The function $f(x)=\cos{\alpha x}$ has period $2\pi$ and is defined by
$f(x)=\cos{\alpha x}$
$(-\pi <x \le \pi), \alpha \notin \mathbf{Z}$
Obtain the Fourier series of $f(x)$ and hence show that
$\sum^{\infty}_{n=1}\frac{1}{n^2-\alpha^2}=\frac{1-\pi\alpha \cot{\pi\alpha}}{2\alpha^2}$
What result is obtained by letting $\alpha \rightarrow 0$ in this formula?
You may assume that $x\cot{x}=1-\frac{1}{3}x^2$ when $x$ is small.
Deduce that
$\sum^{\infty}_{n=1} \frac{1}{n^2(n^2-\alpha^2)}=\frac{1}{2\alpha^4}(1-\frac{\pi^2\alpha^2}{3}-\pi \alpha \cot{\pi \alpha}), \alpha \notin \mathbf{Z}$
The Fourier series is all fine
$\cos{\alpha x}=\frac{1}{\pi \alpha} \sin{\alpha \pi}-\frac{2\alpha \sin{\alpha \pi}}{\pi} \sum^{\infty}_{n=1}\frac{(-1)^n}{n^2-\alpha^2} \cos{nx}$
Putting $x=\pi$ (pt of cty)
$\cos{\alpha \pi}-\frac{1}{\pi \alpha} \sin{\alpha \pi}=-\frac{2\alpha \sin{\alpha \pi}}{\pi} \sum^{\infty}_{n=1}\frac{1}{n^2-\alpha^2}$
$\sum^{\infty}_{n=1}\frac{1}{n^2-\alpha^2}=\frac{1-\pi \alpha \cot{\pi \alpha}}{2\alpha^2}$
As $\alpha \rightarrow 0$,
$\sum^{\infty}_{n=1}=\lim_{\alpha\rightarrow 0}[\frac{1-\pi \alpha \cot{\pi \alpha}}{2\alpha^2}]$
$=\lim_{\alpha\rightarrow 0}[\frac{1-\pi(1-\frac{1}{3}\alpha^2}{2\alpha^2}]\frac{\pi}{6}$
And then , how do I prove the above product?
Do you know of a function with Fourier coefficients $\hat{f}(n) = \frac{1}{n^2}$? If so, how would it be related to a function with Fourier coefficients $\hat{F}(n) = \frac{1}{n^2(n^2-\alpha^2)}$?
Hint: