Show a function satisfies the diffusion equation 
Show $u(x,t) = \int_0^{x/t^{1/2}} e^{-0.25b^2}db$ satisfies $\dfrac{\partial u}{\partial t} = \dfrac{\partial ^2 u}{\partial x^2}$

How do I go about doing this? Particularly because $e^{-x^2}$ has a nasty antiderivative.
 A: Nicolas already answered without the antiderivative using the fundamental theorem of calculus and this is, from far away, the most elegant solution.
It is not  so nasty if you already know the error function $$\int e^{-x^2}\,dx=\frac{1}{2} \sqrt{\pi }\,\, \text{erf}(x)$$ $$\int_0^a e^{-x^2}\,dx=\frac{1}{2} \sqrt{\pi }\,\, \text{erf}(a)$$ So, $$u=\int_0^{\frac{x}{\sqrt{t}}} e^{-\frac {b^2} 4}db=\sqrt{\pi }\,\, \text{erf}\left(\frac{x}{2 \sqrt{t}}\right)$$ $$\dfrac{\partial u}{\partial t} = -\frac{x }{2 t^{3/2}}\,e^{-\frac{x^2}{4 t}}$$ $$ \dfrac{\partial  u}{\partial x}=\frac{e^{-\frac{x^2}{4 t}}}{\sqrt{t}}$$ $$ \dfrac{\partial ^2 u}{\partial x^2}=-\frac{x }{2 t^{3/2}}\,e^{-\frac{x^2}{4 t}}$$
A: We have
$$\frac{\partial u}{\partial t}\left(x,t\right)=-\frac{x}{2t^{3/2}}\mathrm{e}^{-\frac{x^2}{4t}},$$
$$\frac{\partial u}{\partial x}\left(x,t\right)=\frac{1}{\sqrt{t}}\mathrm{e}^{-\frac{x^2}{4t}},$$
$$\frac{\partial^2 u}{\partial x^2}\left(x,t\right)=-\frac{x}{2t^{3/2}}\mathrm{e}^{-\frac{x^2}{4t}}.$$
The following propertie has been used for any smooth functions $a$ and $f$ :
$$\frac{\partial}{\partial t}\int_{0}^{a\left(x,t\right)}f\left(s\right)\mathrm{d}s=\frac{\partial a}{\partial t}\left(x,t\right)f\left(a\left(x,t\right)\right).$$
The same for the partial derivative in $x$.
