Midterm Bonus Assignment

Abstract

This serves as a proof to the Basel Problem, more commonly known as the sum of inverse squares. The purpose of this paper is to prove that this sum converges to $\frac{{\pi }^2}{6}$, using methods from single and multivariable calculus.

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Problem 1

Prove that the double integral and the infinite series shown below are equal:

$${\int }_0^1{\int }_0^1\frac{1}{1-xy}dxdy=\sum _{n=1}^{\infty }\frac{1}{n^2}$$

First, we expand the integral out to an infinite series:

Since our sum will not deal with numbers $>1$, we can express $\frac{1}{1-xy}$ as a sum of $(xy{)}^n$.

$$\frac{1}{1-xy}=\sum _{n=0}^{\infty }(xy{)}^n$$

Then, we can substitute this expanded series into our double integral.

$${\int }_0^1{\int }_0^1\frac{1}{1-xy}dxdy={\int }_0^1{\int }_0^1\sum _{n=0}^{\infty }(xy{)}^{n}dxdy$$

Changing the order of summation/integration

Assuming that the series converges uniformly for all $x,y\in [0,1]$, we can then exchange the order of summation/integration.

$${\int }_0^1{\int }_0^1\sum _{n=0}^{\infty }(xy{)}^{n}dxdy=\sum _{n=0}^{\infty }{\int }_0^1{\int }_0^1(xy{)}^{n}dxdy$$

Evaluating the Integral

From our new double integral, we can identify the inner integral as:

$${\int }_0^1{\int }_0^1(xy{)}^{n}dxdy={\int }_0^1\left({\int }_0^1{x}^{n}dx\right)y^{n}dy$$

Evaluating Inner Integral:

$${\int }_0^1{x}^{n}dx={\frac{x^{n+1}}{n+1}|}_0^1=\frac{1}{n+1}$$

Substituting:

$${\int }_0^1{\int }_0^1(xy{)}^{n}dxdy={\int }_0^1\frac{1}{n+1}{y}^{n}dy$$

Evaluating $w\cdot r\cdot t$:

$${\int }_0^1{y}^{n}dy={\frac{y^{n+1}}{n+1}|}_0^1=\frac{1}{n+1}$$

Putting this all together:

$${\int }_0^1{\int }_0^1(xy{)}^{n}dxdy=\frac{1}{n+1}\cdot \frac{1}{n+1}=\frac{1}{(n+1{)}^2}$$

Applying Sum to Evaluated Integral:

Plugging back in:

$${\int }_0^1{\int }_0^1\frac{1}{1-xy}dxdy=\sum _{n=0}^{\infty }\frac{1}{(n+1{)}^2}$$

Now we let $k=n+1$, and plug in:

$$\sum _{n=0}^{\infty }\frac{1}{(n+1{)}^2}=\sum _{k=1}^{\infty }\frac{1}{k^2}$$

And thus, we have proven that:

$${\int }_0^1{\int }_0^1\frac{1}{1-xy}dxdy=\sum _{n=1}^{\infty }\frac{1}{n^2}$$

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Problem 2

Show that the double integral in Problem 1 equals $\frac{{\pi }^2}{6}$ by making the change of variables:

$$x=\frac{1}{\sqrt{2}}(u-v),y=\frac{1}{\sqrt{2}}(u+v)$$

I most definitely did this incorrect, but I hope some of it is right!

Sketch of the UV Plane:

Change of variables

with the substitution:

$$x=\frac{1}{\sqrt{2}}(u-v),y=\frac{1}{\sqrt{2}}(u+v)$$

we seek to calculate the Jacobian determinant, $\frac{\partial (x,y)}{\partial (u,v)}$:

$$\frac{\partial x}{\partial u}=\frac{1}{\sqrt{2}},\frac{\partial x}{\partial v}=-\frac{1}{\sqrt{2}},\frac{\partial y}{\partial u}=\frac{1}{\sqrt{2}},\frac{\partial y}{\partial v}=\frac{1}{\sqrt{2}}$$

this yields the matrix:

$$Jacobian=\left[\begin{array}{cc}\frac{1}{\sqrt{2}}& -\frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}}& \frac{1}{\sqrt{2}}\end{array}\right]$$

leading us to the jacobian determinant:

$$\det (Jacobian)=\left(\frac{1}{\sqrt{2}}\right)\left(\frac{1}{\sqrt{2}}\right)-\left(-\frac{1}{\sqrt{2}}\right)\left(\frac{1}{\sqrt{2}}\right)=\frac{1}{2}+\frac{1}{2}=1$$

Since $\det (Jacobian)=1$, the change of variables does not scale the integration area.

New limits for integration

we start by substituting the limits $0\le x,y\le 1$ into the variables:

$$\begin{array}{rl}For:& \\ x& =0\to u=v\\ y& =0\to u=-v\\ x& =1andy=1\to squarerotatedby45degrees\end{array}$$

we map the square $[0,1]\times [0,1]$ to a diamond in the uv-plane.

This yields the new bounds for $u$ and $v$:

$$-\sqrt{2}\le u\le \sqrt{2},|v|\le u$$

Transforming the Integral

we now get:

$${\int }_0^1{\int }_0^1\frac{1}{1-xy}dxdy=\int {\int }_D\frac{1}{1-f(u,v)}|J|dudv$$

After this we would try to sub in $f(u,v)=\frac{1}{2}(u^2-v^2)$, to compute the integral, but as much as I have tried am doing something wrong.