How to compute the integral $\int_{-\infty}^\infty e^{-x^2/2}\,dx$? Yes, I know that this is very similar to  $\int_{-\infty}^\infty e^{-x^2}\,dx$, which has been answered a million times, but I still don't know how to apply the technique from that integration to mine.
I don't want to do this using polar coordinates, or "erf". I'd like to use the Gamma function (which I assume is possible..). 
Is this correct? : 
$$\int_{-\infty}^\infty e^{-x^2/2}\,dx=2\int_{0}^\infty e^{-x^2/2}\,dx$$
Let $u = x^2/2  \implies x = \sqrt{2u}$, $du = x \, dx \implies dx = (2u)^{-1/2}$
So, 
$$=2\int_{0}^\infty (2u)^{-1/2}e^{-u}\,dx=\sqrt{2}\int_0^\infty (u)^{-1/2}e^{-u}\,dx=\sqrt{2} \Gamma(1/2)= \sqrt{2\pi}$$
 A: Actually, you can use sum expression for $e^{-x^2}$. Then you can get which function you want in this way and then you can integrate.
A: $$
=2\int_{0}^\infty (2u)^{-1/2}e^{-u}\underbrace{\,dx\,}_{\text{error}}=\sqrt{2}\int_0^\infty (u)^{-1/2}e^{-u}\,dx=\sqrt{2} \Gamma(1/2)= \sqrt{2\pi}
$$
If you put $du$ where $dx$ is, then its correct.  Where you wrote $dx = (2u)^{-1/2}$, you need $dx=(2u)^{-1/2}\,du$.
If you're only trying to reduce the integral $\displaystyle\int_{-\infty}^\infty e^{-x^2/2}\, dx$ to the integral $\displaystyle\int_{-\infty}^\infty e^{-x^2}\,dx$, it can be done as follows:
\begin{align}
u & = \frac x {\sqrt{2}} \\[8pt]
u^2 & = \frac{x^2}2 \\[8pt]
du & = \frac{dx}{\sqrt{2}} \\[8pt]
\sqrt{2}\,du & = dx \\[8pt]
\int e^{-x^2/2}\,dx & = \int e^{-u^2}\sqrt 2\,du = \sqrt 2\int e^{-u^2}\,du.
\end{align}
Then notice that as $x$ goes from $-\infty$ to $+\infty$, so does $u$, since $1/\sqrt 2$ is positive.  You get
$$
\int_{-\infty}^\infty e^{-x^2/2}\,dx = \sqrt 2 \int_{-\infty} e^{-x^2}\,dx = \sqrt 2\sqrt \pi.
$$
