How to prove gaussian-like integral equation true. Integrals is definitely not my strong point, and I'm having trouble proving that:
$${\int_{-\infty}^\infty (e^{\large{\pi}n})^{-\large{x}^2} dx = {1\over\sqrt{n}}}$$
It has similarities to the gaussian integral, but I have no idea on how to prove it correct.
What could be used to attack this?
Thanks.
 A: Here are two useful things. First, do you know the value of
$$\int_{-\infty}^\infty e^{-x^2}\, dx?$$
If you do, you can carry out a simple $u$-sub to do the job.
A: Notice that $(\operatorname{e}^{\pi n})^{-x^2} \equiv \operatorname{e}^{-\pi n x^2}$. We can apply the same method:
$$J^2 = \left(\int_{-\infty}^{\infty}\operatorname{e}^{-\pi n x^2}\, \operatorname{d}\!x\right)\left(\int_{-\infty}^{\infty} \operatorname{e}^{-\pi n y^2} \, \operatorname{d}\!y \right) = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \operatorname{e}^{-\pi n (x^2+y^2)}\, \operatorname{d}\!x\, \operatorname{d}\!y$$
If we make the substitution $x = r\cos\theta$ and $y=r\sin\theta$ we see that $-\infty < x < \infty$ and $-\infty < y < \infty$ is replaced by $0 \le r < \infty$ and $0 \le \theta < 2\pi$. Moreover, $x^2+y^2 = r^2$ and $\operatorname{d}\!x \wedge \operatorname{d}\!y = r \, \operatorname{d}\!r \wedge \operatorname{d}\!\theta$. Meaning that:
\begin{array}
JJ^2 &=& \int_0^{2\pi}\int_0^{\infty} r\operatorname{e}^{-\pi n r^2} \, \operatorname{d}\!r \, \operatorname{d}\!\theta \\ \\
 &=& \int_0^{2\pi} \left[ \frac{1}{2\pi n}\operatorname{e}^{-\pi n r^2}\right]_0^{\infty}\operatorname{d}\!\theta \\ \\
 &=& \int_0^{2\pi} \frac{1}{2\pi n} \, \operatorname{d}\!\theta \\
& = & \left[\frac{1}{2\pi n}\right]_0^{2\pi} \\ \\
&=& \frac{1}{n}
\end{array}
Since $J^2 = \frac{1}{n}$ it follows that $J = \frac{1}{\sqrt{n}}$.
A: The integral of $e^{-\alpha x^2}$, $\alpha > 0$, over $R$ is very well known; we have
$\int_{- \infty}^{\infty} e^{-\alpha x^2}dx = \sqrt{(\frac{\pi}{\alpha})}$;
now setting $\alpha = \pi n$, this yields
$\int_{- \infty}^{\infty} e^{-\pi n x^2}dx = \sqrt{(\frac{1}{n})}$.
