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

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

• As an aside, for $x=2k\pi$, we have $\displaystyle\sum_{n=-\infty}^{\infty} \frac{a}{a^2+n^2} ~=~ \pi\coth(a\pi)$, and for $x=(2k+1)\pi$, we have $\displaystyle\sum_{n=-\infty}^{\infty} \frac{a(-1)^n}{a^2+n^2} ~=~ \pi~\text{csch}(a\pi)$. This can be easily shown by differentiating the natural logarithm of Euler's infinite product expansion for the sine and cosine functions. – Lucian Oct 14 '15 at 20:56

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 $y \rightarrow x$ :

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

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

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.

• - I am sorry for calculation errors, they will be corrected – Machinato Jul 20 '16 at 22:05
• Alternately, we could have tried solving $~\dfrac{S''(x)}a-aS(x)=\dfrac12,~$ where the last term stems from the value of the infinite series symbolically written as $~\displaystyle\sum_{n=1}^\infty\cos(nx).$ – Lucian Jul 24 '16 at 14:04

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 $f(-x)$.