Chapter 14, Problem 5PS

Deriving a Sum Derive Euler’s famous result that was mentioned in Section 9.3, ${\displaystyle \underset{n=1}{\overset{\infty }{\sum }}\frac{1}{n^2}=\frac{{\pi }^2}{6}}$ by completing each step.

(a) Prove that ${\displaystyle \int \frac{dv}{2-u^2+v^2}=\frac{1}{\sqrt{2-u^2}}\arctan \frac{v}{\sqrt{2-u^2}}+C}$

(b) Prove that $I_1={\displaystyle {\int }_0^{\sqrt{2}/2}{\displaystyle {\int }_{-u}^u\frac{2}{2-u^2+v^2}}dvdu=\frac{{\pi }^2}{18}}$ by using the substitution $u=\sqrt{2}\sin \theta$.

(c) Prove that $\begin{array}{l}{I}_2={\displaystyle {\int }_{\sqrt{2}/2}^{\sqrt{2}}{\displaystyle {\int }_{u-\sqrt{2}}^{-u+\sqrt{2}}\frac{2}{2-u^2+v^2}dvdu}}\\ =4{\displaystyle {\int }_{\pi /6}^{\pi /2}\arctan \frac{1-\sin \theta }{\cos \theta }d\theta }\end{array}$ by using the substitution $u=\sqrt{2}\sin \theta$.

(d) Prove the trigonometric identity

$\frac{1-\sin \theta }{\cos \theta }=\tan \left[\frac{\left(\pi /2\right)-\theta }{2}\right]$.

(e) Prove that $I_2={\displaystyle {\int }_{\sqrt{2/2}}^{\sqrt{2}}{\displaystyle {\int }_{u-\sqrt{2}}^{-u+\sqrt{2}}\frac{2}{2-u^2+v^2}dvdu=\frac{{\pi }^2}{9}}}$.

(f) Use the formula for the sum of an infinite geometric series to verify that

${\displaystyle \underset{n=1}{\overset{\infty }{\sum }}\frac{1}{n^2}={\displaystyle {\int }_0^1{\displaystyle {\int }_0^1\frac{1}{1-xy}dxdy}}}$.

(g) Use the change of variables

$u=\frac{x+y}{\sqrt{2}}$ and $v=\frac{y-x}{\sqrt{2}}$ to prove that

${\displaystyle \underset{n=1}{\overset{\infty }{\sum }}\frac{1}{n^2}={\displaystyle {\int }_0^1{\displaystyle {\int }_0^1\frac{1}{1-xy}dxdy=I_1+I_2=\frac{{\pi }^2}{6}}}}$.