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Problem of the Week #72 - August 12th, 2013

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Chris L T521

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Jan 26, 2012
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Thanks again to those who participated in last week's POTW! Here's this week's problem!

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Problem: The double integral
\[\int_0^1\int_0^1 \frac{1}{1-xy}\,dx\,dy\]
is an improper integral and could be defined as the limit of double integrals over the rectangle $[0,t]\times[0,t]$ as $t\to 1^{-}$.


  1. Expand the integrand as a geometric series to show that
    \[\int_0^1 \int_0^1 \frac{1}{1-xy}\,dx\,dy = \sum_{n=1}^{\infty}\frac{1}{n^2}\]
  2. Leonhard Euler proved that
    \[\sum_{n=1}^{\infty} \frac{1}{n^2}=\frac{\pi^2}{6}\]
    Prove this fact by evaluating the integral found in (1).

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Hints:
For (2), start by making the change of variables
\[x=\frac{u-v}{\sqrt{2}},\qquad y=\frac{u+v}{\sqrt{2}}.\]
It may be ideal to plot the corresponding region in the $uv$-plane. If, in evaluating the integral, you encounter either of the expressions
\[\frac{1-\sin\theta}{\cos\theta}\text{ or }\frac{\cos\theta}{1+\sin\theta}\]
you might want to consider using the identity
\[\cos\theta= \sin\left(\frac{\pi}{2}-\theta\right)\]
and the corresponding identity for $\sin\theta$.

Remember to read the POTW submission guidelines to find out how to submit your answers!
 
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Chris L T521

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Jan 26, 2012
995
This week's problem was correctly answered by MarkFL. You can find his very detailed (and great) solution below.

1.) Expanding the integrand as a geometric series, we may write:

\(\displaystyle \frac{1}{1-xy}=\sum_{n=0}^{\infty}(xy)^n\)

Hence the integral becomes:

\(\displaystyle \int_0^1\int_0^1 \sum_{n=0}^{\infty}(xy)^n\,dx\,dy=\int_0^1\left[\sum_{n=0}^{\infty}\frac{x^{n+1}y^n}{n+1} \right]_0^1\,dy=\)

\(\displaystyle \int_0^{1}\sum_{n=1}^{\infty}\frac{y^{n-1}}{n}\,dy=\left[\sum_{n=1}^{\infty}\frac{y^{n}}{n^2} \right]_0^1=\sum_{n=1}^{\infty}\frac{1}{n^2}\)

2.) Using the change of variables:

\(\displaystyle (x,y)=(u-v,u+v)\)

we obtain:

\(\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^2}=\iint_{R}\frac{1}{1-u^2+v^2}\left|\frac{\partial (x,y)}{\partial (u,v)} \right|\,du\,dv\)

Calculating the Jacobian matrix, we find:

\(\displaystyle \frac{\partial (x,y)}{\partial (u,v)}=\begin{vmatrix}\frac{\partial x}{\partial u}&\frac{\partial x}{\partial v}\\\frac{\partial y}{\partial u}&\frac{\partial y}{\partial v}\\\end{vmatrix}=\begin{vmatrix}1&-1\\1&1\\\end{vmatrix}=1(1)-(-1)(1)=2\)

Thus, we have:

\(\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^2}=2\iint_{R}\frac{1}{1-u^2+v^2}\,du\,dv\)

Remapping the boundaries in terms of the new variables, we find $R$ is a square in the $uv$-plane with vertices:

\(\displaystyle (0,0),\,\left(\frac{1}{2},-\frac{1}{2} \right),\,\left(\frac{1}{2},\frac{1}{2} \right),\,(1,0)\)

Reversing the order of integration and using the symmetry of the square, we obtain:

\(\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^2}=4\left(\int_0^{ \frac{1}{2}}\int_0^u \frac{dv\,du}{1-u^2+v^2}+\int_{ \frac{1}{2}}^1\int_0^{1-u} \frac{dv\,du}{1-u^2+v^2} \right)\)

Next, we may compute:

\(\displaystyle \int_0^u\frac{dv}{1-u^2+v^2}=\left[\frac{1}{\sqrt{1-u^2}}\tan^{-1}\left(\frac{v}{\sqrt{1-u^2}} \right) \right]_0^u=\frac{1}{\sqrt{1-u^2}}\tan^{-1}\left(\frac{u}{\sqrt{1-u^2}} \right)=\frac{1}{\sqrt{1-u^2}}\sin^{-1}(u)\)

\(\displaystyle \int_0^{1-u}\frac{dv}{1-u^2+v^2}=\left[\frac{1}{\sqrt{1-u^2}}\tan^{-1}\left(\frac{v}{\sqrt{1-u^2}} \right) \right]_0^{1-u}=\frac{1}{\sqrt{1-u^2}}\tan^{-1}\left(\frac{1-u}{\sqrt{1-u^2}} \right)=\frac{1}{\sqrt{1-u^2}}\tan^{-1}\left(\sqrt{\frac{1-u}{1+u}} \right)\)

If we let:

\(\displaystyle \tan(\theta)=\sqrt{\frac{1-u}{1+u}}\)

Squaring, we obtain:

\(\displaystyle \tan^2(\theta)=\frac{1-u}{1+u}\)

Add through by 1:

\(\displaystyle \tan^2(\theta)+1=\frac{1-u}{1+u}+1\)

Apply a Pythagorean identity on the left and combine terms on the right:

\(\displaystyle \sec^2(\theta)=\frac{2}{1+u}\)

Invert both sides:

\(\displaystyle \cos^2(\theta)=\frac{u+1}{2}\)

Solving for $u$, we find:

\(\displaystyle u=2\cos^2(\theta)-1=\cos(2\theta)\)

Hence, we find:

\(\displaystyle \tan^{-1}\left(\sqrt{\frac{1-u}{1+u}} \right)=\theta=\frac{1}{2}\cos^{-1}(u)\)

Using the identity \(\displaystyle \sin^{-1}(u)+\cos^{-1}(u)=\frac{\pi}{2}\) we finally have:

\(\displaystyle \tan^{-1}\left(\sqrt{\frac{1-u}{1+u}} \right)=\frac{1}{2}\left(\frac{\pi}{2}-\sin^{-1}(u) \right)\)

Utilizing these results, we now may state:

\(\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^2}=4\left(\int_0^{ \frac{1}{2}}\frac{1}{\sqrt{1-u^2}}\sin^{-1}(u)\,du+\frac{1}{2}\int_{ \frac{1}{2}}^1 \frac{1}{\sqrt{1-u^2}}\left(\frac{\pi}{2}-\sin^{-1}(u) \right)\,du \right)\)

Now, let's look at the first integral:

\(\displaystyle \int_0^{ \frac{1}{2}}\frac{1}{\sqrt{1-u^2}}\sin^{-1}(u)\,du\)

Using the substitution:

\(\displaystyle \alpha=\sin^{-1}(u)\,\therefore\,d\alpha=\frac{1}{\sqrt{1-u^2}}\,du\)

we now have:

\(\displaystyle \int_0^{\frac{\pi}{6}}\alpha\,d\alpha=\frac{1}{2} \left[\alpha^2 \right]_0^{\frac{\pi}{6}}= \frac{1}{2}\left(\frac{\pi}{6} \right)^2=\frac{\pi^2}{72}\)

Next, let's break the second integral into two parts:

i) \(\displaystyle \frac{\pi}{4}\int_{\frac{1}{2}}^1 \frac{1}{\sqrt{1-u^2}}\,du=\frac{\pi}{4}\left[\sin^{-1}(u) \right]_{\frac{1}{2}}^1=\frac{\pi}{4}\left(\frac{\pi}{2}-\frac{\pi}{6} \right)=\frac{\pi^2}{12}\)

ii) \(\displaystyle -\frac{1}{2}\int_{\frac{1}{2}}^1 \frac{1}{\sqrt{1-u^2}}\sin^{-1}(u)\,du\)

Using the substitution:

\(\displaystyle \alpha=\sin^{-1}(u)\,\therefore\,d\alpha=\frac{1}{\sqrt{1-u^2}}\,du\)

we now have:

\(\displaystyle -\frac{1}{2}\int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \alpha\,d\alpha=-\frac{1}{4}\left[\alpha^2 \right]_{\frac{\pi}{6}}^{\frac{\pi}{2}}=-\frac{1}{4}\left(\left(\frac{\pi}{2} \right)^2-\left(\frac{\pi}{6} \right)^2 \right)=-\frac{\pi^2}{18}\)

Thus, putting these results together, there results:

\(\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^2}=4\left(\frac{\pi^2}{72}+\frac{\pi^2}{12}-\frac{\pi^2}{18} \right)=4\left(\frac{\pi^2}{24} \right)\)

And, we may then state:

\(\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}\)

Shown as desired.
 
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