complex-analysiscontour-integrationlaplace transform
I want to find the inverse Laplace transform of
$$F(s)=\frac{π\cosh(\sqrt s)}{2 s^{3/2} \sinh(\sqrt s)}$$
using the Bromwich integral
$$f(t)= \frac{1}{2\pi i} \int_{c-i\infty}^{c+i\infty} \frac{π\cosh(\sqrt s)}{2 s^{3/2} \sinh(\sqrt s)}e^{st}\, ds$$
The residue at the pole $s=-(n\pi)^2$ is
$$\frac{-1}{\pi}\sum_{n=1}^{\infty}\frac{1}{n^2}e^{-(n\pi)^2t}$$
My progress so far has been stunted by the fact that we have a branch point at $s=0$. Any help is appreciated.
The result below assumes $t>0$ (as usual for Laplace transforms).
In the Bromwich contour $\gamma$ has to be chosen large enough that it is to the right of all singularities (poles and branch points) so $\gamma = 0^+$ is perfectly valid. The singularities of $\log s/(1+s)$ are a pole at $s=-1$ and two branch points at $s=0$ and $s=\infty$ which we connect via a branch cut along the negative real line.
Then we can deform the contour further to a path which starts at $-\infty -i 0^+$. Runs along the negative real line just below the branch cut. Ends at $0-i 0^+$ in a little semi-circle and then runs back from $0+i 0^+$ to $-\infty +i0^+$ just above the branch cut.
The Bromwich integral thus is given by $$f(t)=\frac1{2\pi i} \int_{-\infty}^0\!dx\, \left( \frac{\log (x-i0^+)}{1+x-i0^+ } - \frac{\log (x+i0^+)}{1+x+i0^+ } \right) e^{x t} $$ as the small circle around the branch point at $0$ does not contribute ($|z|\log z \to 0$ for $|z|\to0$).
In the remaining integral, we use $\log(x \pm i 0^+) = \log |x| \pm i \pi$ valid for $x<0$: $$\begin{align} f(t) &= \frac1{2\pi i} \int_{-\infty}^0\!dx\, \left( \frac{\log |x|-i\pi}{1+x-i0^+ } - \frac{\log |x|+i\pi}{1+x+i0^+} \right) e^{x t}\\ &= \frac1{2\pi i} \overbrace{\int_{-\infty}^0\!dx\, \log |x| \underbrace{\left( \frac1{1+x-i0^+ } - \frac1{1+x+i0^+}\right)}_{2\pi i \delta(x+1)} e^{x t}}^{=0}\\ &\quad -\frac12 \int_{-\infty}^0\!dx \underbrace{\left( \frac1{1+x-i0^+ } + \frac1{1+x+i0^+}\right)}_{2\mathcal{P}\,(1+x)^{-1}} e^{x t} \\ &= -\int_{-\infty}^0\!dx \,\mathcal{P} \frac{e^{x t}}{1+x} =- e^{-t} \int_{-\infty}^t\!ds \,\mathcal{P} \frac{e^{s}}{s}\\ &=- e^{-t} \mathop{\rm Ei}(t) \end{align}$$ with $s=(1+x)t$ and Ei the exponential integral.
There is no pole at $z=-1$; it is merely a branch point. The Bromwich contour from which the ILT may be found must be deformed so as to avoid this branch point, like this:
You may show that the integrals over $C_2$, $C_4$, and $C_6$ all vanish. The result is, letting $z=-1+e^{i \pi} u$ on $C_3$ and $z=-1+e^{-i \pi} u$ on $C_5$,
$$\int_{c-i \infty}^{c+i \infty} ds \frac{e^{s t}}{s \sqrt{1+s}} + e^{-i \pi/2} \int_{\infty}^0 du \frac{e^{-(1+u) t}}{(1+u) \sqrt{u}} \\ + e^{i \pi/2} \int_0^{\infty} du \frac{e^{-(1+u) t}}{(1+u) \sqrt{u}} = i 2 \pi$$
as the residue at the pole $z=0$ is $1$.
From this, you may rearrange to get that the ILT is
$$\frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \frac{e^{s t}}{s \sqrt{1+s}} = 1-\frac1{\pi} \int_0^{\infty} du \frac{e^{-(1+u) t}}{(1+u) \sqrt{u}} $$
The integral may be evaluated by differentiating with respect to $t$ and subbing $u=v^2$. The result, which I leave to the reader, is
$$\frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \frac{e^{s t}}{s \sqrt{1+s}} = \operatorname{erf}{\sqrt{t}} $$
ADDENDUM
A little more detail on the evaluation of the integral on the RHS above. Let
$$I(t) = \int_0^{\infty} du \frac{e^{-(1+u) t}}{(1+u) \sqrt{u}} = \int_{-\infty}^{\infty} dv \frac{e^{-t (1+v^2)}}{1+v^2}$$
Then
$$I'(t) = -\int_{-\infty}^{\infty} dv\, e^{-t (1+v^2)} = \sqrt{\pi} t^{-1/2} e^{-t} $$
$$\implies I(t) = I(0) - \sqrt{\pi} \int_0^t dt' \, t'^{-1/2} e^{-t'} = \pi - 2 \sqrt{\pi} \int_0^{\sqrt{t}} du \, e^{-u^2} = \pi - \pi \, \operatorname{erf}{\sqrt{t}}$$
The result follows.
Best Answer
Inasmuch as the function $F(s)=\frac{\pi\cosh(\sqrt s)}{2s^{3/2}\sinh(\sqrt s)}$ is an even function of $\sqrt s$, there is no branch point at $s=0$. Rather, there is a second order pole at $s=0$.
Hence, we have for $t>0$
$$\begin{align} \mathscr{L}^{-1}\{F\}(t)&=\frac1{2\pi i}\int_{c-i\infty}^{c+i\infty}F(s)e^{st}\,ds\\\\ &=\text{Res}\left(F(s)e^{st}, s=0\right)+\sum_{n=1}^\infty\text{Res}\left(F(s)e^{st}, s=-n^2\pi^2\right) \end{align}$$
For the residue at $s=0$ we have for $t>0$
$$\text{Res}\left(F(s)e^{st}, s=0\right)=\frac\pi2\,\lim_{s\to 0}\frac{d}{ds}\left(\sqrt s\coth(\sqrt s)e^{st}\right)=\frac\pi6+\frac\pi2 t$$
For the residues at $-n^2\pi^2$, $n\ne0$, we have for $t>0$
$$\text{Res}\left(F(s)e^{st}, s=-n^2\pi^2\right)=\lim_{s\to -n^2\pi^2}(s+n^2\pi^2)\frac{\pi \cosh(\sqrt s)}{2s^{3/2}\sinh(\sqrt s)}\,e^{st}=-\frac{e^{-n^2\pi^2t}}{n^2\pi}$$
Hence, we find that
$$\mathscr{L}^{-1}\{F\}(t)=\frac\pi6+\frac{\pi}{2}t-\sum_{n=1}^\infty \frac{e^{-n^2\pi^2t}}{n^2\pi}$$