Integral solution, kind of Laplace transfom $\int_{0}^{T_0} \frac{e^{-st-\frac{x^2}{4t}}}{\sqrt{\pi t}}dt$.

Question

I try to find a solution to the folowing problem:

$$\int_{0}^{T_0} \frac{e^{-st-\frac{x^2}{4t}}}{\sqrt{\pi t}}dt$$

with

$ x \in \mathbb{R}$, $0<x<1$

$ s \in \mathbb{C}$,

$T_0 \in \mathbb{R}$, $T_0 >0$

but without sucess. I thus tried with Maple and it gives me the folowing solution

Could someone tell me how such a solution is obtained ? Thank you!

Answer 1

For $\,T_0\rightarrow\infty$ (the Laplace transform), there's a closed-form expression for the result. Here's how it works manually:

\begin{align} \int_0^\infty\frac{e^{-st-\frac{x^2}{4t}}}{\sqrt{\pi t}}dt&=\frac{2}{\sqrt{\pi}}\int_0^\infty e^{-st-\frac{x^2}{4t}}d\sqrt{t}=\frac{2}{\sqrt{\pi}}\int_0^\infty e^{-st^2-\frac{x^2}{4t^2}}dt\\ &=\frac{2}{\sqrt{\pi}}e^{-|x|\sqrt{s}}\int_0^\infty e^{-(\sqrt{s}t-\frac{|x|}{2t})^2}dt. \end{align}

To do the last integral, use the formula

$$\boxed{\int_0^\infty e^{-(At-\frac{B}{t})^2}dt=\frac{\sqrt{\pi}}{2A},\quad A, B>0.}$$

This formula can be derived using the change of variable $\,u=At-\frac{B}{t}\in(-\infty,\infty)$. And then

$$\int_0^\infty e^{-(At-\frac{B}{t})^2}dt=\frac{1}{2A}\int_{-\infty}^\infty e^{-u^2}\left(1+\frac{u}{\sqrt{u^2+4AB}}\right)du.$$

The first term $\,1\,$ gives a Gaussian integral that integrates to $\sqrt{\pi}\,$ and the second term vanishes because it's odd while $\,e^{-u^2}$ is even. Using this integration formula, we obtain

\begin{align} \int_0^\infty\frac{e^{-st-\frac{x^2}{4t}}}{\sqrt{\pi t}}dt=\frac{1}{\sqrt{s}}e^{-|x|\sqrt{s}},\quad s>0. \end{align}

For finite $T_0$, the answer is a lot more complicated. The Gaussian integral becomes incomplete and gives an error function. The second term is no longer canceled by symmetry, but can be shown to also give an error function using change of variable.

Answer 2

Thanks to @ZhuoranHe answer, let $f(t)=\dfrac{1}{\sqrt{\pi t}}e^{-x^2/4t}$ be a periodic function with period $T_0$. We know $\displaystyle{\cal L}(f)=\dfrac{1}{\sqrt{s}}e^{-|x|\sqrt{s}}$, and in the other side the Laplace transform of a periodic function is $${\cal L}(f)=\dfrac{\int_0^{T_0}e^{-st}\dfrac{1}{\sqrt{\pi t}}e^{-x^2/4t}dt}{1-e^{-sT_0}}$$ then $$\int_0^{T_0}\dfrac{1}{\sqrt{\pi t}}e^{-st-x^2/4t}\,dt=\dfrac{1-e^{-sT_0}}{\sqrt{s}}e^{-|x|\sqrt{s}}$$