Making use of Fourier series to evaluate an infinite sum

Solution 1:

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 $$\sum_{k=1}^{\infty} \frac{ (-1)^{k-1} k \sin(kx)}{a^2 + k^2} = \frac{\pi}{2} \frac{\sinh(ax)}{\sinh(a \pi)}$$