Making use of Fourier series to evaluate an infinite sum Show that $$\sum_{k=1}^{\infty}\frac{(-1)^{k-1}k \sin(ax)}{a^{2}+k^{2}}=\frac{\pi}{2}\frac{\sinh(ax)}{\sinh(\pi a)}, \;\ x\in (-\pi,\pi)$$
It appears to me this series is crying out for the use of Fourier series.  It seems to me I get close, but am failing to put the pieces all together.
Thus, I tried using the Fourier series for $f(x)=e^{ax}$.  $\qquad \cosh(ax)$ gives the same thing. 
$$a_{k}=\frac{1}{\pi}\int_{-\pi}^{\pi}e^{ax} \cos(kx)dx=\frac{e^{a\pi}(a \cos(k\pi)+k \sin(k\pi))}{\pi (a^{2}+k^{2})}-\frac{e^{-a\pi}(a \cos(k\pi)-k \sin(k \pi))}{\pi (a^{2}+k^{2})}$$
$b_{k}=0$
The given series is evident amongst $a_{n}$, but I kind of get hung up.
$\displaystyle a_{0}=\frac{2 \sinh(\pi a)}{\pi a}$
Now, using $\displaystyle e^{ax}=\frac{a_{0}}{2}+\sum_{k=1}^{\infty}a_{k} \cos(kx)\Rightarrow e^{ax}=\frac{\sinh(\pi a)}{\pi a}+\sum_{k=1}^{\infty}a_{k} \cos(kx)$
I even tried equating real and imaginary parts. Where the definite integral becomes
$$\left(\frac{(a \cos(k\pi)+k \sin(k\pi))e^{a \pi}}{a^{2}+k^{2}}+\frac{(a \sin(k\pi)-k  \cos(k\pi))e^{a\pi}}{a^{2}+k^{2}}i\right)$$ $$-\left(\frac{(a \cos(k\pi)-k \sin(k\pi))e^{-\pi a}}{a^{2}+k^{2}}-\frac{(a \sin(k\pi)+k \cos(k\pi))e^{-a\pi}}{a^{2}+k^{2}}i\right)$$
I think I can equate the imaginary and real parts: 
$$e^{ax}=\frac{e^{\pi a}-e^{-\pi a}}{2}\sum_{k=1}^{\infty}..............$$
I get mixed up here.  The identity in the previous line is $ \sinh(\pi a)$, so it looks like I am onto something.  If that $e^{ax}$ on the left of the equals sign were $ \sinh(ax)$, then a little algebra and I would practically be done. 
I tried to get to the known sum, but can't quite get there.  
There is something I  am overlooking. Can anyone point me in the right direction, please?. 
 A: Consider $\sinh(ax)$. Note that it is an odd function and hence if we expand it in sine's and cosine's the coefficient of cosines will be zero.
Hence, we can write $\displaystyle \sinh(ax) = \sum_{k=1}^{\infty} a_k \sin(kx)$
$$a_k = \frac1{\pi} \int_{-\pi}^{\pi} \sinh(ax) \sin(kx) dx$$
$$I = \int_{-\pi}^{\pi} \sinh(ax) \sin(kx) dx = \int_{-\pi}^{\pi} \left(\frac{e^{ax} - e^{-ax}}{2} \right) \left( \frac{e^{ikx} - e^{-ikx}}{2i} \right) dx $$
$$I = \frac1{4i} \left( \frac{e^{(a+ik)x}}{a+ik} - \frac{e^{(a-ik)x}}{a-ik} - \frac{e^{(-a+ik)x}}{-a+ik} - \frac{e^{-(a+ik)x}}{a+ik} \right)_{-\pi}^{\pi}$$
$$I = \frac{2}{4i} \left( \frac{(-1)^k e^{a \pi}}{a+ik} - \frac{(-1)^k e^{a \pi}}{a-ik} - \frac{(-1)^k e^{-a \pi}}{-a+ik} - \frac{(-1)^k e^{-a \pi}}{a+ik} \right)$$
$$I = \frac{(-1)^k}{2i} \left( \frac{-2 i k e^{a \pi}}{a^2+k^2} + \frac{2 i k e^{-a \pi}}{a^2+k^2} \right) = (-1)^{k-1} \frac{2 k }{a^2 + k^2} \sinh(a \pi)$$
Hence, we have
$$\sinh(ax) = \frac{ 2 \sinh(a \pi) }{\pi} \sum_{k=1}^{\infty} (-1)^{k-1} \frac{k}{a^2 + k^2} \sin(kx)$$
Rewriting, we get the desired result, namely
$$\sum_{k=1}^{\infty} \frac{ (-1)^{k-1} k \sin(kx)}{a^2 + k^2} = \frac{\pi}{2} \frac{\sinh(ax)}{\sinh(a \pi)}$$
