# Math Help - infinite series

1. ## infinite series

In a proof of $\int_{-\infty}^{\infty} f(x) \ \frac{ \sin x}{x} \ dx = \int^{\pi}_{0} f(x) \ dx$ when $f(x)$ is $\pi$ - periodic on the entire real line, it is asserted that $\csc x = \sum_{k=-\infty}^{\infty} \frac{(-1)^{k}}{x + k \pi}$ .

Neither Maple nor Mathematica recognize this series, even for specific values of $x$. But numerically at least it appears to equal $\csc x$ for some specific values of $x$.

Any idea about the origins of this series?

EDIT: It's probably from a complex Fourier series.

2. ## Re: infinite series

Originally Posted by Random Variable
In a proof of $\int_{-\infty}^{\infty} f(x) \ \frac{ \sin x}{x} \ dx = \int^{\pi}_{0} f(x) \ dx$ when $f(x)$ is $\pi$ - periodic on the entire real line, it is asserted that $\csc x = \sum_{k=-\infty}^{\infty} \frac{(-1)^{k}}{x + k \pi}$ .

Neither Maple nor Mathematica recognize this series, even for specific values of $x$. But numerically at least it appears to equal $\csc x$ for some specific values of $x$.

Any idea about the origins of this series?

EDIT: It's probably from a complex Fourier series.
The problem has been 'attacked' here...

http://www.mathhelpforum.com/math-he...is-179123.html

Kind regards

$\chi$ $\sigma$

3. ## Re: infinite series

Another way to sum the series, which didn't occur to me when I started this thread, is to let $f(z) = \frac{\pi \csc \pi z}{x + \pi z}$.

EDIT: Then $\sum_{k= -\infty}^{\infty} \frac{(-1)^{k}}{x + \pi k} = -Res[f,-\frac{x}{\pi}]$ .

4. ## Re: infinite series

CoffeeMachine

It says that your private message quota has been exceeded, so I'll post the solution here.

$\int_{-\infty}^{\infty} f(x) \ \frac{\sin x}{x} \ dx = \sum_{k = -\infty}^{\infty} \int_{k \pi}^{(k+1) \pi} f(x) \ \frac{\sin x}{x} \ dx$

let $x = u +k \pi$

$= \sum_{k = -\infty}^{\infty} \int_{0}^{\pi} f(u + k \pi) \ \frac{\sin (u+ k \pi)}{u+k \pi} \ du$

$= \sum_{k = -\infty}^{\infty} \int^{\pi}_{0} f(u) \ (-1)^{k} \frac{\sin u}{u+ k \pi} \ du$

$= \int^{\pi}_{0} f(u) \sin u \sum_{k = -\infty}^{\infty} (-1)^{k} \frac{1}{u+ k \pi} \ du$

$= \int^{\pi}_{0} f(u) \ du$