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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.

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1 Answer 1

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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:

pic

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,

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

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

$$\begin{align}\frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \, s^{-3/2} e^{-s^{1/2}} e^{s t} &= \operatorname{erf}{\left ( \frac1{2 \sqrt{t}} \right )} + \frac{2}{\sqrt{\pi}} \sqrt{t} e^{-\frac1{4 t}} - 1 \\ &= \frac{2}{\sqrt{\pi}} \sqrt{t} e^{-\frac1{4 t}} - \operatorname{erfc}{\left ( \frac1{2 \sqrt{t}} \right )}\end{align}$$

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  • $\begingroup$ May I ask how you made that awesome diagram? $\endgroup$
    – Braindead
    Mar 4, 2015 at 14:30
  • $\begingroup$ @Braindead: Good question...it's been a while and I've reused it so many times I've lost count. I think in Microsoft Visio, although since then I have made much better diagrams in Mathematica. $\endgroup$
    – Ron Gordon
    Mar 4, 2015 at 16:09
  • $\begingroup$ Thanks very much for your very detailed and helpful answer. I will be studying it carefully. Apologies for my late acknowledgement of your answer! $\endgroup$
    – A. Kennard
    Mar 22, 2015 at 20:08
  • $\begingroup$ Can you explain why the integration of the contour C2 and C6 is zero? Jordan's lemma? it seems that the integrand is not zero when $z\to \infty$. I don't know how to treat the term $e^{-z^a}$ when $z\to \infty$, what is its value? $\endgroup$ Jun 1, 2015 at 13:10
  • $\begingroup$ @buzhidao: keep in mind that $a \in (0,1)$ so that $e^{-z^a}$ varies more slowly than $e^{t z}$. $e^{z t}$ vanishes more rapidly on $C_2$ and $C_6$ than $e^{-z^a}$ could ever increase. $\endgroup$
    – Ron Gordon
    Jun 1, 2015 at 13:13

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