How to establish the identity of the infinite sum

You may notice that $\int_{0}^{+\infty}\frac{\sin(ax)}{a}e^{-bx}\,dx = \frac{1}{a^2+b^2}$, replace $b$ with $z+n$ and sum over $n\geq 0$ to get an integral that is easy to compute through the residue theorem. An alternative is to apply the Poisson summation formula.

The trick here is to use $$\frac{1}{(z+w)^2+a^2} \times \pi\cot(\pi w)$$ as shown at the following MSE link. The sum term is also quadratric in $n$ so the estimates of the integrals presented there apply to the present case as well.

We get for the residues at $w = -z \pm ia$ the closed form

$$\left.\frac{1}{2(z+w)} \pi\cot(\pi w)\right|_{w=-z\pm ia}.$$

Recall that

$$\cot(v) = i \frac{\exp(iv)+\exp(-iv)}{\exp(iv)-\exp(-iv)}.$$

Introducing $x=\exp(\pi i z)$ and $y=\exp(\pi a)$ we get for the two residues

$$\frac{\pi}{2a} \frac{1/x/y+xy}{1/x/y-xy} - \frac{\pi}{2a} \frac{y/x+x/y}{y/x-x/y} = \frac{\pi}{2a} \left(\frac{1+x^2y^2}{1-x^2y^2} - \frac{y^2+x^2}{y^2-x^2}\right) \\ = \frac{\pi}{2a} \frac{y^2+x^2y^4-x^2-x^4y^2-y^2+x^2y^4-x^2+x^4y^2} {(1-x^2y^2)(y^2-x^2)} \\ = \frac{\pi}{2a} \frac{2x^2y^4-2x^2}{(1-x^2y^2)(y^2-x^2)} = \frac{\pi}{a} \frac{y^2-1/y^2}{(y^2-x^2-x^2y^4+x^4y^2)/y^2/x^2} \\ = \frac{\pi}{a} \frac{y^2-1/y^2}{1/x^2 - 1/y^2 - y^2 + x^2}.$$

Flip the sign to get

$$\bbox[5px,border:2px solid #00A000]{ \frac{\pi}{a} \frac{\sinh(2\pi a)}{\cosh(2\pi a)-\cos(2\pi z)}.}$$

Observe that when $z = q \mp ia$ with $q$ an integer we have

$$\cos(2\pi z) = \cos(2\pi q \mp 2\pi i a) = \cosh(2\pi a)$$

and the formula becomes singular. This is correct however since in this case the sum term $$\frac{1}{(z+n)^2+a^2}$$ is singular as well namely when $n = -q$ and the sum is undefined.