I am trying to solve a Laplace transform problem that has gotten way over my head in terms of complex analysis knowledge. I would like to solve the Inverse Laplace Transform $(s\rightarrow t)$ of
$$\frac{1}{s^{\alpha+1}}\exp(-s^{\alpha}),\qquad \alpha\in(0,1)$$
I have tried a few things so far and have run into the limits of my knowledge. This will involve an integral of the form
$$\frac{1}{2\pi i}\lim_{T\rightarrow\infty}\int_{c-iT}^{c+iT}\,ds\,\frac{1}{s^{\alpha+1}}\exp(-s^{\alpha}+st)$$
I feel like some kind of residue evaluation is the only way to approach this. Naively, to calculate the residue at $s=0$, I'd expand the exponential and look for the term proportional to $s^{-1}$ (there is only one such term), and the proportionality factor is my residue.
However, I am concerned about the branching induced by the fractional power of $\alpha$, and I am not sure how to treat that in my evaluation of the residue.
I also tried the substitution $u=s^{-\alpha}$, which leads to the integral
$$\int_{\gamma}\,du\,\exp(-\frac{1}{u}+t(\frac{1}{u})^{1/\alpha}),$$
where the contour integral is now around the circle of radus 1/2 centered at $u=1/2$. This doesn't help much with the branching issue, and I don't know how to deal with the essential singularity at the origin–is the contour just deformed to include (or not include) this singularity?
I have a bad feeling that the fractional power might make my desired integrals undefined. In that case I have tried some method of converting the Laplace transform of a function $LT[f(t)](s)$ to the Mellin transform $MT[f(t)](q)$ by the integral
$$MT[f(t)](q) = \frac{1}{\Gamma(1-q)}\int_{0}^{\infty}\,ds\, s^{-q}LT[f(t)](s),$$
though I am worried about convergence issues here, too. However, it might be the only way to proceed in order to get some reasonably tractable/numerically solvable analytic expression or integral.
Best Answer
The way to attack a problem like this is via Cauchy's theorem on a properly distorted Bromwich contour. Here, we want our contour to avoid the branch point at $z=0$. This, we consider
$$\oint_C dz \, z^{-a-1} e^{-z^a} e^{z t} $$
where $a \in (0,1)$ and $C$ is the following contour:
We will define $\text{Arg}{z} \in (-\pi,\pi]$, so the branch is the negative real axis. There are $6$ pieces to this contour, $C_k$, $k \in \{1,2,3,4,5,6\}$, as follows.
$C_1$ is the contour along the line $z \in [c-i R,c+i R]$ for some large value of $R$.
$C_2$ is the contour along a circular arc of radius $R$ from the top of $C_1$ to just above the negative real axis.
$C_3$ is the contour along a line just above the negative real axis between $[-R, -\epsilon]$ for some small $\epsilon$.
$C_4$ is the contour along a circular arc of radius $\epsilon$ about the origin.
$C_5$ is the contour along a line just below the negative real axis between $[-\epsilon,-R]$.
$C_6$ is the contour along the circular arc of radius $R$ from just below the negative real axis to the bottom of $C_1$.
When $t \gt 0$, the integral over the contours $C_2$ and $C_6$ vanish in the limit as $R \to \infty$.
The contour integral is thus equal to, in this limit,
$$\int_{c-i \infty}^{c+i \infty} ds \, s^{-a-1} e^{-s^a} e^{s t} + e^{-i \pi a} \int_{\infty}^{\epsilon} dx \, x^{-a-1} e^{-e^{i \pi a} x^a} e^{-x t} \\ + i \epsilon^{-a} \int_{\pi}^{-\pi} d\phi \, e^{-i a \phi} e^{-\epsilon^a e^{i a \phi}} e^{\epsilon t e^{i \phi}} + e^{i a \pi} \int_{\epsilon}^{\infty} dx \, x^{-a-1} e^{-e^{-i \pi a} x^a} e^{-x t}$$
Note that there is an apparent singularity at $\epsilon = 0$; however, the divergences cancel in the limit as $\epsilon \to 0$.
In this limit, the third integral has the following leading behavior:
$$-i \frac{2}{a} \epsilon^{-a} \sin{\pi a} +i 2 \pi$$
Rescaling and combining the second and fourth integrals, we get for the contour integral:
$$\int_{c-i \infty}^{c+i \infty} ds \, s^{-a-1} e^{-s^a} e^{s t} -i 2 t^a \operatorname{Im}{\left [e^{-i \pi a}\int_{\epsilon t}^{\infty} \frac{du}{u^{1+a}} e^{-u} e^{-e^{i \pi a} t^{-a} u^a} \right]}-i \frac{2}{a} \epsilon^{-a} \sin{\pi a} +i 2 \pi $$
We may Taylor expand the second exponential in the integrand because it is subdominant to the first exponential (at least for the first $n$ terms, where $n$ is that largest integer such that $\lfloor n a \rfloor = 0$). We need only expand to the first two terms to treat the limit as $\epsilon \to 0$. Note that
$$\int_{\epsilon t}^{\infty} \frac{du}{u^{1+a}} e^{-u} = \frac{t^{-a}}{a} \epsilon^{-a} + \Gamma(-a) + O \left ( \epsilon^a \right ) $$
The second term produces
$$-e^{i \pi a} t^{-a}\int_{\epsilon t}^{\infty} \frac{du}{u} e^{-u} $$
Because the exponentials outside the integral cancel, the imaginary part of the term is zero. Thus, we now take the limit as $\epsilon \to 0$. Because the contour integral is zero by Cauchy's theorem, we get for the ILT,
ADDENDUM
Even though the above result is suitable for numerical calculation, we can illustrate the above result with an analytical example. Consider the case $a=1/2$. Subbing $u=x^2$ and taking the imaginary part of the integral, we end up with
$$\operatorname{Im}{\left [e^{-i \pi a}\int_{0}^{\infty} \frac{du}{u^{1+a}} e^{-u} \left ( e^{-e^{i \pi a} t^{-a} u^a} - 1 + e^{i \pi a} t^{-a} u^a \right ) \right]} = 2 \int_{-\infty}^{\infty} dx \, e^{-x^2} \frac{\sin^2{\beta x}}{x^2}$$
where $\beta = 1/(2 \sqrt{t})$.
The latter integral may be evaluated using Parseval's theorem, because the individual factors of the integrand are inverse Fourier transforms of simple functions. For example,
$$\int_{-\infty}^{\infty} dx \, e^{-x^2} e^{i k x} = \sqrt{\pi} e^{-k^2/4} $$ $$\int_{-\infty}^{\infty} dx \, \frac{\sin^2{\beta x}}{x^2} e^{i k x} =\begin{cases} \pi \beta \left ( 1-\frac{|k|}{2 \beta} \right ) & |k| \lt 2 \beta \\ 0 & |k| \gt 2 \beta \end{cases}$$
The integral is then equal to
$$2 \frac1{2 \pi} \sqrt{\pi} \pi \beta \int_{-2 \beta}^{2 \beta} dk \, \left ( 1-\frac{|k|}{2 \beta} \right ) e^{-k^2/4} = \sqrt{\pi} \beta \int_0^{2 \beta} dk \, \left ( 1-\frac{k}{2 \beta} \right ) e^{-k^2/4}$$
The evaluation is fairly straightforward using the definition of the error function. The result is, for the integral,
$$2 \pi \beta \operatorname{erf}{\beta} - 2 \sqrt{\pi} \left (1-e^{-\beta^2}\right ) $$
Now plugging this back into the main result above and using $\beta = 1/(2 \sqrt{t})$, we get that
$$\begin{align}\frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \, s^{-3/2} e^{-s^{1/2}} e^{s t} &= \operatorname{erf}{\left ( \frac1{2 \sqrt{t}} \right )} + \frac{2}{\sqrt{\pi}} \sqrt{t} e^{-\frac1{4 t}} - 1 \\ &= \frac{2}{\sqrt{\pi}} \sqrt{t} e^{-\frac1{4 t}} - \operatorname{erfc}{\left ( \frac1{2 \sqrt{t}} \right )}\end{align}$$