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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,

$$=2\int_{0}^\infty (2u)^{-1/2}e^{-u}\,dx=\sqrt{2}\int_0^\infty (u)^{-1/2}e^{-u}\,dx=\sqrt{2} \Gamma(1/2)= \sqrt{2\pi}$$

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    $\begingroup$ \implies is $\implies$ and \sqrt is $\sqrt{}$ and \Gamma is $\Gamma$ and use $$...$$ to center equations with big and beautiful equations,etc. $\endgroup$ – RE60K Mar 25 '15 at 14:12
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    $\begingroup$ It's easier just to substitute $x = \sqrt{2} u$ and use the known result for $\int_{-\infty}^{\infty} e^{-x^2} dx$. $\endgroup$ – Travis Mar 25 '15 at 14:14
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    $\begingroup$ Change of variables $u=x/\sqrt{2}$. $\endgroup$ – RTJ Mar 25 '15 at 14:15
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    $\begingroup$ Yes it is correct. $\endgroup$ – Bman72 Mar 25 '15 at 14:15
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    $\begingroup$ @ADG : The poster wrote "x dx", clearly intending a space between "x" and "dx", but that doesn't happen unless you code it as x\,dx or x~dx or the like, thus: $x dx$ versus $x\,dx$. ${}\qquad{}$ $\endgroup$ – Michael Hardy Mar 25 '15 at 14:41
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$$ =2\int_{0}^\infty (2u)^{-1/2}e^{-u}\underbrace{\,dx\,}_{\text{error}}=\sqrt{2}\int_0^\infty (u)^{-1/2}e^{-u}\,dx=\sqrt{2} \Gamma(1/2)= \sqrt{2\pi} $$ 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 $$ \int_{-\infty}^\infty e^{-x^2/2}\,dx = \sqrt 2 \int_{-\infty} e^{-x^2}\,dx = \sqrt 2\sqrt \pi. $$

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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.

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