Find the first term in the asymptotic expansion of $$f(t) = \frac1{2 \pi i}\int_{\alpha - i \infty}^{\alpha + i \infty} e^{tz} z^{-1} \tanh(z)dz$$ as $t \to \infty$
I am finding this problem difficult, as there are an infinite number of (simple) poles from $g(z) = z^{-1} \tanh(z)$, so that determining the way to deform the contour is unclear.
The poles of $g$ are at $z_n = \frac{\pi i}2 + n \pi i$ for $n\in \mathbb{Z}$, and the residues are $R_n = \text{Res}_{z = z_n} g(z) = \frac{-2i}{(2n+1)\pi}$
I figured I needed to determine which of these poles would have the largest asymptotic contribution to the integral, at which point I could expand $g$ around this pole as:
$$g(z) = \frac{R_n}{z - z_n} + O(1)$$
Which has a single simple pole at $z = z_n$. Hence, I could deform the contour into a circle around this pole, and from the Residue Theorem, I'd have that
$$f(t) \sim R_n e^{z_nt} \ \ \text{ as } \ \ t \to \infty$$
From this, it would appear that I should choose the $z_n$ such that $e^{z_nt}$ is asymptotically largest, however, such a $z_n$ doesn't exist, as $\lim_{n \to \infty} z_n = \infty$
So it would appear there is a flaw in my argument that I am not seeing. Any help is appreciated.
To check convergence of this integral, I found that $$|e^{tz}z^{-1} \tanh(z)| = \frac{e^{xt}(e^{4x} - 2e^{2x}\cos(2y) + 1)}{(x^2 + y^2)(e^{4x} + 2e^{2x}\cos(2y) + 1)}$$
And since $x$ is fixed on the contour from $\alpha - i \infty$ to $\alpha + i\infty$, the integral will converge for any $\alpha$.
We can deform the contour through the left half-plane, requiring $\alpha > 0$ for the method I used above.
All poles are located on the imaginary axis, thus the sum over the residues $$f(t) = -\frac {2i} \pi \sum_{k = -\infty}^\infty \frac {e^{(2k + 1)i \pi t/2}} {2k + 1}$$ is convergent by Dirichlet's test. This shows that $f(t + 4) = f(t)$; I'm not sure that the question about the asymptotic of $f$ is meaningful. A closed form for $f$ is $$f(t) = \operatorname{sgn} \sin \frac {\pi t} 2.$$