While doing some 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


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 substitution $x\mapsto1+x$ 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!

  • $\begingroup$ 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


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?

  • $\begingroup$ 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

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

Playing around with series expansions.

$\begin{array}\\ I_1 &=\int_0^1 \frac{\ln(1+x)}x dx\\ &=\int_0^1 \sum_{n=0}^{\infty} \dfrac{(-1)^{n}x^n}{n+1}dx\\ &=\sum_{n=0}^{\infty} \dfrac{(-1)^{n}x^{n+1}}{(n+1)^2}|_0^1\\ &=\sum_{n=0}^{\infty} \dfrac{(-1)^{n}}{(n+1)^2}\\ I_2 &=\int_0^1 \frac{\ln x}{1-x}dx\\ &=\int_0^1 \frac{\ln (1-x)}{1-(1-x)}dx\\ &=\int_0^1 \frac{\ln (1-x)}{x}dx\\ &=-\int_0^1 \sum_{n=0}^{\infty}\dfrac{x^n}{n+1}dx\\ &=-\sum_{n=0}^{\infty}\int_0^1 \dfrac{x^n}{n+1}dx\\ &=-\sum_{n=0}^{\infty}\dfrac1{(n+1)^2}\\ 2I_1+I_2 &=\sum_{n=0}^{\infty} \dfrac{2(-1)^{n}-1}{(n+1)^2}\\ &=\sum_{n=0}^{\infty} \dfrac{2(-1)^{2n}-1}{(2n+1)^2} +\sum_{n=0}^{\infty} \dfrac{2(-1)^{2n+1}-1}{(2n+2)^2}\\ &=\sum_{n=0}^{\infty} \dfrac{1}{(2n+1)^2} +\sum_{n=0}^{\infty} \dfrac{-3}{(2n+2)^2}\\ &=\sum_{n=0}^{\infty} \dfrac{1}{(2n+1)^2} +\sum_{n=0}^{\infty} \dfrac{1}{(2n+2)^2} +\sum_{n=0}^{\infty} \dfrac{-4}{(2n+2)^2}\\ &=\sum_{n=0}^{\infty} \dfrac{1}{(n+1)^2} -\sum_{n=0}^{\infty} \dfrac{4}{4(n+1)^2}\\ &=\sum_{n=0}^{\infty} \dfrac{1}{(n+1)^2} -\sum_{n=0}^{\infty} \dfrac{1}{(n+1)^2}\\ &=0\\ \end{array} $

And it works!


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