$\frac{\pi}{sin(a \pi)}=\frac{1}{a}+2a\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^2-a^2}$

Question

Why does $$\frac{\pi}{\sin\left(a\pi\right)} = \frac{1}{a} + 2a\sum_{n = 1}^{\infty} \frac{\left(-1\right)^{n + 1}}{n^{2} - a^{2}} \quad\mbox{for}\ a >0\ \mbox{and}\ a\ \mbox{is not an}\ Integer. $$ Why is this the case I've been trying to figure out why this is the case but can't make head or tail of it can anyone help $?$.

Answer 1

Via a Fourier expansion of $\cos(\pi a x)$ on the interval $[-1,1]$, we can show that $$ \cos(\pi a x) = \frac{\sin(a\pi)}{a\pi}+\sum_{n=1}^{\infty}\frac{1}{\pi}\frac{2a(-1)^{n+1}}{n^2-a^2}\sin(\pi a)\cos(n \pi x)\,. $$ This holds everywhere inside $(-1,1)$, so it holds in particular at $x=0$ where the cosine functions evaluate to 1, in which case this becomes $$ 1 = \frac{\sin(a\pi)}{a\pi}+\sum_{n=1}^{\infty}\frac{1}{\pi}\frac{2a(-1)^{n+1}}{n^2-a^2}\sin(\pi a)\,. $$ The result follows by multiplying both sides by $a\,/\sin(a\pi)$.