Inverse Laplace transform of exponentials and Incomplete gamma functions I came to this final problem to be solved. I would like to understand a way to tackle this problem:
Inverse Laplace transform of
$$A(s)=\frac{1}{s}\exp{(s^{\beta}z)}\Gamma(0,s^{\beta}z)$$
$$B(s)=\frac{1}{s}-\frac{s^{\beta}}{s}\exp{(s^{\beta}z)}\Gamma(0,s^{\beta}z)$$
where $z$ is a complex number and $\beta<1$ a real positive number.
I cannot find an easy way of finding an analytical solution, if it exists.
 A: \begin{equation}
A(s)=\frac{1}{s} e^{s^\beta z} \Gamma(0, s^\beta z)= \frac{1}{s} \int_0^{+\infty} \frac{e^{-u}}{s^\beta z+u} du
\end{equation}
Now we can write:
\begin{equation}
e^{-u}= \sum_{k=0}^{+\infty} \frac{(-u)^k}{k!}
\end{equation}
Then
\begin{equation}
A(s)=\frac{1}{s} \int_0^{+\infty}\sum_{k=0}^{+\infty} \frac{(-u)^k}{k!(s^\beta z+u)} du=\frac{1}{s} \sum_{k=0}^{+\infty} \frac{1}{k!}\int_0^{+\infty} \frac{(-u)^k}{s^\beta z+u} du=\sum_{k=0}^{+\infty} \frac{(-s^\beta z)^k}{s k!} \int_{0}^{+\infty} \frac{x^k}{1+x} dx
\end{equation}
By using the inverse Laplace transform
\begin{equation}
a(t)=L^{-1}\{A(s)\}=\sum_{k=0}^{+\infty}\frac{t^{-k\beta} (-z)^k}{k! \Gamma(1-k\beta)} \int_{0}^{+\infty} \frac{x^k}{1+x} dx
\end{equation}
As Sary pointed out, in my previous calculation I did a mistake.The integral converge to $\frac{-\pi}{\sin{\pi k}}$ is and only if the $k$ is a real number ranging in the interval (-1,0). In my previous computation I choosen a rectangular contour such that $u= u_r+i u_i$ and I choosen $u_i$ to include the pole.Unfortunately when doing that the integral along the side of the rectangle vanishes converge only if we assume that $k$ ranges between -1 and 0. Thank you Sary.
\begin{equation}
b(t)=L^{-1}\{B(s)\}=1-\frac{t^{-(1+\beta)}}{\Gamma(-\beta)} \star a(t)
\end{equation}
where $\star$ is the convolution. Now can we say more about the sum? I'm not familiar with the Hankel contour path so maybe this is the right way to find a compact solution.
A: The following is not really an answer but was too long for a comment ; your function is messy and I must admit I don't really want to go into details. But you could theoretically proceed as follows. Start from the same expression Upak gave :
$$ A(s) = \frac1s \int_0^\infty \frac{e^{-u}{\rm d}u}{s^\beta z + u}. $$
Assume at first to simplify that $z$ is not a non-positive real (so that the denominator above never vanishes for $\Re(s)>0$.) You have that $A(s)$ stays bounded for, say, $\Re(s) \geq 1$. The inverse Laplace transform is, for $\sigma \geq 1$ (say) and $t\in{\bf R}$,
$$ a(t) = \frac1{2\pi i}\int_{\sigma - i\infty}^{\sigma + i\infty} A(s) e^{st}{\rm d} s = \frac1{2\pi i}\int_{\sigma - i\infty}^{\sigma + i\infty} \int_0^\infty \frac{e^{-u}{\rm d}u}{s^\beta z + u} \frac{e^{st}{\rm d}u}{s}$$
Here you can interchange integrals. You need to compute
$$ W(t, u) = \frac1{2\pi i}\int_{\sigma - i\infty}^{\sigma + i\infty}\frac{e^{st}{\rm d}s}{s(s^\beta z + u)} $$
This is done by shifting the contours to the left or the right, depending on the sign of $t$ (to the right when $t<0$, to the left when $t>0$, so that you use the decay of $e^{st}$). Here you may have a pole when $s^\beta z + u = 0$ (or not, depending on $u$ and $z$ !). But more importantly you have a branch cut at $s=0$. So that upon shifting the contour to the left, you should go towards a contour of this shape (a Hankel contour)

See also that PDF file on the $\Gamma$ function. What you end up with is another integral expression for $W(t, u)$. You may think that this serves no purpose, but the resulting integral may depend in $u$ and $t$ in a nicer way, giving you insight on asymptotic behaviour for instance.
A reference on singularity analysis is chapter VI of late Flajolet and Sedgewick's "Analytic Combinatorics", but they work things out specifically with the asymptotic analysis of combinatorial objects in mind.
