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Monday Math 142

Find \int_0^1\int_0^1\frac1{1-x^2y^2}\,dx\,dy:
1). Using an infinite series and the Riemann zeta function
2). Using neither infinite series nor polylogarithms


1).Infinite Series
The infinite geometric series \frac1{1-z}=1+z+z^2+z^3+\cdots holds for |z|<1, which, except for the singularity at (x,y)=(1,1), (xy)^2 satisfies for our region of integration; thus
\int_0^1\int_0^1\frac1{1-x^2y^2}\,dx\,dy=\int_0^1\int_0^1\sum_{n=0}^{\infty}\left(x^2y^2\right)^n\,dx\,dy;
we can exchange the order of integration and summation
\int_0^1\int_0^1\sum_{n=0}^{\infty}\left(x^2y^2\right)^n\,dx\,dy=\sum_{n=0}^{\infty}\int_0^1\int_0^1x^{2n}y^{2n}\,dx\,dy=\sum_{n=0}^{\infty}\left(\int_0^1x^{2n}\,dx\right)\left(\int_0^1y^{2n}\,dy\right)
and performing the integral \int_0^1x^{2n}\,dx=\left[\frac{x^{2n+1}}{2n+1}\right]_{0}^1=\frac{1}{2n+1}, we see
\int_0^1\int_0^1\frac1{1-x^2y^2}\,dx\,dy=\sum_{n=0}^{\infty}\frac{1}{(2n+1)^2}.
Now,
\zeta(2)=\sum_{n=1}^{\infty}\frac1{n^2};
we can separate this sum into sums over the odd and even integers
\zeta(2)=\sum_{n=1}^{\infty}\frac1{(2n)^2}+\sum_{n=0}^{\infty}\frac1{(2n+1)^2};
But the sum over the even integers is
\sum_{n=1}^{\infty}\frac1{(2n)^2}=\sum_{n=1}^{\infty}\frac1{4n^2}=\frac14\sum_{n=1}^{\infty}\frac1{n^2}=\frac14\zeta(2), and so
\sum_{n=0}^{\infty}\frac1{(2n+1)^2}=\frac34\zeta(2)=\frac34\frac{\pi^2}6=\frac{\pi^2}8
and so \int_0^1\int_0^1\frac1{1-x^2y^2}\,dx\,dy=\frac{\pi^2}8.

2). Without series or polylogarithms.
If we try to integrate directly over one variable, say x, we see that \int\frac1{1-x^2y^2}\,dx=\frac{\tanh^{-1}xy}y,
so we obtain single integral
\int_0^1\frac{\tanh^{-1}y}y\,dy,
and since \tanh^{-1}z=\frac12\ln\frac{1+z}{1-z}, we have
\int_0^1\frac{\ln(1+y)-\ln(1-y)}{2y}\,dy,
which requires polylogarithms to integrate.
So instead, we need to make a change of variables. The key is to define new variables u and v, with x=\frac{\sin{u}}{\cos{v}} and y=\frac{\sin{v}}{\cos{u}}. To require a one-to-one mapping to positive x and y, we need 0\le{u}\le\frac{\pi}2, 0\le{v}\le\frac{\pi}2. Further, we see x≤1 gives us
\begin{array}{rcl}\frac{\sin{u}}{\cos{v}}&\le&1\\\sin{u}&\le&\cos{v}\\\sin{u}&\le&\sin\left(\frac{\pi}2-v\right)\\u&\le&\frac{\pi}2-v\\u+v&\le&\frac{\pi}2\end{array},
and similarly, the upper limit on y gives
\begin{array}{rcl}\frac{\sin{v}}{\cos{u}}&\le&1\\\sin{v}&\le&\cos{u}\\\sin{v}&\le&\sin\left(\frac{\pi}2-u\right)\\v&\le&\frac{\pi}2-u\\u+v&\le&\frac{\pi}2\end{array}.
Thus, we see the region of integration R in the uv plane is the isoceles right triangle bounded by u=0, v=0, and u+v=\frac{\pi}2.
Now, we need to find the Jacobian determinant, since
dx\,dy=\left|\frac{\partial(x,y)}{\partial(u,v)}\right|\,du\,dv.
We find
\begin{array}{rcl}\left|\frac{\partial(x,y)}{\partial(u,v)}\right|&=&\left|\begin{array}{cc}\frac{\partial{x}}{\partial{u}}&\frac{\partial{x}}{\partial{v}}\\\frac{\partial{y}}{\partial{u}}&\frac{\partial{y}}{\partial{v}}\end{array}\right|\\&=&\left|\begin{array}{cc}\frac{\cos{u}}{\cos{v}}&\frac{\sin{u}\sin{v}}{\cos^2{v}}\\\frac{\sin{u}\sin{v}}{\cos^2{u}}&\frac{\cos{v}}{\cos{u}}\end{array}\right|\\&=&\left(\frac{\cos{u}}{\cos{v}}\right)\left(\frac{\cos{v}}{\cos{u}}\right)-\left(\frac{\sin{u}\sin{v}}{\cos^2{v}}\right)\left(\frac{\sin{u}\sin{v}}{\cos^2{u}}\right)\\&=&1-\frac{\sin^2{u}\sin^2{v}}{\cos^2{u}\cos^2{v}}\\\left|\frac{\partial(x,y)}{\partial(u,v)}\right|&=&1-\tan^2{u}\tan^2{v}\end{array}.
Thus, we have
\begin{array}{rcl}\iint\frac1{1-x^2y^2}\,dx\,dy&=&\iint\frac1{1-\left(\frac{\sin{u}}{\cos{v}}\right)^2\left(\frac{\sin{v}}{\cos{u}}\right)^2}\left|\frac{\partial(x,y)}{\partial(u,v)}\right|\,du\,dv\\&=&\iint\frac1{1-\frac{\sin^2{u}\sin^2{v}}{\cos^2{v}\cos^2{u}}}\left(1-\tan^2{u}\tan^2{v}\right)\,du\,dv\\&=&\iint\frac{1-\tan^2{u}\tan^2{v}}{1-\tan^2{u}\tan^2{v}}\,du\,dv\\&=&\iint\,du\,dv\end{array},
and our integral is just equal to the area of the triangle R, which is
A=\frac12\left(\frac{\pi}2\right)\left(\frac{\pi}2\right)=\frac{\pi^2}8.

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