I have what I think is a relatively simple contour integral involving arctan, but it is giving me difficulty. I would really appreciate any help.

The integral itself is, with $\tau$, $\lambda$, and $k$ all real and positive,

$\frac{1}{2 \pi i} \int_{\gamma - i \infty}^{\gamma + i \infty} e^{s t} \frac{1 + \tau s}{\tau k^{2}} \Bigg\{ 1 + \tau s - \tau s \Bigg[ 1 - \frac{1}{\lambda k } \arctan\bigg( \frac{\lambda k}{1 + \tau s} \bigg) \Bigg]^{-1} \Bigg\} \, ds.$

This is, of course, just an inverse Laplace transform (Bromwich integral) with $\gamma$ chosen so that all singularities lie to its left. The only singularity I see is at

$s = \frac{1}{\tau} \bigg( \frac{\lambda k}{\tan(\lambda k)} - 1 \bigg)$.

From this, it seems that the contour integral should be equivalent to applying the residue theorem with a closed arc at infinity and ($\gamma - i \infty$, $\gamma + i \infty$). But this gives me results that don't make sense.

Is this because, depending on the parameters $k$ and $\lambda$, the arc can cross the imaginary axis (at which arctan has a branch cut)? Do I need to do some fancy contour to avoid this?

Again, thanks!