Evaluation of an integral associated with integral kernel of resolvent of Laplacian I came across evaluating the following sort of integral when I was considering the integral kernel for resolvent of Laplacian $(I-\Delta)^{-1}$:
$$
K(x)=\int_0^{\infty}\frac{\exp(-t-\frac{|x|^2}{4t})}{t^{\frac{n}{2}}}dt
$$
However, what I did know is that in the 1-dimensional case, i.e. $n=1$, the operator $(1-\partial_x^2)^{-1}$ has an integral kernel representation of the form:
$$
(1-\partial_x^2)^{-1}f=\int_{-\infty}^{\infty}\frac{1}{2}e^{-|x-y|}f(y)dy
$$
i.e. the integral kernel is$$
K(x)=\frac{1}{2}e^{-|x|}
$$
My question is, how can one evaluate the first integral for the case $n=1$ to obtain exactly the integral kernel in the third equation? 
 A: Plugging in $n=1$, the integral becomes
$$ K(x) = \int_0^{\infty} t^{-\frac{1}{2}} e^{-t-\frac{x^2}{4t}} \,dt.$$
Taking a Fourier transform, we have
$$\mathcal{F}K(y) = \frac{1}{\sqrt{2\pi}} \int_0^{\infty} t^{-\frac{1}{2}} e^{-t}\int_{-\infty}^{\infty} e^{-ixy} e^{-\frac{x^2}{4t}}\,dx\,dt.$$
The inner integral is the Fourier transform of the Gaussian with a dilation. In general, if $\mathcal{D}_{\alpha}f(x) = f(\alpha x)$ for nonzero $\alpha$, then a change of variable gives
$$\mathcal{F}\mathcal{D}_{\alpha}f(y) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} e^{-ixy}f(\alpha x)\,dx = \frac{1}{|\alpha|\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{-ix\frac{y}{\alpha}} f(x)\,dx.$$
Which is nothing more than $\frac{1}{|\alpha|}\mathcal{D}_{\alpha^{-1}} \mathcal{F}f(y)$, meaning if you dilate in one domain, you get the opposite dilation in the other domain (with a scale factor). In our case $\alpha = (2t)^{-1/2}$. Moreover, the Fourier transform of a Gaussian ($\exp(-x^2/2)$) is another Gaussian since the Gaussian is an eigenfunction of the Fourier transform, so we get that
$$ \mathcal{F}K(y) = \int_0^{\infty} t^{-\frac{1}{2}}e^{-t} \cdot (2t)^{\frac{1}{2}} e^{-ty^2}\,dt = 2^{\frac{1}{2}}\int_0^{\infty} e^{-(1+y^2)t}\,dt. $$
Letting $u = (1+y^2)t$, we get that $du = (1+y^2)\,dt$ and so
$$ \mathcal{F}K(y) = \frac{\sqrt{2}}{1+y^2} \int_0^{\infty} e^{-u}\,du = \frac{\sqrt{2}}{1+y^2}.$$
The Fourier transform is injective and the only function whose Fourier transform is $\frac{\sqrt{2}}{1+y^2}$ is $\sqrt{\pi} e^{-|x|}$.
I will leave it to you to argue that $K$ (as given in the first line) is in $L^1(\Bbb R)$ and that you can interchange the integrals as I did in the second line so that the use of the Fourier transform is justified.
Edit: Our constants do not quite match up. My guess is that your initial definition of $K$ has a constant factor out front which should take care of the factor of $\sqrt{\pi}$.
My reason for using the Fourier transform here is twofold. The main problem in your integral is the $e^{-\frac{x^2}{4t}}$ part. This is a Gaussian in $x$ - not $t$. So my thought was along the lines of: in the $t$-integral, the $\frac{1}{t}$ part is very problematic and hard to work with. Moreover the dilation property of the Fourier transform means that if I did a Fourier transform, the $\frac{1}{t}$ would appear in the numerator afterward which is infinitely easier to handle. Coupled with the Gaussian being an eigenfunction of the Fourier transform, it was clear that the Fourier transform was the way to go.
