For fixed momentum put
$$
E_k=\sqrt{v^2k^2+m(T)^2}.
$$
In the contour formula, $F(z)=1/(E_k^2-z^2)$ has poles at $z=\pm E_k$. Deforming the contour onto those poles and using the oddness of $\coth(z/(2T))$ gives the standard <Bosonic Matsubara sum>
$$
\boxed{
T\sum_{n\in\mathbb Z}
\frac1{\omega_n^2+E_k^2}
=\frac1{2E_k}\coth\left(\frac{E_k}{2T}\right).}
$$
Equivalently,
$$
\frac1{2E_k}\coth\left(\frac{E_k}{2T}\right)
=\frac{1+2n_B(E_k)}{2E_k},
$$
where $n_B(E)=1/(e^{E/T}-1)$ is the <Bose-Einstein distribution>. The first term is the zero-point fluctuation and the second is its thermal occupation.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 337 / 1 / d / Solution

{c}

For $E>0$, contour summation over bosonic Matsubara frequencies gives
$$
T\sum_{n\in\mathbb Z}\frac1{\omega_n^2+E^2}
=\frac1{2E}\coth\frac{E}{2T}.
$$
It separates into the zero-point term $1/(2E)$ and a thermal Bose--Einstein occupation term.

= Matsubara sum
{c}
{synonym}


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 Incoming links: Bosonic Matsubara sum