Will someone check my work....I think this is right

$\displaystyle \int_0^{\infty}e^{-a^2x^2}dx,\forall{a}>0$

First $\displaystyle Let\text{ }u=a^2x^2\Rightarrow{dx=\frac{1}{2a\sqrt{u}}}$

and $\displaystyle u(0)=a^2\cdot{0^2}=0$

and $\displaystyle u(\infty)=a^2\cdot{\infty}=\infty$

So this would be $\displaystyle \frac{1}{2a}\int_0^{\infty}\frac{e^{-u}}{\sqrt{u}}du=\frac{1}{2a}\int_0^{\infty}u^{\fra c{-1}{2}}e^{-u}du=\frac{\Gamma\bigg(\frac{1}{2}\bigg)}{2a}=\fra c{\sqrt{\pi}}{2a}$

This doesnt seem right...I mean I think I did my calculations right but this was in a very simple book and this seems uncharcteristic of such a book

....

Could someone verify?