TITLE: How to show that the integral is finite for $(x,t)$ such that $x>0$ and $t>0$?
QUESTION [7 upvotes]: Given $g:[0,+\infty) \rightarrow R$ continuous and bounded, let
$$u(x,t)=\frac{x}{\sqrt{4 \pi}}\int \limits_0^t \frac{1}{(t-s)^{3/2}}e^{\frac{-x^2}{4(t-s)}}g(s)ds$$
How to show that the integral is finite for $(x,t)$ such that $x>0$ and $t>0$?
I tried to use $e^{-a} \le \frac{m!}{a^m}$ for arbitrary $m$. But I still don't know how to get it is finite. Am I on the right track?

REPLY [2 votes]: An elegant way to show that the integral is finite, is to change your integration variable from $s$ to $q = \frac 1{t-s}$. Note that $ds = dq  (t-s)^2$, so you get rid of the only term that might cause divergence: $(t-s)^{-3/2}$. The resulting integral looks neat.