Laplace inverse of $\frac{e^{-\sqrt{s+2}}}{s}$

Question

I want to find out $$\mathcal{L^{-1}}\{\frac{e^{-\sqrt{s+2}}}{s}\}$$ How do you find the inverse Laplace?

thanks

Answer 1

Thank you for the interesting question. Here is a rather brute force solution, which may add a few steps to Jan Erland's solution.

First, let us recall that, if $$ \mathcal L(f) = \int_0^\infty f(t) \, e^{-st} \, dt = F(s) $$ Then $$ \mathcal L(f') = \int_0^\infty f'(t) \, e^{-st} \, dt = s \, F(s) - f(0). $$ In our case, $F(s) = e^{-\sqrt{s+2}}/s$, so, we shall seek a function $f'(t)$ whose Laplace transform is $sF(s) = e^{-\sqrt{s+2}}$. Then $f(t)$ can be found from integrating $f'(t)$ from $t = 0$, where $f(0) = 0$.

Using the inverse formula, we have $$ f'(t) = \frac{1}{2\, \pi \, i}\int_{\gamma-i\infty}^{\gamma+i\infty} e^{-\sqrt{s+2} + st} \, ds, $$ where $\gamma$ is a large positive number, such that all poles, if any, lie on the left side of the line $y = \gamma$.

For our case, there is no pole, but only a branch cut at $s = -2$. We shall let the branch cut extend to $-\infty$ from $s = -2$, and wrap the contour around the branch cut from $-\infty + 0^- \, i$ to $-2 + 0^- \, i$ (lower half) and then from $-2 + 0^+ \, i$ to $-\infty + 0^+ \, i$ (higher half).

Let $s = -2 + A\,e^{-i\pi}$ with $A \ge 0$, then $$ \begin{aligned} f'(t) &= \frac{e^{-2t}}{\pi} \int_0^{+\infty} \sin(\sqrt{A}) \, e^{- A t} dA \\ &= \frac{e^{-2t}}{\pi} \int_{-\infty}^{+\infty} u \, \sin(u) \, e^{-u^2 t} du \\ &= \frac{e^{-2t}}{\pi} \mathrm{Im} \left(\frac{\partial}{\partial v} \int_{-\infty}^{+\infty} e^{-u^2 t + v u} d u \right)_{v = i} \\ &= \frac{e^{-2t}}{\pi} \mathrm{Im} \left[\frac{\partial}{\partial v}\left( \sqrt{\frac{\pi}{t}} \, \exp{\frac{v^2}{4t}} \right)\right]_{v = i} \\ &= \frac{1}{2 \, \sqrt{\pi \, t^3}} \, \exp\left(-2 \, t-\frac{1}{4\,t}\right). \end{aligned} $$ Here, we have used the fact that $\mathrm{Im} \, e^{iu} = \sin(u)$,

Finally, let us integrate $f'(t)$. $$ \begin{aligned} f(t) &= \int_0^t f'(\tau) \, d\tau \\ &=\frac{1}{\sqrt\pi} \int_{1/\sqrt{t}}^\infty \exp\left(-\frac{x^2}{4}-\frac{2}{x^2}\right) \, dx \\ &=\frac{1}{\sqrt\pi} \int_{1/\sqrt{t}}^\infty \exp\left(-\frac{x^2}{4}-\frac{2}{x^2}\right) \, d\left( \frac{x}{2}+\frac{\sqrt{2}}{x} \right) \\ &\quad+ \frac{1}{\sqrt\pi} \int_{1/\sqrt{t}}^\infty \exp\left(-\frac{x^2}{4}-\frac{2}{x^2}\right) \, d\left( \frac{x}{2}-\frac{\sqrt{2}}{x} \right) \\ &=\frac{1}{\sqrt\pi} \left[ \int_{\frac{1}{2\sqrt{t}}+\sqrt{2t}}^\infty \exp\left(\sqrt{2} -y^2\right) \, dy + \int_{\frac{1}{2\sqrt{t}}-\sqrt{2t}}^\infty \exp\left(-\sqrt{2} -z^2\right) \, dz \right] \\ &= \frac{e^{\sqrt 2}}{2} \, \mathrm{erfc}\left( \frac{1}{2\sqrt{t}}+\sqrt{2\, t} \right) +\frac{e^{-\sqrt 2}}{2} \, \mathrm{erfc}\left( \frac{1}{2\sqrt{t}}-\sqrt{2\, t} \right). \end{aligned} $$ Here, we have changed variables $$ \begin{aligned} x &\equiv \frac{1}{\sqrt{\tau}}, \\ y &\equiv \frac{x}{2} + \frac{\sqrt{2}}{x}, \\ z &\equiv \frac{x}{2} - \frac{\sqrt{2}}{x}. \end{aligned} $$ Our result agrees with Jan Erland's.

Answer 2

Mathematica 10.0 gives this output:

$$\mathcal{L}_{s}^{-1}\left[\frac{e^{-\sqrt{s+2}}}{s}\right]_{(t)}=\frac{1}{2}e^{-\sqrt{2}}\left(\text{erfc}\left(\frac{1-2t\sqrt{2}}{2\sqrt{t}}\right)+e^{e\sqrt{2}}\space\text{erfc}\left(\frac{1+2t\sqrt{2}}{2\sqrt{t}}\right)\right)$$