# Inverse Laplace Transform of $e^{c \cdot s^2}$

I am trying to find the Inverse Laplace Transform of the function

$$F(s)=e^{c \cdot s^2}$$

where $c > 0$.

• Are you sure it is $s^2$ and not $e^{c\cdot s}$? Commented Jan 30, 2015 at 18:17
• Unfortunately, yes. I know that $e^{cs}$ would give the delta function, but this one beats me so far. Commented Jan 30, 2015 at 18:20
• Did it come from applications? Commented Jan 30, 2015 at 18:31
• Yes, I am trying to solve the one-dimensional linear viscous wave equation with given initial conditions. Commented Jan 30, 2015 at 18:34

I don't know if this will be of any use but note that (My original thought before I finished): $$\int e^{x^2}dx = \frac{\sqrt{\pi}}{2}\text{Erfi}[x]$$ Now, if we proceed with your problem, we have \begin{align} \frac{1}{2\pi i}\lim_{u\to\infty}\int_{-iu}^{iu}\exp[cs^2 + st]ds &= \frac{1}{2\pi i}\lim_{u\to\infty}\int_{-iu}^{iu}\exp\biggl[c\Bigl(s^2 + \frac{t}{c}s + \frac{t^2}{4c^2}\Bigr)-\frac{t^2}{4c}\biggr]ds\\ &= \frac{e^{-\frac{t^2}{4c}}}{2\pi i}\lim_{u\to\infty}\int_{-iu}^{iu}\exp\biggl[c\Bigl(s+\frac{t}{2c}\Bigr)^2\biggr]ds\\ &= \frac{e^{-\frac{t^2}{4c}}}{4 i\sqrt{c\pi}}\lim_{u\to\infty}\biggl[\text{Erfi}\Bigl(\frac{2iu+t}{2\sqrt{c}}\Bigr)-\text{Erfi}\Bigl(\frac{-2iu+t}{2\sqrt{c}}\Bigr)\biggr]\\ &= \frac{e^{-\frac{t^2}{4c}}}{4\sqrt{c\pi}}\lim_{u\to\infty}\biggl[\text{Erf}\Bigl(\frac{2u-it}{2\sqrt{c}}\Bigr)-\text{Erf}\Bigl(\frac{-2u-it}{2\sqrt{c}}\Bigr)\biggr] \end{align} For $t\geq 0$, $\text{Erf}[\infty - it] \to 1$ and $\text{Erf}[-\infty - it] \to -1$ so $$\mathcal{L}^{-1}\{e^{cs^2}\}=\frac{e^{-\frac{t^2}{4c}}}{2\sqrt{c\pi}}$$
• I went through your derivation, and I could not find any mistake. However, I checked it with both tables and Mathematica, and the laplace transform of $e^{-t^2/4c}/2\sqrt{c \pi}$ is $e^{cs^2} ~ erfc(s)$ Commented Jan 30, 2015 at 20:36
• @bkosztin I will have to think about it. I don't see how $\text{Erfc}(s)$ will show up. I can delete the answer if you like. Commented Jan 30, 2015 at 20:39