The Laplace transform of a function $\phi(t)$ can be written as $$\hat{\phi}(s) = \cosh (2x\sqrt{s})/(\sqrt{s}\sinh\sqrt{s}).$$ Determine $\phi(t)$ where $x$ is real.
So I know that $$\cosh(at) = \frac{s}{s^2 - a^2}$$ and I know that $$\sinh(at) = \frac{a}{s^2 - a^2}$$ but I'm stuck on how to tet to that point where I can take the inverse Laplace function and determine my answer.
The inverse Laplace Transform, $\phi(t)$, is given by the Bromwich integral
$$\begin{align} \phi(t)&=\frac1{2\pi i }\int_{c-i\infty}^{c+i\infty}\frac{\cosh(2x\sqrt s)}{\sqrt s\sinh(\sqrt{s})}e^{st}\,ds\\\\ \end{align}$$
The integrand has no branch point singularities since $\frac{\cosh(2xz)}{z\sinh(z)}$ is an even function of $z$. There are simple poles of the integrand at $s=0$ and $s=-n^2\pi^2$. Then, we can write for $t>0$
$$\begin{align} \phi(t)&=\sum_{n=0 }^\infty \text{Res}\left(\frac{\cosh(2x\sqrt s)}{\sqrt s\sinh(\sqrt{s})}e^{st}, s=-n^2\pi^2\right)\\\\ &=\sum_{n=0 }^\infty\lim_{s\to -n^2\pi^2}\left((s+n^2\pi^2)\frac{\cosh(2x\sqrt s)}{\sqrt s\sinh(\sqrt{s})}e^{st}\right)\\\\ &=\sum_{n=0 }^\infty (-1)^n2\cosh(i2xn\pi)e^{-n^2\pi^2t}\\\\ &=2\sum_{n=0}^\infty (-1)^n \cos(2n\pi x)e^{-n^2\pi^2 t} \end{align}$$