Density function for a normal distribution

1. Mar 9, 2015

Calpalned

1. The problem statement, all variables and given/known data
I have to prove that $\int e^{\frac{x^2}{-2}}dx$ from +∞ to -∞ = $\sqrt{2\pi}$

2. Relevant equations
N/A

3. The attempt at a solution
My GSI went from
1) $\int e^{\frac{x^2}{-2}}dx$ from +∞ to -∞ = $\sqrt{2\pi}$
to
2) $(\int e^{\frac{x^2}{-2}}dx)(\int e^{\frac{y^2}{-2}}dy)$ is equal to $2\pi$
Where did the red part of the function come from?
This leads to another question, how do we convert a single integral to multiple integrals? Thank you.

2. Mar 9, 2015

$$I = \int_{-\infty}^\infty e^{-\frac{x^2} 2} \, dx$$
you can also write it as
$$I = \int_{-\infty}^\infty e^{-\frac{y^2} 2} \, dy$$
since the variable of integration is immaterial. Multiplying I with itself gives
$$I^2 = \left(\int_{-\infty}^\infty e^{-\frac{x^2} 2} \, dx\right) \left(\int_{-\infty}^\infty e^{-\frac{y^2} 2} \, dy\right)$$

This product of two integrals can be written as a double integral. If you can do that, and then show that $I^2 = 2 \pi$, you will
essentially be done, no?

3. Mar 9, 2015