '18 at 12:41

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