Short integral question Can anyone here just tell me this is true? I just need a YES/NO, because I am a bit confused right now...
\begin{align}
\int\limits_{-\infty}^{\infty}\exp\left[{-\frac{x^2}{a}}\right]dx = \left.\left( -\frac{a}{2x} \right)\exp\left[{-\frac{x^2}{a}}\right]\right|_{-\infty}^{\infty}
\end{align}
 A: Hint
To find the value of the integral:
Multiply the  integral by $\int\limits_{-\infty}^{\infty}\exp\left[{-\frac{y^2}{a}}\right]dy$ then use the polar coordinates.
A: Here's a way to find out the simplest case (understand and explain each step):
$$I:=\int\limits_{-\infty}^\infty e^{-x^2}dx\implies I^2=\left(\int\limits_{-\infty}^\infty e^{-x^2}dx\right)^2=\int\limits_{-\infty}^\infty e^{-x^2}dx\int\limits_{-\infty}^\infty e^{-y^2}dy=$$
$$=\int\limits_{-\infty}^\infty\int\limits_{-\infty}^\infty e^{-(x^2+y^2)}dxdy\stackrel{\text{polar coord.}}=\int\limits_0^\infty\int\limits_0^{2\pi}re^{-r^2}d\theta dr=$$
$$=\left.-\pi\int\limits_0^\infty(-2r\,dr)e^{-r^2}=-\pi e^{-r^2}\right|_0^\infty=-\pi(0-1)=\pi$$
and from here
$$I=\sqrt\pi$$
Now your integral, assuming $\,a>0\,$:
$$J:=\int\limits_{-\infty}^\infty e^{-x^2/a}dx\;\ldots\;\;\text{substitution}:\;\;u:=\frac x{\sqrt a}\;,\;dx=\sqrt a\,du\implies$$
$$J=\sqrt a\int\limits_{-\infty}^\infty e^{-u^2}du=\sqrt{a\pi}$$
A: evaluate $$ \large{ \int_{-\infty}^{\infty} e^{\frac{-x^2}{a}} \ dx} $$
now we have $$ \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-\frac{x^2+y^2}{a}} \ dx \ dy = 4\int_0^{\infty} \int_0^{\infty} e^{-\frac{x^2+y^2}{a}} \ dx \ dy $$
(ok i like 0 to inf ) to polar coordinate
$$4\int_0^{\frac{\pi}{2}} \int_0^{\infty} re^{-\frac{r^2}{a}} \ dr \ d\theta$$
now it became easy and note that
$$\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-\frac{x^2+y^2}{a}} \ dx \ dy = \left(\large{ \int_{-\infty}^{\infty} e^{\frac{-x^2}{a}} \ dx} \right)^2 $$
