# A Gaussian distribution characteristic function

1. Jan 5, 2017

### senobim

Hello, guys. I am trying to solve for characteristic function of normal distribution and I've got to the point where some manipulation has been made with the term in integrands exponent. And a new term of t2σ2/2 has appeared. Could you be so kind and explain that to me, please.

$$=Ae^{it\mu}\int_{-\infty}^{\infty}e^{-\frac{1}{c^2}(\alpha^{2}-i2t\sigma ^{2}\alpha)}d\alpha=Ae^{(it\mu-\frac{t^{2}\sigma^{2}}{2})}\int_{-\infty}^{\infty}e^{-\frac{(\alpha-it\sigma^{2})^{2}}{c^{2}}} d\alpha$$

2. Jan 5, 2017

What was done is called "completing the square". e.g. take $x^2+6x$. You divide the x coefficient by 2 and square that result to complete the square: $x^2+6x=x^2+6x+9-9 =(x+3)^2-9$. ( $(6/2)^2=9$). You add it and also subtract it from the expression. editing ... In the case you presented, $\alpha=x$. $\\$ Additional editing: On closer inspection, the term should have $\sigma^4$ and not $\sigma^2$, and it should have a $c^2$ in the denominator instead of a 2. No wonder it puzzled you !!!

Last edited: Jan 5, 2017
3. Jan 6, 2017

### senobim

Nice! Thank you very much!