Finding Inverse Laplace Transform $\exp\left(-\sqrt{\frac{s}{a}}\right)$

Question

Can somebody help me to find the inverse Laplace transform of $$F(s)\exp\left(-\sqrt{\frac{s}{a}}\right)$$ or at least $$\exp\left(-\sqrt{\frac{s}{a}}\right)?$$

I'll be so grateful.

Answer 1

Here is what Wolfram Alpha gives (I couldn't think of a way to do this by hand since inverse Laplace transforms are highly nontrivial). You could then use the convolution theorem for the Laplace transform to write the general expression.

Answer 2

According to the definition of Error function, we can find the following strong identity: $$\mathcal{L}\left(\frac{k}{2\sqrt{\pi~t^3}}e^{-k^2/4t}\right)=e^{-k\sqrt{s}},~~k>0$$ It seems that here, we have $$k=\frac{1}{\sqrt{a}}$$

Answer 3

Let's work with the exponential only first; I will assume that $a>0$ for simplicity. You can use a contour integration without that substitution as follows by deforming the Bromwich contour about the negative real axis and exploiting a branch cut of $\sqrt{z/a}$ about that axis. So, consider the integral

$$\oint_C dz \: e^{-\sqrt{z/a}} e^{z t}$$

where $C$ is a keyhole contour about the negative real axis, as pictured below.

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 $C_1$.

We will show that the integral along $C_2$,$C_4$, and $C_6$ vanish in the limits of $R \rightarrow \infty$ and $\epsilon \rightarrow 0$.

On $C_2$, the real part of the argument of the exponential is

$$R t \cos{\theta} - \sqrt{R/a} \cos{\frac{\theta}{2}}$$

where $\theta \in [\pi/2,\pi)$. Clearly, $\cos{\theta} < 0$ and $\cos{\frac{\theta}{2}} > 0$, so that the integrand exponentially decays as $R \rightarrow \infty$ and therefore the integral vanishes along $C_2$.

On $C_6$, we have the same thing, but now $\theta \in (-\pi,-\pi/2]$. This means that, due to the evenness of cosine, the integrand exponentially decays again as $R \rightarrow \infty$ and therefore the integral also vanishes along $C_6$.

On $C_4$, the integral vanishes as $\epsilon$ in the limit $\epsilon \rightarrow 0$. Thus, we are left with the following by Cauchy's integral theorem (i.e., no poles inside $C$):

$$\left [ \int_{C_1} + \int_{C_3} + \int_{C_5}\right] dz \: e^{-\sqrt{z/a}} e^{z t} = 0$$

On $C_3$, we parametrize by $z=e^{i \pi} x$ and the integral along $C_3$ becomes

$$\int_{C_3} dz \: e^{-\sqrt{z/a}} e^{z t} = e^{i \pi} \int_{\infty}^0 dx \: e^{-i \sqrt{x/a}} e^{-x t}$$

On $C_5$, however, we parametrize by $z=e^{-i \pi} x$ and the integral along $C_5$ becomes

$$\int_{C_5} dz \: e^{-\sqrt{z/a}} e^{z t} = e^{-i \pi} \int_0^{\infty} dx \: e^{i \sqrt{x/a}} e^{-x t}$$

We may now write

$$-\frac{1}{i 2 \pi} \int_0^{\infty} dx \: e^{- x t} \left ( e^{i \sqrt{x/a}} - e^{-i \sqrt{x/a}} \right ) + \frac{1}{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \: e^{-\sqrt{s/a}} e^{s t} = 0$$

Therefore, the ILT of $\hat{f}(s) = e^{-\sqrt{s/a}}$ is given by

$$\begin{align}\frac{1}{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \: e^{-\sqrt{s/a}} e^{s t} &= \frac{1}{i 2 \pi} \int_0^{\infty} dx \: e^{- x t} \left ( e^{i \sqrt{x/a}} - e^{-i \sqrt{x/a}} \right )\\ &= \frac{a}{\pi} \int_{-\infty}^{\infty} du\: u \,e^{-a t u^2} \sin{u}\end{align}$$

The last step involved substituting $x=a u^2$ and exploiting the evenness of the integrand. This integral may be evaluated as follows:

$$\begin{align}\frac{a}{\pi} \int_{-\infty}^{\infty} du\: u \,e^{-a t u^2} \sin{u} &= \frac{a}{\pi} \Im{\left [\int_{-\infty}^{\infty} du\:u\, e^{-a t u^2} e^{i u} \right]}\\ &= \frac{a}{\pi} \Im{\left [\int_{-\infty}^{\infty} du\:u\, e^{-a t (u-i/(2 a t))^2} e^{-1/(4 a t)}\right ]}\\ &= \frac{a}{\pi} e^{-1/(4 a t)} \Im{\left [\int_{-\infty}^{\infty} dv \: \left ( v + \frac{i}{2 a t} \right ) e^{-a t v^2} \right]}\\ &= \frac{a}{\pi} e^{-1/(4 a t)} \frac{1}{2 a t} \sqrt{\frac{\pi}{ at}} \end{align}$$

Therefore the result is that

$$\mathcal{L}^{-1}[e^{-\sqrt{z/a}}](t) = \frac{1}{i 2 \pi} \int_{c-i \infty}^{c+i \infty} dz \: e^{-\sqrt{z/a}} e^{z t} = \frac{1}{2 \sqrt{\pi a}} t^{-3/2} e^{-\frac{1}{4 a t}}$$

Now, to get the ILT of $F(s) e^{-\sqrt{z/a}}$, you may use the convolution theorem for Laplace transforms:

$$\mathcal{L}^{-1}[F(s) e^{-\sqrt{z/a}}](t) = \frac{1}{2 \sqrt{\pi a}} \int_0^t dt' \, f(t') \, (t-t')^{-3/2}\, e^{-\frac{1}{4 a (t-t')}}$$

where $f$ is the ILT of $F$.