How to partially differentiate the integral $\int_{0}^{x/\sqrt{t}}\exp(-\xi^2/4)d\xi$ w.r.t $x$ and $t$?

Question

How do I partially differentiate the following integral:

$$ u(x,t) := \int_{0}^{x/\sqrt{t}}e^{-\xi^2/4} \,\mathrm d \xi $$

I would like to calculate $\frac{\partial u}{\partial x}(x, t)$ and $\frac{\partial u}{\partial t}(x, t)$.

I have tried to solve the indefinite integral:

$$ \int e^{-\xi^2/4} \,\mathrm d \xi $$

hoping to get a nicer formula, but even the CAS only spit out

$$ \int e^{-\xi^2/4} \,\mathrm d \xi = \sqrt{\pi}\operatorname{erf}\left(\frac{\xi}{2}\right)=2\int_{0}^{\frac{\xi}{2}}e^{-\tau^2}\,\mathrm d\tau $$

which is of not much help, since it is basically just a slight rearrangement of the orginal function.

I have also tried to apply the fundamental theorem of calculus, by rewriting

$$ u(x,t) := \int_{0}^{g(x,t)}h(\xi)d\xi = H(\xi)\Big|_0^{g(x,t)}=H(g(x,t))-H(0) $$

but I'm also stuck here.

Answer 1

$$u(x,t) = \int_{0}^{x/\sqrt{t}}e^{-\xi^2/4} d \xi$$

Use Leibnitz rule for integration

$$ \frac{\partial u}{\partial x}(x, t) = \exp\left(-\frac{x^2}{4t}\right) \frac{d}{dx} \left[\frac{x}{\sqrt t}\right] = \exp\left(-\frac{x^2}{4t}\right) \frac{1}{\sqrt t } $$

Similarly can you calculate $\frac{\partial u}{\partial t}(x, t)$?

Answer 2

HINT

Think of $$ f(z) = \int_0^z e^{-s^2/4}ds $$ with the property $$f'(z) = e^{-z^2/4}.$$

Note that your $u(x,t) = f(x/\sqrt{t})$. Based on this and using the chain rule, can you figure out the partials you are asking for?