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I've read somewhere on this site that if you consider:

$$\int_0^1 \int_0^1 \frac{1}{1-xy} \,dy\,dx$$

Then using the power series, we have this is equal to $\sum_{n=1}^{\infty} \frac{1}{n^2}$ which I decided to try and was able to show.

Apparently we can show this is equal to $\frac{\pi^2}{6}$, and using what little I know about double integrals from a few khan academy videos (I haven't taken multivariable calculus yet), I tried to evaluate this double integral by techniques of single variable calculus $u=xy$..and I got this:

$$-\int_{0}^{1} \frac{\ln (1-x)}{x}dx$$

The usual way I would evaluate this is with a Taylor series, but that just that just leads us in circles.

So I want to know how can I evaluate this, so we can prove $\sum_{n=1}^{\infty} \frac{1}{n^2}=\frac{\pi^2}{6}$.

Ahmed S. Attaalla
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1 Answers1

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Here, we present an approach the relies on integration methodologies only, including transformation of coordinates. To that end we proceed.

We can transform coordinates by setting $x=s+t$ and $y=s-t$. Then, $dx\,dy\to 2\,ds\,dt$ and the transformed integral domain is the square-shaped region with vertices in the $s-t$ plane at $(0,0)$, $(1/2,1/2)$, $(1/2,-1/2)$, and $(1,0)$. Then, we can write

$$\begin{align} \int_0^1\int_0^1 \frac{1}{1-xy}\,dx\,dy&=\int_0^{1/2}\int_{-s}^{s}\frac{2}{(1-s^2)+t^2}\,dt\,ds+\int_{1/2}^{1}\int_{s-1}^{1-s}\frac{2}{(1-s^2)+t^2}\,dt\,ds\\\\ &=\int_0^{1/2}\frac{4}{\sqrt{1-s^2}}\arctan\left(\frac{s}{\sqrt{1-s^2}}\right)\,ds\\\\&+\int_{1/2}^{1}\frac{4}{\sqrt{1-s^2}}\arctan\left(\sqrt{\frac{1-s}{1+s}}\right)\,ds\\\\ &=4\int_0^{1/2}\frac{\arcsin(s)}{\sqrt{1-s^2}}\,ds+4\int_{1/2}^1\frac{\arccos(s)}{2\sqrt{1-s^2}}\,ds\\\\ &=2\arcsin^2(1/2)+\arccos^2(1/2)\\\\ &=2\left(\frac{\pi}{6}\right)^2+\left(\frac{\pi}{3}\right)^2\\\\ &=\frac{\pi^2}{6} \end{align}$$

as expected!

Mark Viola
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  • Substitution is kind of weird in two variables. I guess I'll have to dig deeper in it. And $\arctan(\frac{s}{\sqrt{1-s^2}})$ can be simplified by setting it equal to $\theta$, taking the tangent of both sides, and considering the right triangle with opposite $s$ and adjacent $\sqrt{1-s^2}$ if I remember right, correct? And thanks for your answer which is a little bit different than Christian Blatters comment. – Ahmed S. Attaalla Aug 07 '16 at 15:33
  • Integration under coordinate transformation requires calculation of the Jacobian. And yes, your observation is consistent with $\arctan(s/\sqrt{1-s^2}=\arcsin(s)$ and $\arctan(\sqrt{(1-s)/(1+s)}=\frac12 \arccos(s)$. I had not read the comment from Christian Blatters before posting this. But I had a brief look and think it is a really solid reference! -Mark – Mark Viola Aug 07 '16 at 15:42
  • You're welcome. My pleasure. -Mark – Mark Viola Aug 07 '16 at 15:42
  • I believe you have a mistype in line where you you first go from double integrals to single @Dr.MV . – Ahmed S. Attaalla Aug 07 '16 at 15:45
  • I see the same integral twice. @Dr.MV – Ahmed S. Attaalla Aug 07 '16 at 16:01
  • But what about $\arctan (\sqrt{\frac{1-s}{1+s}})$ you never right than anywhere in your answer. @Dr.MV – Ahmed S. Attaalla Aug 07 '16 at 16:44
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    Ahmed, good catch on the typo. Thank you. I've edited. I guess this highlights the danger of cutting and pasting. ;-)) – Mark Viola Aug 07 '16 at 17:00
  • In the upper right integral the upper $s$ limit ($2$) and the numerator of the integrand ($1$) may have been switched. – robjohn Aug 07 '16 at 18:39
  • @robjohn Yes, indeed. I've edited. Thank you for the catch. -Mark – Mark Viola Aug 07 '16 at 21:00
  • Would the down voter care to comment? – Mark Viola Aug 07 '16 at 22:33
  • (+1) - but I think you have mini typos in the third double integral of the first line... The $2$ in the upper $s$ limit should be $1$? And you're missing the $2$ of the Jacobian transformation (probably it migrated to the limit)? – peter a g Nov 01 '16 at 12:48
  • @peterag Great catch! Much appreciated. I've edited accordingly. -Mark – Mark Viola Nov 01 '16 at 15:42