I want to calculate the ILT of this function with the residue theorem: $$F(s) = \frac{\cosh{(\alpha \sqrt{s})} \operatorname{csch}(\beta \sqrt{s})}{\sqrt{s}(s-\theta)}.$$
The branch cut is along the negative real axis, but I didn't get the right result. Would you like to help me with this issue? Thanks a lot.
Inasmuch as $\frac{\cosh(\alpha \sqrt s)\text{csch}(\beta \sqrt s)}{\sqrt{s}}$ is an even function of $\sqrt{s}$, there is no branch cut required. There are, however, poles at $s=\theta$, $s=0$, and $s=-n^2\pi^2/\beta^2$ for all integers $|n|\ge 1$.
Therefore, the Residue Theorem guarantees that
$$\begin{align} \mathscr{L}^{-1}\{F\}(t)&=\frac1{2\pi i}\int_{\sigma-i\infty}^{\sigma+i\infty}\frac{\cosh(\alpha \sqrt s)\text{csch}(\beta \sqrt s)}{\sqrt{s}(s-\theta)}e^{st}\,ds\\\\ &=\underbrace{\frac{\cosh(\alpha \sqrt \theta)\text{csch}(\beta \sqrt \theta)}{\sqrt{\theta}}\,\,e^{\theta t}}_{\text{Residue at }\theta}+\underbrace{\frac1{-\theta \beta}}_{\text{Residue at }0}\\\\ &+\sum_{|n|\ge1} \text{Res}\left(\frac{\cosh(\alpha \sqrt s)\text{csch}(\beta \sqrt s)}{\sqrt{s}(s-\theta)}e^{st},s=-n^2\pi^2/\beta^2\right)\\\\ \end{align}$$
Can you finish now?