How to show that the integral is finite for $(x,t)$ such that $x>0$ and $t>0$? 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?
 A: Hint. Since $g:[0,+\infty) \rightarrow R$ is continuous and bounded, the potential problems are near $s=0^+$ and near $s=t^-$.


*

*For $s$ near $0^+$, we have
$$
\frac{1}{(t-s)^{3/2}}e^{\frac{-x^2}{4(t-s)}}g(s) \sim \frac{1}{t^{3/2}}e^{\frac{-x^2}{4t}}g(0)
$$   and the integral '$ \displaystyle
\int_{0}\frac{1}{t^{3/2}}e^{\frac{-x^2}{4t}}g(0)\:ds
$' is finite.

*For $s$ near $t^-$, we have  $$
    \frac{1}{(t-s)^{3/2}}e^{\frac{-x^2}{4(t-s)}}g(s) \sim
    \frac{1}{(t-s)^{3/2}}e^{\frac{-x^2}{4(t-s)}}g(t) $$   and the integral '$
    \displaystyle \int^{t}\frac{1}{(t-s)^{3/2}}e^{\frac{-x^2}{4(t-s)}}\:ds $'
is finite, using for example the fact that
$$
0<\frac{1}{(t-s)^{3/2}}e^{\frac{-x^2}{4(t-s)}}< \frac{1}{(t-s)^{1/2}}
$$ for $s$ sufficiently near $t^-$.

A: 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.   
