Lagrange Multipliers for the derivation of Maxwell-Boltzmann distribution
Ignoring multiplicative constants, the required integral is (in general form)

$$ \int \sqrt{x} e^{ax} \mathrm{d} x = \frac{1}{a}\sqrt{x}e^{ax}+\frac{i\sqrt{\pi}}{2a^{3/2}} \mathrm{erf}(i\sqrt{ax})$$ 

where $$ \mathrm{erf}(x) = \frac{2}{\sqrt{\pi}} \int_0^x e^{-t^2}\mathrm{d}t$$

Setting $x=E$, $a=\beta$, and the integration limits as $0$ and $\infty$, gives

$$ \int_0^\infty \sqrt{E} e^{\beta E} \mathrm{d} E = \frac{1}{\beta}\sqrt{E}e^{\beta E}\Bigg|_{0}^\infty + \frac{i\sqrt{\pi}}{2\beta^{3/2}} \mathrm{erf}(i\sqrt{\beta E})\Bigg|_{0}^\infty = \frac{1}{\beta}\sqrt{E}e^{\beta E}\Bigg|^\infty +\frac{i\sqrt{\pi}}{2\beta^{3/2}} \mathrm{erf}(i\sqrt{\beta E})\Bigg|^\infty $$

since $\mathrm{erf}(0)=0$.
To make further progress, consider that the function $\mathrm{erf}(x)$ must have a real input $x$ (otherwise it will "blow up" to $\infty$), which implies that $\beta$ must be negative, or, explicitly, that  $\beta = -|\beta| $ (we could also argue that the second term on the r.h.s must be real and come to the same conclusion). With negative $\beta$ we have that

$$\mathrm{erf}(i\sqrt{\beta E})=\mathrm{erf}(i\sqrt{-|\beta|E})=\mathrm{erf}(-\sqrt{|\beta|E})=-\mathrm{erf}(\sqrt{|\beta|E})$$

With this condition we also see that the first term on the r.h.s. is zero and as a result that

$$ \int_0^\infty \sqrt{E} e^{-|\beta| E} \mathrm{d} E =  \frac{\sqrt{\pi}}{2|\beta|^{3/2}} $$