Integral of $m$ products of identical normal probability density functions

Question

I am looking for a closed form solution of the integral from $m$ products of identical normal distribution probability density functions (PDF's). The mean value $\bar{x}=0$ and $\sigma$ identical for all distributions. Giving:

$$p(m)=\int_{-\infty}^{\infty} \left[ {\frac {1}{\sigma {\sqrt {2\pi }}}}e^{-{\frac {1}{2}}\left({\frac {x}{\sigma }}\right)^{2}} \right]^m dx$$

Empirical with Wolfram Alpha [WA] after puzzling I found the following closed function:

$$p(m)= \frac{1}{\sqrt{m} \cdot (\sigma \sqrt{2 \pi})^{(m-1)} }$$

I am looking for proof of this expression. The question is related to: SE, this function fit's perfect my Monte Carlo simulations.

Answer 1

I was thinking to complicated initially. With help of the derivation normal distribution I was able to find the answer see: Wiki.

$$\int _{-\infty }^{\infty }e^{-x^{2}}\,dx=\sqrt{\pi}$$

My question basically is:

$$\frac{1}{(\sqrt{\pi})^m} \int _{-\infty }^{\infty } e^{-mx^{2}}\,dx$$

Define:

$$u=\sqrt{m}x$$ $$dx=\frac{1}{\sqrt{m}} du$$

$$\frac{1}{\sqrt{m}(\sqrt{\pi})^m} \int _{-\infty }^{\infty } e^{-u^{2}}\,du$$

$$\frac{\sqrt{\pi}}{\sqrt{m}(\sqrt{\pi})^m}$$

$$\frac{1}{\sqrt{m}(\sqrt{\pi})^{(m-1)}}$$