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$$ \int_{[0,1] \times [0,1]} \frac{dx \, dy}{1-xy}=\frac{\pi^2}{6}, $$ there is a hint for substitution given but I can't seem to get anywhere with it. $$ \begin{cases} x = \frac{u-v}{\sqrt{2}} \\ y= \frac{u+v}{\sqrt{2}} \end{cases} $$ and evaluate by letting $u = \sqrt{2}\sin t$. There should also encounter the term $(1-\sin t)/\cos t$ (Which I couldn't obtain)

I have managed to compute the jacobian of transformation, which is exactly 1, and did the substitution for u and v, obtaining: $$ \int_{[0,\sqrt{2}]\times [-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}]} \frac{2}{2-u^2 +v^2} du\, dv $$ doing the subtitution for $u$, we get: $$ \int \frac{2\sqrt{2}\cos t}{2\cos^2 t+v^2}dt\, dv $$ It is then not obvious how to integrate with respect to $t$ now, so change order of integration and integrate with respect to v.

which will be equal to $\int 4\arctan \frac{1}{2\cos t} \, dt$

Now this is a rather complicated integral, how to proceed from here? or is there an alternative method which I overlooked in the previous steps?

Could I have some insight on this problem please??

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Welsome to MSE. For formatting math, see here for a quick tutorial. – Daryl Nov 5 '12 at 12:03
Welsome to MSE. For formatting math, see here for a quick tutorial. – Daryl Nov 5 '12 at 12:03
A direct method is to expand 1/(1-xy) into the series of general term x^ny^n and, since everything is positive, to integrate this series termwise. – Did Nov 5 '12 at 12:06

Instead of doing the $u$ substitution before evaluating the integral in $v$, it's easier to do it after computing that integral. Here is a detailed computation.

(I've assumed that you want to evaluate this integral and not simply to show that it is equal to $\zeta(2)$. This exercise appears in several references: Tom Apostol's A Proof that Euler Missed: Evaluating $\zeta(2)$ the Easy Way, Martin Aigner and Günter Ziegler's Proofs from The BOOK and as an exercise in a number theory text by LeVeque .)

By the substitution $x=\frac{\sqrt{2}}{2}\left( u-v\right) ,y=\frac{\sqrt{2}}{2}\left( u+v\right) $, whose Jacobian $J=\frac{\partial (x,y)}{\partial (u,v)}=1$, the region of integration becomes the blue square in the $u,v$-plane with vertices

$$(0,0),\left(\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2}\right),(\sqrt{2},0),\left(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right), $$ as shown in the following figure.

enter image description here

Observing that

$$ \frac{1}{1-xy}=\frac{2}{2-u^{2}+v^{2}} $$

is symmetric in $v$, we get

$$\begin{eqnarray*} I &=&\int_{0}^{1}\int_{0}^{1}\frac{1}{1-xy}\,dx\,dy & x=\frac{\sqrt{2}}{2 }\left( u-v\right) ,\quad y=\frac{\sqrt{2}}{2}\left( u+v\right) \\ &=&2\int_{u=0}^{\sqrt{2}/2}\int_{v=0}^{u}\frac{2}{2-u^{2}+v^{2}}\times 1\,du\,dv \\ &&+2\int_{u=\sqrt{2}/2}^{\sqrt{2}}\int_{v=0}^{\sqrt{2}-u}\frac{2}{ 2-u^{2}+v^{2}}\times 1\,du\,dv \\ &=&4\int_{0}^{\sqrt{2}/2}\left( \int_{0}^{u}\frac{dv}{2-u^{2}+v^{2}}\right) \,du\, \\ &&+4\int_{\sqrt{2}/2}^{\sqrt{2}}\left( \int_{0}^{\sqrt{2}-u}\frac{dv}{ 2-u^{2}+v^{2}}\right) \,du \\ &=&4\int_{0}^{\sqrt{2}/2}\frac{1}{\sqrt{2-u^{2}}}\arctan \frac{u}{\sqrt{ 2-u^{2}}}\,du, & u=\sqrt{2}\sin t \\ &&+4\int_{\sqrt{2}/2}^{\sqrt{2}}\frac{1}{\sqrt{2-u^{2}}}\arctan \frac{\sqrt{2 }-u}{\sqrt{2-u^{2}}}\,du\, & u=\sqrt{2}\cos \theta \\ &=&4\int_{0}^{\pi /6}\arctan \left( \tan t\right)\, dt\\&&+4\int_{0}^{\pi /3}\arctan \left( \frac{1-\cos \theta}{\sin \theta}\right) d\theta, \\ &=&4\int_{0}^{\pi /6}t\,dt+4\int_{0}^{\pi /3}\arctan \left( \tan \frac{\theta}{2} \right) d\theta \\ &=&\frac{\pi ^{2}}{18}+\frac{\pi ^{2}}{9}=\frac{\pi ^{2}}{6}. \end{eqnarray*} $$

One of the substitutions is a new one. After the first pair of substitutions $x,y$, we have done two additional ones in the resulting integrals, as indicated above: in the 1st, $u=\sqrt{2}\sin t,du=\sqrt{2}\cos t\,dt$, and in the 2nd, $u=\sqrt{2}\cos \theta,du=-\sqrt{2}\sin \theta\,d\theta$.

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Also the first proof in this list of proofs of $\zeta(2)=\pi^2/6$ by Robin Chapman. – Américo Tavares Nov 5 '12 at 16:00

As far as I understand you question, you would like to solve the integral via substitution. (otherwise did's comment to your question might provide a simpler solution)

You are right that the Jacobian is 1 however you did some mistakes with the integration bounds after the substitution. The correct solution is $$\int_{[0,1] \times [0,1]} \frac{dx \, dy}{1-xy}=\underbrace{\int_0^{\sqrt{2}/2}\!du \int_{-u}^u\!dv \frac{2}{2-u^2+v^2}}_{I_1}+ \int_{\sqrt{2}/2}^\sqrt{2}\!du \int_{-\sqrt{2}+u}^{\sqrt{2}-u}\!dv\frac{2}{2-u^2+v^2} $$ which you obtain by rewriting the conditions $0\leq x,y \leq 1$ in terms of $u$ and $v$.

Now, I show you how to integrate the first of the two integrals, $I_1$ (the other is very similar). Calculating the integral over $u$, we obtain $$ I_1= \int_{0}^{\sqrt{2}/2}\!du\, \frac{4\arctan(u/\sqrt{2-u^2})}{\sqrt{2-u^2}}.$$ Substituting $x=\arctan(u/\sqrt{2-u^2})$ yields $$ I_1 = 4 \int_0^{\pi/6} dx\,x= \frac{\pi^2}{18}. $$ The last step is equivalent to the maybe more intuitive substitution $u=\sqrt{2} \sin t$ as indicated in your post.

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since $\ln(1-x)=-\sum_{n=1}^{\infty}\frac{x^{n}}{n}$, then $\int_{0}^{1}\frac{\ln(1-x)}{-x}dydx=\int_{0}^{1}\sum_{n=1}^{\infty}\frac{x^{n-1}}{n}dx=\sum_{n=1}^{\infty}\int_{0}^{1}\frac{x^{n-1}}{n}dx=\sum_{n=1}^{\infty}\frac{1}{n^2}=....$

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