Computation of the Laplace transform of the Gaussian heat kernel

Question

Currently I'm interested in the Laplace transform of the Gaussian heat kernel $$ k_t(x):=\frac{1}{(4\pi t)^\frac{d}{2}}e ^{-\frac{|x|^2}{4t}}. $$ Using the Laplace transform $$ G_\lambda(x)=\int_0^\infty e^{-\lambda t}p_t(x)\mathrm{d}t=\int_0^\infty e^{-\lambda t}\frac{1}{(4\pi t)^\frac{d}{2}}e^{-\frac{|x|^2}{4t}}\mathrm{d}t $$ Via square addition on the exponent, I ended up trying to compute $$ G_\lambda(x)=(4\pi)^{-\frac{d}{2}}e^{\sqrt{\lambda}|x|}\int_0^\infty t^{-\frac{d}{2}}e^{-\bigl(\frac{|x|}{\sqrt{4t}}+\sqrt{\lambda t}\bigr)^2}\mathrm{d}t. $$ From there I'm not able to go any further, since I don't know how to get rid of the term $t^{-\frac{d}{2}}$. I actually looked up in the already discussed threads on the Laplace transform of the Gaussian but I'm not able to follow the final computation. I would be deeply thankful for any hints how to go on from the last expression. Best regards, Philipp

Answer 1

What you are calculating is known as the resolvent which is the fundamental solution of the equation $$ \Delta f-\lambda f=-\rho\, $$ for the the whole domain $\mathbb R^d.$ In [1] it is shown that, for $d\ge 2,$ \begin{align}\boxed{\quad \int_0^\infty e^{-\lambda t}\frac{1}{(4\pi t)^{d/2}}\exp\Big(-\frac{|x-y|^2}{4t}\Big)\,dt=\frac{\lambda^\frac{d-2}{4} K_\frac{d-2}{2}\Big(\sqrt{\lambda}\,|x-y|\Big)}{2\pi^{d/2}\Big(2\,|x-y|\Big)^\frac{d-2}{2}}\,,\quad} \end{align} where $K_\nu(z)$ is the Bessel function of the third kind.

[1] R.L. Schilling, L. Partzsch, Brownian Motion. An Introduction to Stochastic Processes. de Gruyter Graduate, Berlin 2012.