So here's what I got and I don't know how to move forward, or if I can move forward:
Notice that $f(-x)=\sum_{m=1}^\infty\frac{\cos(-mx)}{1+m^2}=\sum_{m=1}^\infty\frac{\cos(mx)}{1+m^2}=f(x)$ so $f(x)$ is even, so we are only considering the fourier series of the form $f(x)=\sum_{m=1}^\infty\frac{\cos(mx)}{1+m^2}=\frac{a_0}{2}+\sum_{n=1}^\infty a_n\cos(nx)=Sf(x)$, where $a_0=\frac{2}{\pi}\int_{-\pi}^\pi\sum_{m=1}^\infty\frac{\cos(mx)}{1+m^2}=\frac{2}{\pi}\sum_{m=1}^\infty\int_{-\pi}^{\pi}\frac{\cos(mx)}{1+m^2}=\frac{2}{\pi}\sum_{m=1}^\infty\frac{2\sin(m\pi)}{m^3+m}=0$ since $\sin(m\pi)=0$ for $m\in\mathbb{N}$
and $a_n=\frac{2}{\pi}\int_{-\pi}^{\pi}\sum_{m=1}^{\infty}\frac{\cos(mx)}{1+m^2}dx=\frac{2}{\pi}\sum_{m=1}^{\infty}\int_{0}^{\pi}\frac{\cos(mx)\cos(nx)}{1+m^2}dx$ (point where I'm stuck).
Update: $a_n=\frac{2}{\pi}\sum_{m=1}^{\infty}\int_{0}^{\pi}\frac{\cos(mx)\cos(nx)}{1+m^2}dx=\frac{2}{\pi}\sum_{m=1}^{\infty}\frac{1}{1+m^2}\int_{0}^{\pi}\frac{\cos(m-n)x+\cos(m+n)x}{2}dx$
$=\frac{1}{\pi}\sum_{m=1}^{\infty}\frac{1}{1+m^2}(\frac{\sin(m-n)\pi}{m-n}+\frac{\sin(m+n)\pi}{m+n})dx=0$ since $\sin(n\pi)=0\,\forall n\in\mathbb{N}$ which is still wrong, I know it to be true
Hint: $$\frac 1 {1+m^2}=\frac 1 2 \left (\frac 1 {1+im} + \frac 1 {1-im}\right)=\frac {(-1)^m} {4\pi(e^{\pi}-e^{-\pi})} \left (\int_{-\pi}^{\pi}e^{imx}e^{x}dx + \int_{-\pi}^{\pi}e^{imx}e^{-x}dx\right)$$
Can you see this being the Fourier coefficient of a function?