Mathoverflow,

I'd like to carry out the following integral,

$$f(t) = \int_{- \infty}^{\infty}\frac{-i\Omega e^{i \Omega t}}{1-\sqrt{-i\Omega}\coth(\sqrt{-i\Omega})} d\Omega.$$

Here's what I've done so far. First, there is a removable singularity at the origin that is of no import. Second, there are infinitely many poles along the negative imaginary axis. These singularities are simple poles of order one and correspond to roots of the denominator of the integrand. We label these roots $\Omega_j=-ir_j^2$ with $r_j>0$. The locations of the simple poles are therefore obtained by solving for the $r_j$ such that $\tan(r_j) + r_j = 0$.

I take the contour $\gamma$ to be in the negative half plane, including the real axis and around all of these singularities.

We evaluate the residue of $f(\sqrt{\Omega})$ as $p/q^{\prime}$ where $p/q \equiv f$. This results in $$ \text{Res}\left(f,\Omega_j\right) = \frac{-2 i (-i \Omega_j)^{3/2} e^{i\Omega_j t}}{\coth(\sqrt{-i \Omega_j})-\sqrt{-i \Omega_j}\text{csch}^2(\sqrt{-i \Omega_j})}$$ $$= \frac{-2 i (-i r_j)^{3} e^{r_j^2 t}}{i\cot r_j-ir_j\csc^2r_j}$$ $$= -2i\frac{ r_j^{3} e^{r_j^2 t}}{\cot r_j-r_j\csc^2r_j}$$ $$= 2i\frac{ r_j^{3} e^{r_j^2 t}}{1/r_j - r_j - 1/r_j}$$ $$= -2i r_j^2 e^{r_j^2 t}$$

This gives the integral as,

$$f(t) = 4\pi\sum_{j=0}^{\infty} r_j^2 e^{r_j^2 t}$$

My troubles are:

- This series does not converge. I would like the exponential to have a negative power, but I need the integral to be in the negative half plane since this is where all of the singularities are. I'm not sure how to resolve this.
- In the frequency domain, I take the limit as $\Omega\to0$ and find the integrand to be $-3$, suggesting the inverse to be $f(t) = -3\delta(t)$. I'm not able to recover this limit.