Calculate $\sum\limits_{n=1}^{\infty} \frac{a\cos(nx)}{a^2+n^2}$

Question

I have to calculate $\sum\limits_{n=1}^{\infty} \frac{a\cos(nx)}{a^2+n^2}$ for $x\in(0,2\pi)$.

I have used the function $f(x)=e^{ax}$ and I have calculated the Fourier coefficients which are: $$a_0=\dfrac 1{2a} \dfrac {e^{2\pi a}-1}{\pi}$$ $$a_n=\dfrac {e^{2\pi a}-1}{\pi} \dfrac{a}{a^2+n^2}$$ $$b_n=\dfrac {e^{2\pi a}-1}{\pi} \dfrac{-n}{a^2+n^2}$$

In the end, when it is written as Fourier series: $$e^{ax}=\frac{e^{2\pi a}-1}{\pi}\left(\frac{1}{2a}+\sum^\infty_{n=1}\frac{a\cos(nx)-n\sin(nx)}{a^2+n^2}\right),\text{ for }x\in(0,2\pi).$$

My question is how can I use all these facts to calculate $\sum\limits_{n=1}^{\infty} \frac{a\cos(nx)}{a^2+n^2}$?

Answer 1

Or equivallently, we can choose : $f(x)=\cosh ax$, then

$$f(x)=\frac{a_0}{2}+\sum_{n=1}^\infty a_n\cos nx + b_n \sin nx$$

where, since $f(x)$ is symmetric, $b_n =0$ and also

$$a_n=\frac{2}{\pi}\int_0^\pi\cosh ax \cos nx \;\mathrm{d}x=\frac{2a}{\pi}\frac{\sinh\pi a}{a^2+n^2}\,\cos{\pi n}$$

hence

$$\cosh ax= \frac{2}{\pi}\frac{\sinh\pi a}{2a}+a\sinh\pi a\sum_{n=1}^\infty\frac{2}{\pi}\frac{\cos{\pi n}\cos n x}{a^2+n^2}$$

For $x=\pi-y$ we get :

$$\cosh a(\pi-y)= \frac{2}{\pi}\frac{\sinh\pi a}{2a}+a\sinh\pi a\sum_{n=1}^\infty\frac{2}{\pi}\frac{\cos n y}{a^2+n^2}$$

Rearanging and renaming $y \rightarrow x$ :

$$\sum_{n=1}^\infty\frac{a \cos n x}{a^2+n^2}= \frac{\pi\cosh a(\pi-x)}{2\sinh \pi a}-\frac{1}{2a} \quad\quad ;\text{for}\quad x\in [0,2\pi]$$

Addendum :

$$\sum_{n=-\infty}^\infty\frac{a \cos n x}{a^2+n^2}= \frac{\pi\cosh a(\pi-x)}{\sinh \pi a}\quad\quad ;\text{for}\quad x\in [0,2\pi]$$

Substituing $x=0$

$$\sum_{n=-\infty}^\infty\frac{a}{a^2+n^2}= \pi\coth \pi a$$

and $x=\pi$

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

which coincides with @Lucian's commentary.

Answer 2

Your function $f$ is the periodic function of period $2\pi$, that is equal to $\exp(ax)$ for $x\in (0,2\pi)$. Hence for $x\in(0, 2\pi)$, you have $f(-x)=f(2\pi-x)=\exp(a(2\pi-x))$. You can calculate this by your Fourier expansion; and now add the expressions for $f(x)$ and $f(-x)$.

Answer 3

$\newcommand{\d}{\,\mathrm{d}}\newcommand{\res}{\operatorname{Res}}$We can calculate this using complex analysis. Fix a real $a\neq0$ and $0<\tau<2\pi$. Define: $$f:\Bbb C\to\Bbb C,\,z\mapsto\frac{e^{i\tau z}}{e^{2\pi iz}-1}\cdot\frac{1}{z^2+a^2}$$

Two observations:

• $z\mapsto z\cdot f(z)$ vanishes at infinity (when bounded away from its poles, of course)
• $f$ has simple poles at $\pm ia$ and at the integers, $\Bbb Z$

The bounds on $\tau$ are essential for the first observation.

From integrating on the box contours with vertices $\pm(N+1/2)(1\pm i)$ where $N\in\Bbb N$ is large, we find the integrals vanish and thus that: $$2\pi i\sum_{n\in\Bbb Z}\res(f;n)=-2\pi i(\res(f;ia)+\res(f;-ia))$$The left hand side equals exactly: $$\sum_{n\in\Bbb Z}\frac{e^{i\tau n}}{n^2+a^2}$$And the right hand side equals: $$\begin{align}-2\pi i\left(\frac{e^{-\tau a}}{2ia(e^{-2\pi a}-1)}-\frac{e^{\tau a}}{2ia(e^{2\pi a}-1)}\right)&=\frac{\pi}{a}\cdot\frac{e^{\tau a}(e^{-2\pi a}-1)-e^{-\tau a}(e^{2\pi a}-1)}{(e^{-2\pi a}-1)(e^{2\pi a}-1)}\\&=\frac{\pi}{a}\cdot\frac{2\sinh((\tau-2\pi)a)+2\sinh(-\tau a)}{2-2\cosh(2\pi a)}\\&=\frac{\pi}{a}\cdot\frac{\sinh(\tau a)+\sinh((2\pi-\tau)a)}{\cosh(2\pi a)-1}\end{align}$$Observe that the imaginary parts of the series is zero. It follows that: $$\sum_{n=1}^\infty\frac{\cos(n\tau)}{n^2+a^2}=\frac{\pi}{2a}\left(\frac{\sinh(\tau a)+\sinh((2\pi-\tau)a)}{\cosh(2\pi a)-1}-\frac{1}{\pi a}\right)$$