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:

  1. 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.
  2. 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.
  • 3
    $\begingroup$ you are closing the integral in the negative half plane, so you have assumed $t<0$ and then your sum over $j$ converges; if $t>0$, you will close the integral in the positive half plane, resulting in zero. $\endgroup$ – Carlo Beenakker Sep 27 '13 at 7:34

Carlo Beenakker's answer is right. When t>0, you cannot close the contour in the lower half-plane, because the exp in the numerator is large in the lower half-plane. You must close the contour in the upper half-plane. Then your integral is 0, for $t>0$. When $t<0$ your series computed with residues converges.

  • $\begingroup$ I disagree (or I'm wrong and am confused): the origin is not a pole because $$\lim_{x\to0} \hat F(x) = \frac{-i x e^{i x t}}{1-\sqrt{-i x} \coth(\sqrt{-i x})} = 3$$ Furthermore, in the Taylor series expansion of $\hat F(x)$, the root in the denominator at the origin cancels. $\endgroup$ – DieLuftDerFreiheit Sep 27 '13 at 19:35
  • $\begingroup$ I agree that the contour should be closed in the upper half plane. I now realize that there is a pole at infinity of order 1/2. How do I account for this pole at infinity when summing residues? $\endgroup$ – DieLuftDerFreiheit Sep 27 '13 at 19:36
  • $\begingroup$ Sorry, 0 is indeed removable. Therefore, the integral is 0. $\endgroup$ – Alexandre Eremenko Sep 28 '13 at 2:10
  • $\begingroup$ There is no "pole at infinity". Infinity is an essential singularity. But the function under the integral decreases sufficiently fast in the upper half-plane to close the contour. $\endgroup$ – Alexandre Eremenko Sep 28 '13 at 2:11
  • $\begingroup$ I'm a bit confused: for $t>0$ the integrand is indeed zero, but why would it be zero for $t<0$? Then the contour has to be closed in the lower half-plane, resulting in this sum over residues $r_j$ (which converges for $t<0$). $\endgroup$ – Carlo Beenakker Sep 28 '13 at 13:48

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