Making use of Fourier series to evaluate an infinite sum

Question

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?.

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

Answer 2

A solution, we consider the function $$ f(z) = \frac{z \sin(z \theta)}{z^2 + a^2} $$ which has poles in ai,-ai the then known theorem $$\displaystyle \sum_{n=-\infty}^{+\infty}{\left(-1 \right)^n f\left(n \right)}=-\sum_{i=1}^{m}{Res\left(h,z_{i} \right)}$$ where m is the set of its poles f

And $\displaystyle h\left(z \right)=\frac{\pi f\left(z \right)}{\sin \pi z}$ so here: $$\displaystyle Res\left(h,ai \right)=\frac{\pi z \sin\left(z \theta \right)}{\left[\left(z^2+a^2 \right)\sin \pi z \right]'}=\frac{\pi a i \sin \left(a i \theta \right)}{2z\sin \pi z+\pi \cos \pi z \left(z^2+a^2 \right)}=$$ $$\displaystyle \frac{\pi ai \sinh \left(a\theta \right)}{2ai \sin \left(\pi ai \right)}=\frac{\pi ai \sinh \left(a \theta \right)}{2iai \sinh \left(\pi a \right)}=\frac{\pi}{2}\frac{\sinh \left(a\theta \right)}{\sinh \left(\pi a \right)}$$

and $$\displaystyle Res\left(h,ai \right)=\frac{\pi z \sin\left(z \theta \right)}{\left[\left(z^2+a^2 \right)\sin \pi z \right]'}=\frac{-\pi a i \sin \left(-a i \theta \right)}{2z\sin \pi z+\pi \cos \pi z \left(z^2+a^2 \right)}=$$ $$\displaystyle \frac{\pi ai \sinh \left(a\theta \right)}{-2ai \sin \left(-\pi ai \right)}=\frac{\pi ai \sinh \left(a \theta \right)}{2iai \sinh \left(\pi a \right)}=\frac{\pi}{2}\frac{\sinh \left(a\theta \right)}{\sinh \left(\pi a \right)}.$$

Consequently $$\displaystyle \sum_{k-=\infty}^{+\infty}{\frac{\left(-1 \right)^{k} k \sin \left(k \theta \right)}{k^2+a^2}}=-\sum_{i=1}^{2}{Res\left(h,z_{i} \right)}=-\pi \frac{\sinh \left(a\theta \right)}{\sinh \left(\pi a \right)}.$$

It is easy to see if $\displaystyle k=2m,k,=2m+1, m \in \mathbb{Z}$ then $\displaystyle k,-k$ they are contemporaneous or superfluous, so it will apply: $$\displaystyle \frac{\left(-1 \right)^{-k}\left(-k \right)\sin \left(-k\theta \right)}{\left(-k \right)^2+a^2}=\frac{\left(-1 \right)^{k}\left(k \right)\sin \left(k\theta \right)}{k^2+a^2}$$, $\forall k \in \mathbb{Z}.$

Therefore $$\displaystyle \sum_{k-=\infty}^{+\infty}{\frac{\left(-1 \right)^{k} k \sin \left(k \theta \right)}{k^2+a^2}}=\frac{1}{2}\sum_{k=0}^{+\infty}{\frac{\left(-1 \right)^{k} k \sin \left(k \theta \right)}{k^2+a^2}}$$ because for $k=0$ the corresponding term is 0, it will be

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

so multiplying and me $(-1)^{-1}=-1$ appears the $k-1$ and so $$\displaystyle \boxed{\sum_{k=1}^{+\infty}{\frac{\left(-1 \right)^{k-1} k \sin \left(k \theta \right)}{k^2+a^2}}=\frac{\pi}{2}\frac{\sinh \left(a\theta \right)}{\sinh \left(\pi a \right)}}$$