## Inverse Laplace Transform with a coinciding pole and branch point

Recently, the following Laplace transform was asked to be inverted:

$$F(s) = s^{-a-1} e^{-s^a}$$

where $a \in (0,1)$. This is a tough problem for two reasons. One is that there is very little chance of there being an analytical result for arbitrary values of $a$. The other, however, is more subtle: there is a coinciding pole and branch point. These two facts will collide to create a very difficult situation as we figure out just what is an acceptable end result. It turns out, however, that there is such a result: even though it is not in closed form, it at least may be evaluated using some numerical scheme if need be. Further, there is at least one special case that does yield a simple result (although that result is known to Mathematica and the tables).

The goal here is to address the subtlety in formulating a solution to such a problem. We will encounter divergences along the way, and we will have to know that the divergences must cancel if there is to be a solution. Thus, rather than be put off by the divergences, we must let them be a guide to the correct solution.

Because of the branch point, 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,

$$\frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \, s^{-a-1} e^{-s^a} e^{s t} = \frac{t^a}{\pi} \operatorname{Im}{\left [e^{-i \pi a}\int_{0}^{\infty} \frac{du}{u^{1+a}} e^{-u} \left ( e^{-e^{i \pi a} t^{-a} u^a} – 1 + e^{i \pi a} t^{-a} u^a \right ) \right]} + \frac{t^a}{\Gamma(1+a)} – 1$$

Actually, I have been able to go a little further with this, although I am not sure how practical it really will turn out to be. Basically, I expanded out the integrand and took the imaginary part. Then, keeping in mind Parseval’s theorem as I employ below, I split up this integrand and set up Fourier transforms of the individual pieces. Unfortunately, one of those pieces is the one-sided exponential of an arbitrary power, which does not have an analytical transform. The other piece, however, does and, further, has finite support. Thus, we may in fact use such a formulation to produce a simple numerical scheme for evaluating the integral for arbitrary $a$.

Now, even though the above result is suitable for numerical calculation, we can illustrate the above result with an analytical example that also illustrates my above reasoning with regard to Parseval. 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}

This is a known result, and the interested reader should play around with the behavior of this ILT near $t=0$. An exercise for the reader is to contemplate higher-order poles for the LT.