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 τ, λ, 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 γ 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!