Finding function given Fourier coefficients

Question

I am trying to find a function with Fourier coefficients $\frac{1}{ 1 + (2 \pi n)^2}$. How do I go about this problem?

Answer 1

We can explicitly sum up $\displaystyle S(x)=\sum_{n \in \mathbb{Z}} \frac{e^{2 \pi i x n}}{ 1 + (2 \pi n)^2}$, using integration in the complex plane.

First, we note that $S(x)$ is a periodic function with the period $1$, so we can consider only $x\in[0;1]$.

Next, we construct the function $\displaystyle g(z)=\frac{2\pi i}{e^{2\pi iz}-1}\,\frac{e^{2\pi ixz}}{1+(2\pi z)^2}$ and it integrate in the complex plane along the circle with the radius $R\to\infty$.

On the one hand, for $x\in [0;1]$ the integral along the circle $\to0$, as $R\to\infty$.

On the other hand, $$\oint g(z)dz=\frac{i}{2\pi}\oint_{C_R}\frac{1}{e^{2\pi iz}-1}\,\frac{e^{2\pi ixz}}{(z+\frac i{2\pi})(z-\frac i{2\pi})}dz=2\pi i\sum Res \,g(z)$$ $$=2\pi i\left(S(x)+\frac i{2\pi}\frac{e^{- x}}{e^{-1}-1}\frac{2\pi}{2i}-\frac i{2\pi}\frac{e^{ x}}{e^{1}-1}\frac{2\pi}{2i}\right)=0$$ $$\Rightarrow\,S(x)=\frac12\left(\frac{e^{ x-\frac12}}{e^\frac12-e^{-\frac12}}-\frac{e^{- x+\frac12}}{e^{-\frac12}-e^\frac12}\right)=\frac{\cosh(x-\frac12)}{2\sinh\frac12}$$

Answer 2

given the series $$\sum_{n=-\infty}^{\infty} \frac{e^{2\pi i n x}}{1+(\pi n)^2}$$ we can immediately turn this into the cosine series (since the sine terms cancel each other out):

$$\sum_{n=-\infty}^{\infty} \frac{\cos({2\pi n x})}{1+(\pi n)^2}$$ now since cos is even: $$1+2\sum_{n=1}^{\infty} \frac{\cos({2\pi n x})}{1+(\pi n)^2}$$

now using the fourier series of cosh(ax), we can evaluate this to be: $$\frac{\cosh(2x-1)}{\sinh(1)}$$

for x $\in (0,1)$

in case anyone is reading this, I'm wondering about the sine series: $\sum_{n=1}^{\infty}\frac {sin(2\pi n x)}{(a^2+n^2)}$ and what it evaluates to.