I want to verify that $u(x,t)$ is a solution to the semi-infinite heat equation, i.e. $$ u_t=u_{xx} \ \forall \ x>0, \ t>0 $$ where $$ u(x,t) = \frac{x}{\sqrt{4\pi}}\int_0^t g(t-\tau)\frac{e^{-x^2/4\tau}}{\tau^{3/2}} d\tau $$ To start, I wanted to find the partial derivative with respect to $t$. Since our bounds are dependent on $t$ we have to use Leibniz integral rule. This yields: $$ u_t = \frac{x}{\sqrt{4\pi}}g(0)\frac{e^{-x^2/4t}}{t^{3/2}} + \frac{x}{\sqrt{4\pi}}\int_0^t \frac{\partial}{\partial t} [g(t-\tau)\frac{e^{-x^2/4\tau}}{\tau^{3/2}}] d\tau $$ Now we are given that $g(0)=0$ so the surface term vanishes. I could pretty easily find that partial derivative since the only thing that's dependent on $t$ is the function $g$. However, I don't think that's going to be fruitful later on when I try to equate this to $u_{xx}$. My thought instead was to use some sort of change of variables to relate $\frac{\partial}{\partial t}$ to $\frac{\partial}{\partial \tau}$ but I'm not quite sure where I would get an equation relating the two. Would it be correct to write: $$ s = t - \tau \implies t = s + \tau \implies dt = d\tau $$ I feel like there should be a negative sign to the relation but I don't quite see what's wrong with my logic. I also thought to use the chain rule perhaps where: $$ \frac{\partial}{\partial t}=\frac{\partial}{\partial\tau}\frac{\partial\tau}{\partial t} $$ but then again I need to find what $\frac{\partial\tau}{\partial t}$ is and it's not clear to me how I would do this. Is it just $1$?
Thanks for any advice, help, hints, or links you can provide to clear up my issue!
HINT:
Note that $$\frac{\partial g(t-\tau)}{\partial t}=-\frac{\partial g(t-\tau)}{\partial \tau}$$
Then, integrate by parts to place the partial derivative with respect to $\tau$ on the kernel function.
Can you proceed?