# Normalization Constant

• September 18th 2007, 08:45 PM
myoplex11
Normalization Constant
Calculate normilization constant for f(x) = e^ [1.5(x^2)]
• September 18th 2007, 08:51 PM
ThePerfectHacker
Quote:

Originally Posted by myoplex11
Calculate normilization constant for f(x) = e^ [1.5(x^2)]

I think you mean $e^{-1.5x^2}$ because the former does not convergene when we take the integral.

We need to find a constant ("normalization") $k$ so that:
$k\int_{-\infty}^{\infty} e^{-1.5x^2} dx = 1$
In order to turn it into a probability density function.

Thus,
$k\int_{-\infty}^{\infty} e^{-(\sqrt{1.5} x)^2} dx = 1$
Let $t=\sqrt{1.5}x$ use this substitution to get,
$\frac{k}{\sqrt{1.5}}\int_{-\infty}^{\infty} e^{-t^2} dt = 1$
Thus,
$k \cdot \sqrt{1.5}^{-1} \cdot \sqrt{\pi} = 1$
Thus,
$k = \frac{\sqrt{3}}{\sqrt{2\pi}}$