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?.