Rylan Duncan

2022-01-24

Finding the infinite sum involving $\mathrm{coth}$ function using contour integration
I am looking to show: $\sum _{n=1}^{\mathrm{\infty }}\frac{\mathrm{coth}\left(n\pi \right)}{{n}^{3}}=\frac{7{\pi }^{3}}{180}$

Nevaeh Jensen

Expert

Observe that the integrand has simple poles at the points $z=n\pi$ and $z=\in \pi$, with .
The residues at the poles are computed as :
$\underset{z\to n\pi }{lim}\left(z-n\pi \right)\frac{\mathrm{cot}z\mathrm{coth}z}{{z}^{3}}=\underset{\zeta \to 0}{lim}\zeta \frac{\mathrm{cot}\left(\zeta \right)\mathrm{coth}\left(\zeta +n\pi \right)}{{\left(\zeta +n\pi \right)}^{3}}=\frac{\mathrm{coth}\left(n\pi \right)}{{\left(n\pi \right)}^{3}}$
and
$\underset{z\to \in \pi }{lim}\left(z-\in \pi \right)\frac{\mathrm{cot}z\mathrm{coth}z}{{z}^{3}}=\underset{\zeta \to 0}{lim}\zeta \frac{\mathrm{cot}\left(\zeta +\in \pi \right)\mathrm{coth}\left(\zeta \right)}{{\left(\zeta +\in \pi \right)}^{3}}=\frac{\mathrm{coth}\left(n\pi \right)}{{\left(n\pi \right)}^{3}},$
where we used
$\underset{x\to 0}{lim}x\mathrm{cot}x=\underset{x\to 0}{lim}x\mathrm{coth}x=1$
$\mathrm{cot}\left(x+n\pi \right)=\mathrm{cot}x$
$\mathrm{coth}\left(x+\in \pi \right)=\mathrm{coth}\left(x\right)$
$\mathrm{cot}\left(ix\right)=-i\mathrm{coth}\left(x\right)$
With this and a suitable choice of the integration contour you will obtain:
$4\sum _{n=1}^{\mathrm{\infty }}\frac{\mathrm{coth}\left(n\pi \right)}{{\left(n\pi \right)}^{3}}=\sum _{n=1}^{\mathrm{\infty }}\left[Res\left(f,-n\pi \right)+Res\left(f,n\pi \right)+Res\left(f,-\in \pi \right)+Res\left(f,\in \pi \right)\right]$
$=-Res\left(f,0\right)=\frac{7}{45}$

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