TITLE: Prove $\csc(x)=\sum_{k=-\infty}^{\infty}\frac{(-1)^k}{x+k\pi}$
QUESTION [10 upvotes]: Prove $$\csc(x)=\sum_{k=-\infty}^{\infty}\frac{(-1)^k}{x+k\pi}$$

Hardy uses this fact without proof in a monograph on different ways to evaluate $\int_0^{\infty}\frac{\sin(x)}{x} dx$.

REPLY [2 votes]: We begin by expanding the function $\cos(xy)$ in a Fourier series, 
$$\cos(xy)=a_0/2+\sum_{k=1}^\infty a_k\cos(ky) \tag1$$
for $x\in [-\pi/\pi]$.  The Fourier coefficients  $(1)$ are given by
$$\begin{align}
a_k&=\frac{2}{\pi}\int_0^\pi \cos(xy)\cos(ky)\,dy\\\\
&=\frac1\pi (-1)^k \sin(\pi x)\left(\frac{1}{x +k}+\frac{1}{x -k}\right)\tag2
\end{align}$$
Substituting $(2)$ into $(1)$, setting $y=0$, and dividing by $\sin(\pi x)$ reveals
$$\begin{align}
\pi \csc(\pi x)&=\frac1y +\sum_{n=1}^\infty (-1)^k\left(\frac{1}{x -k}+\frac{1}{x +k}\right)\\\\
&=\sum_{k=-\infty}^\infty \frac{(-1)^k}{x-k}\\\\
&=\sum_{k=-\infty}^\infty \frac{(-1)^k}{x+k}\tag3
\end{align}$$
Finally, enforcing the substitution $x\to x/\pi$ and dividing by $\pi$ in $(3)$ yields the coveted result
$$\csc(x)=\sum_{k=-\infty}^\infty \frac{(-1)^k}{x+k\pi}$$