How to prove this limit Let $g:[0,\infty)\to\mathbb{R}$ with $g(0)=0$. For $x>0$ and $t>0$, define:
\begin{equation}
u(x,t) = \frac{x}{\sqrt{4\pi}} \int_0^t \frac{1}{(t-s)^{3/2}} e^{-\frac{x^2}{4(t-s)}} g(s) \,\mathrm{d}s.
\end{equation}
Show that $\lim\limits_{x\to0^+} u(x,t)=g(t)$.
I tried to split the integration from $0$ to $t-\delta$ and from $t-\delta$ to $t$ for some $\delta>0$. I think the limit of the integral from $0$ to $t-\delta$ vanishes, but what should I do for the second integral?
Thank you.
PS: This formula arises from the heat equation.
 A: Starting from the integral expression for $u$, we have
$$u(x,t)=\frac{x}{\sqrt{4\pi}}\int_0^t g(s) \frac{e^{-\frac{x^2}{4(t-s)}}}{(t-s)^{\frac32}}ds$$
We let $\delta>0$ and write $u$ as
$$u(x,t)=\frac{x}{\sqrt{4\pi}}\int_0^{t-\delta} g(s) \frac{e^{-\frac{x^2}{4(t-s)}}}{(t-s)^{\frac32}}ds+\frac{x}{\sqrt{4\pi}}\int_{t-\delta}^t g(s) \frac{e^{-\frac{x^2}{4(t-s)}}}{(t-s)^{\frac32}}ds$$
We will adopt a formal, heuristic approach to facilitate the development, although the ensuing arguments can be made rigorous.  
As $x \to 0$, it is easy to see that the first term on the right-hand side goes to zero.  Next, we let $\delta$ be small and exploit the continuity of $g$.  Thus, $\lim_{x \to 0} u(x,t)$ becomes 
$$\lim_{x \to 0} u(x,t)=\lim_{x \to 0} \frac{x}{\sqrt{4\pi}}g(t) \int_{t-\delta}^t \frac{e^{-\frac{x^2}{4(t-s)}}}{(t-s)^{\frac32}}ds$$
We now make the substitution $t-s =\frac{1}{y^2}$, $ds=\frac{2}{y^3}dy$, and limits of integration extend from $\delta^{-\frac12}$ to $\infty$. Thus, 
$$\lim_{x \to 0} u(x,t)=\lim_{x \to 0} \frac{2x}{\sqrt{4\pi}}g(t)\int_{\delta^{-\frac12}}^{\infty}  e^{-\frac{x^2y^2}{4}}dy$$
Now, substitute $u=\frac12 xy$, $dy=\frac{2}{x} du$ and the limits of integration go from $\frac12 x\delta^{-\frac12}$ to $\infty$.  Then, we have
$$\lim_{x \to 0} u(x,t)=\lim_{x \to 0} \frac{2}{\sqrt{\pi}}g(t)\int_{\frac12 x\delta^{-\frac12}}^{\infty}  e^{-u^2}du = \frac{2}{\sqrt{\pi}}g(t)\int_{0}^{\infty}  e^{-u^2}du=g(t)$$
where we used $\int_{0}^{\infty}  e^{-u^2}du =\frac{\sqrt{\pi}}{2}$.
A: Let $f(t) = {1 \over \sqrt{\pi}}{ 1\over t \sqrt{t}} e^{-{1 \over t}} 1_{(0,\infty)}(t)$. Then $f(t) \ge 0$ for all $t$ and $f \in L^1(\mathbb{R})$. Using the substitution $t = {1 \over s^2}$ we can see that $\int f  = 1$.
We also see that $f$ is bounded.
Let $f_\epsilon(t) = {1 \over \epsilon} f({t \over \epsilon})$. Then $\{f_\epsilon\}_{\epsilon>0}$ form an approximation to the identity and we have
$\lim_{\epsilon \downarrow 0} (f_\epsilon * g) (t) = g(t)$ at every $t>0$ at
which $g$ is continuous.
Letting $\epsilon = {x^2 \over 4}$ and making the necessary substitutions yields
the desired result.
