Compute the Laplace transform of $\frac{1}{\sqrt{\pi t}} e^{\frac{-a^2}{4t}}$

Question

I'm struggling to find this transform. I tried to complete squares but the expression I had left was too messy. Could you please provide any suggestions/general hints/solution?

Answer 1

You're looking for $$\frac{1}{\sqrt{\pi}}\int_0^\infty \frac{dt}{\sqrt{t}} \exp{\left(-\frac{a^2}{4t} - st\right)}$$ Let $u=\sqrt{t}$, then you obtain $$\frac{1}{\sqrt{\pi}} \int_{-\infty}^\infty \exp{\left(-\frac{a^2}{4u^2} - su^2\right)}\,du.$$ Do some factoring to get $$\frac{1}{\sqrt{\pi}} \exp-|a|\sqrt{s}\int_{-\infty}^\infty du\, \exp{ -s\left(u-\frac{|a|}{2u\sqrt{s}}\right)^2}.$$ By Glasser, the last integral is just equivalent to a gaussian integral and so your initial integral evaluates to $$\frac{\exp-|a|\sqrt{s}}{\sqrt{s}}.$$