# Finding the infinite sum involving $\coth$ function using contour integration

I am looking to show: $$\sum_{n=1}^∞ \frac{\coth(nπ)}{n^3} = \frac{7π^3}{180}$$

There is a hint earlier that you are supposed to be using the function $$f(z)=\frac{\cot z\coth z}{z^3}$$. I have calculated the residue at the pole of order 5 at $$z=0$$ as $$-\frac{7}{45}$$, but I am unsure how to calculate the other residues, so I can use the residue theorem.

I think there is a simple pole whenever $$z=\frac{(2n+1)π}{2}$$, as this is when $$\cot z=0$$ but I just don't know how to find the residue here. I'm presuming my residues will lead to the sum I'm wanting to find coming out in some form when I apply the residue theorem, but I'm just not sure how to get there.

Thanks so much for any help in advance.

• Surely you mean poles at $z=n\pi$, when $\cot z=\infty$, don't you? – user May 6 '19 at 19:36

Observe that the integrand has simple poles at the points $$z=n\pi$$ and $$z=i n\pi$$, with $$n\in\mathbb Z$$, $$n\ne0$$.
The residues at the poles are computed as : $$\lim_{z\to n\pi}(z-n\pi)\frac{\cot z\coth z}{z^3}=\lim_{\zeta\to0}\zeta\frac{\cot (\zeta)\coth(\zeta+n\pi)}{(\zeta+n\pi)^3}=\frac{\coth(n\pi)}{(n\pi)^3},$$ and $$\lim_{z\to in\pi}(z-in\pi)\frac{\cot z\coth z}{z^3}=\lim_{\zeta\to0}\zeta\frac{\cot (\zeta+in\pi)\coth(\zeta)}{(\zeta+i n\pi)^3}=\frac{\coth(n\pi)}{(n\pi)^3},$$ where we used \begin{align} &\lim_{x\to0}x\cot x=\lim_{x\to0}x\coth x=1,\\ &\cot(x+n\pi)=\cot(x),\\ &\coth(x+in\pi)=\coth(x),\\ &\cot(i x)=-i\coth(x). \end{align}
With this and a suitable choice of the integration contour you will obtain: \begin{align} 4\sum_{n=1}^\infty\frac{\coth(n\pi)}{(n\pi)^3}&=\sum_{n=1}^\infty\left[\operatorname{Res}(f,-n\pi)+\operatorname{Res}(f,n\pi)+\operatorname{Res}(f,-in\pi)+\operatorname{Res}(f,in\pi)\right]\\ &=-\operatorname{Res}(f,0)=\frac7{45}. \end{align}
In this answer, it is shown that $$\pi\coth(\pi n)=\frac1n+2n\sum_{k=1}^\infty\frac1{n^2+k^2}\tag1$$ Therefore, since $$\zeta(2)=\frac{\pi^2}6$$, \begin{align} \sum_{n=1}^\infty\sum_{k=1}^\infty\frac1{n^2\!\left(n^2+k^2\right)} &=\sum_{n=1}^\infty\sum_{k=1}^\infty\frac1{k^2}\left(\frac1{n^2}-\frac1{n^2+k^2}\right)\\ &=\sum_{n=1}^\infty\sum_{k=1}^\infty\frac1{k^2n^2}-\sum_{n=1}^\infty\sum_{k=1}^\infty\frac1{k^2\!\left(n^2+k^2\right)}\\ &=\frac12\sum_{n=1}^\infty\sum_{k=1}^\infty\frac1{k^2n^2}\\[3pt] &=\frac12\zeta(2)^2\\[6pt] &=\frac{\pi^4}{72}\tag2 \end{align} Thus, since $$\zeta(4)=\frac{\pi^4}{90}$$ \begin{align} \sum_{n=1}^\infty\frac{\coth(\pi n)}{n^3} &=\frac1\pi\sum_{n=1}^\infty\frac1{n^4}+\frac2\pi\sum_{n=1}^\infty\sum_{k=1}^\infty\frac1{n^2\!\left(n^2+k^2\right)}\\ &=\frac{\pi^3}{90}+\frac{\pi^3}{36}\\[3pt] &=\frac{7\pi^3}{180}\tag3 \end{align}