Question 7.32: Prove that 1/1³ - 1/3³ + 1/5³ - 1/7³ + ··· = π³/32.
Prove that \frac{1}{1^{3}}-\frac{1}{3^{3}}+\frac{1}{5^{3}}-\frac{1}{7^{3}}+\cdots=\frac{\pi^{3}}{32}.
We have
\begin{aligned} F(z) & =\frac{\pi \sec \pi z}{z^{3}}=\frac{\pi}{z^{3} \cos \pi z}=\frac{\pi}{z^{3}\left(1-\pi^{2} z^{2} / 2 !+\cdots\right)} \\ & =\frac{\pi}{z^{3}}\left(1+\frac{\pi^{2} z^{2}}{2}+\cdots\right)=\frac{\pi}{z^{3}}+\frac{\pi^{3}}{2 z}+\cdots \end{aligned}
so that the residue at z=0 is \pi^{3} / 2.
The residue of F(z) at z=n+\frac{1}{2}, n=0, \pm 1, \pm 2, \ldots [which are the simple poles of sec \pi z ], is
\underset{z \rightarrow n+1 / 2}{\lim}\left\{z-\left(n+\frac{1}{2}\right)\right\} \frac{\pi}{z^{3} \cos \pi z}=\frac{\pi}{\left(n+\frac{1}{2}\right)^{3}} \underset{z \rightarrow n+1 / 2}{\lim} \frac{z-\left(n+\frac{1}{2}\right)}{\cos \pi z}=\frac{-(-1)^{n}}{\left(n+\frac{1}{2}\right)^{3}}
If C_{N} is a square with vertices at N(1+i), N(1-i), N(-1+i), N(-1-i), then
\oint\limits_{C_{N}} \frac{\pi \sec \pi z}{z^{3}} d z=-\sum\limits_{n=-N}^{N} \frac{(-1)^{n}}{\left(n+\frac{1}{2}\right)^{3}}+\frac{\pi^{3}}{2}=-8 \sum\limits_{n=-N}^{N} \frac{(-1)^{n}}{(2 n+1)^{3}}+\frac{\pi^{3}}{2}
and since the integral on the left approaches zero as N \rightarrow \infty, we have
\sum\limits_{-\infty}^{\infty} \frac{(-1)^{n}}{(2 n+1)^{3}}=2\left\{\frac{1}{1^{3}}-\frac{1}{3^{3}}+\frac{1}{5^{3}}-\cdots\right\}=\frac{\pi^{3}}{16}
from which the required result follows.