11

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

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

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 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!

mrtaurho
  • 16,103
  • 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. – Isaac Browne Oct 04 '18 at 20:49
  • @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? – mrtaurho Oct 04 '18 at 20:57
  • The equality you just wrote is fine, but in order to change the bounds from the RHS to from $0$ to $1$, you need to multiply by $-1$ again. – Isaac Browne Oct 04 '18 at 21:05
  • I missed the crucial part by a typo... The minus signs of the $du$ and the $u$ in the denominator cancel each other out. But there still remains one minus sign infront, the one with $\frac12$, of the integral which allows me to swap the bounds of integration, or am I mistaken? – mrtaurho Oct 04 '18 at 21:08
  • 1
    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. – Isaac Browne Oct 04 '18 at 21:15
  • To put it in a nutshell: Did I made an error while substitute $x=1+y$ or is the mistake somewhere else ? – mrtaurho Oct 04 '18 at 21:20
  • 1
    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. – Isaac Browne Oct 04 '18 at 21:22
  • Okay. But then something deeply went wrong. Does not my argumentation within the comment section about the vanishing of the minus sign still holds? Hence it cancels out with the one provided by the differential. So or so I come to the conclusion that there will be an extra negative sign in the end... Sorry if I am missing your point right now. I am still quite confused by the whole equality. – mrtaurho Oct 04 '18 at 21:26
  • In your last line the integral would have a + inside instead of a -. And that results in a true equality – Isaac Browne Oct 04 '18 at 21:32
  • 1
    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! – mrtaurho Oct 04 '18 at 23:21

3 Answers3

14

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?

Mark Viola
  • 179,405
2

\begin{align} \int_0^1\frac{\ln(1+x)}{x}dx&=\int_0^1\frac{\ln\left(\frac{1-x^2}{1-x}\right)}{x}dx\\ &=\underbrace{\int_0^1\frac{\ln(1-x^2)}{x}dx}_{1-x^2-\to x}-\underbrace{\int_0^1\frac{\ln(1-x)}{x}dx}_{1-x\to x}\\ &=\frac12\int_0^1\frac{\ln(x)}{1-x}dx-\int_0^1\frac{\ln(x)}{1-x}dx\\ &=-\frac12\int_0^1\frac{\ln x}{1-x}dx. \end{align}

Ali Shadhar
  • 25,498
1

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!

marty cohen
  • 107,799