Apostol's Double Integral to Calculate

$\zeta (2)$

2022-11-14
J. Colliander

Notes for MATH200 Lecture 2022-11-14
Notes for MATH253 Lecture 2022-11-25

These notes are available at: bit.ly/basel-problem

0. Background

The Basel Problem, first posed in the mid-17th century, asks for a closed form expression for the value of the sum of the square reciprocals

$$\sum _{k=1}^{\mathrm{\infty }}\frac{1}{k^2}.$$
The problem remained open for almost a century. Leonhard Euler answered the question in 1734 and showed the sum converges to

$\frac{{\pi }^2}{6}$. The Basel Problem and Euler's work inspired Bernard Riemann to introduce the Riemann zeta function

$$\zeta (s)=\sum _{k=1}^{\mathrm{\infty }}\frac{1}{k^s}$$
and the celebrated Riemann Hypothesis. This video is an intriguing introduction to the Riemann Hypothesis.

Tom Apostol (1983) found an approach to calculating

$\zeta (2)$ using multivariable calculus. Additional information about the Basel Problem is available in these excellent slides by Brendan W. Sullivan. These notes follow Apostol's argument.

1. A Double Integral and Geometric Series

The double integral

$$I={\int }_0^1{\int }_0^1\frac{1}{1-xy}dxdy$$
can be recast using the geometric series as

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

Interchanging the sum and integrals, and carrying out the integrations term-by-term, reveals

$$I=\sum _{n=0}^{\mathrm{\infty }}\frac{1}{n+1}\frac{1}{n+1}=\sum _{k=1}^{\mathrm{\infty }}\frac{1}{k^2}.$$
Therefore, the answer to the Basel Problem is the volume under the surface

$z=\frac{1}{1-xy}$ above the unit square

$Q=\{(x,y):0<x<1,0<y<1\}.$

2. Inspecting the Integral

The integral over the unit square can be decomposed into integrals over the triangles above and below the diagonal

$y=x$

$$I={\int }_0^1{\int }_0^y\frac{1}{1-xy}dxdy+{\int }_0^1{\int }_0^x\frac{1}{1-xy}dydx.$$

Interchanging the symbols

$x\leftrightarrow y$ and recognizing

$xy=yx$ shows these two integrals are equal so

$$I=2{\int }_0^1{\int }_0^y\frac{1}{1-xy}dxdy.$$
The integrand explodes to infinity as

$(x,y)$ approaches

$(1,1)$.

3. Change Variables

We change variables by writing

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

$$u=\frac{1}{\sqrt{2}}(x+y);v=\frac{1}{\sqrt{2}}(y-x).$$
The Jacobian of this change of variables can be calculated

$$\left|\begin{array}{cc}\frac{\partial x}{\partial u}& \frac{\partial x}{\partial v}\\ \frac{\partial y}{\partial u}& \frac{\partial x}{\partial v}\end{array}\right|=\left|\begin{array}{cc}\frac{1}{\sqrt{2}}& -\frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}}& \frac{1}{\sqrt{2}}\end{array}\right|=1.$$
This change of variables is a rotation by

$\frac{\pi }{4}$ and places the hypoteneuse of the upper triangle along the

$u$ axis where

$v=0$.

After calculating

$xy=\frac{1}{2}(u^2-v^2)$, integrand is reexpressed as

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

$$I=4{\int }_0^{\frac{1}{\sqrt{2}}}{\int }_0^u\frac{1}{2-u^2+v^2}dvdu+4{\int }_{\frac{1}{\sqrt{2}}}^{\sqrt{2}}{\int }_0^{\sqrt{2}-u}\frac{1}{2-u^2+v^2}dvdu=I_1+I_2.$$

4. Evaluate Inner Integrals using Inverse Tangent

The change of variables allows us to use the integration formula

$${\int }_0^x\frac{dt}{a^2+t^2}=\frac{1}{a}{\tan }^{-1}(\frac{x}{a}).$$
The inner

$v$-integral in

$I_1$,

$${\int }_0^u\frac{1}{(\sqrt{2-u^2}{)}^2+v^2}dv=\frac{1}{\sqrt{2-u^2}}{\tan }^{-1}(\frac{u}{\sqrt{2-u^2}}).$$
The inner

$v$-integral in

$I_2$,

$${\int }_0^{\sqrt{2}-u}\frac{1}{(\sqrt{2-u^2}{)}^2+v^2}dv=\frac{1}{\sqrt{2-u^2}}{\tan }^{-1}(\frac{\sqrt{2}-u}{\sqrt{2-u^2}}).$$
It remains to calculate

$$I=4{\int }_0^{\frac{1}{\sqrt{2}}}{\tan }^{-1}(\frac{u}{\sqrt{2-u^2}})\frac{du}{\sqrt{2-u^2}}+4{\int }_{\frac{1}{\sqrt{2}}}^{\sqrt{2}}{\tan }^{-1}(\frac{\sqrt{2}-u}{\sqrt{2-u^2}})\frac{du}{\sqrt{2-u^2}}=I_1+I_2.$$

5. Evaluate Outer Integrals by Trigonometric Substitution

Evaluate

$I_1$

Change variables by writing

$u=\sqrt{2}\sin (\theta )$. The integration in

$u$ across

$[0,\frac{1}{\sqrt{2}}]$ corresponds to integration in

$\theta$ across

$[0,\frac{\pi }{6}]$ (since

$\frac{1}{2}=\sin (\frac{\pi }{6})$). Calculating, we find

$du=\sqrt{2}\cos \theta =\sqrt{2-u^2}d\theta .$ It may be helpful to draw a right triangle with hypoteneuse of length

$\sqrt{2}$, an adjacent side of length

$\sqrt{2}\cos (\theta )$, and opposite side of length

$\sqrt{2}\sin (\theta )=u$ to confirm that

$\cos (\theta )=\sqrt{2-u^2}.$ From the triangle,

$$\tan (\theta )=\frac{u}{\sqrt{2-u^2}},$$
so,

$$I_1=4{\int }_0^{\frac{\pi }{6}}\theta d\theta =2(\frac{\pi }{6}{)}^2.$$

Evaluate

$I_2$

Change variables by writing

$u=\sqrt{2}\cos (2\theta )$. The integration in

$u$ across

$[\frac{1}{\sqrt{2}},\sqrt{2}]$ correponds to integrating backwards from

$\frac{\pi }{6}$ to

$0$. We calculate

$$du=-2\sqrt{2}\sin (2\theta )d\theta =-2\sqrt{2}\sqrt{1-{\cos }^2(2\theta )}d\theta =-2\sqrt{2-u^2}d\theta .$$
This change of variable is inspired because

$$\frac{\sqrt{2}-u}{\sqrt{2-u^2}}=\frac{\sqrt{2}(1-\cos (2\theta ))}{\sqrt{2-2{\cos }^2(2\theta )}}=\frac{\sqrt{2}(1-\cos (2\theta ))}{\sqrt{2}\sqrt{(1+\cos (2\theta ))(1-\cos (2\theta ))}},$$
and continuing the calculation using double angle formulae

$$=\frac{\sqrt{1-\cos (2\theta )}}{\sqrt{1+\cos (2\theta )}}=\sqrt{\frac{2{\sin }^2\theta }{2{\cos }^2\theta }}=\tan (\theta )!$$
Therefore,

$$I_2=4{\int }_{\frac{\pi }{6}}^0\theta (-2)d\theta =8{\int }_0^{\frac{\pi }{6}}\theta d\theta =4(\frac{\pi }{6}{)}^2.$$
Combining, we find the answer to the Basel Problem by writing

$$I=I_1+I_2=2(\frac{\pi }{6}{)}^2+4(\frac{\pi }{6}{)}^2=\frac{{\pi }^2}{6}.$$
This argument establishes that

$$\zeta (2)=\sum _{k=1}^{\mathrm{\infty }}\frac{1}{k^2}=\frac{{\pi }^2}{6}.$$