1. ## completing the square?

$\displaystyle \int_{-\infty}^{\infty} \frac{e^{tx}}{\sigma\sqrt{2\Pi}} e^{\frac{-(x-\mu)^2}{2 \sigma{^2}}} dx$

Then i let, $\displaystyle z = \frac{x - \mu}{\sigma}; dz = \frac{1}{\sigma}; x = z\sigma + \mu$

$\displaystyle \int_{-\infty}^{\infty} \frac{e^{tz\sigma + \mu}}{\sqrt{2\Pi}} e^{\frac{z^2}{2}} dz$

hmm! i really don't know what I'm doing but I'll just keep going..
$\displaystyle \int_{-\infty}^{\infty} \frac{1}{\sqrt{2\Pi}} e^{\frac{z^2+ 2t(z\sigma + \mu)}{2}}e^{\frac{t^2}{2}} dz$

yeahh i really don't know what to do but the answer is meant to be:
$\displaystyle e^{t\mu + \frac{t^2\sigma^2}{2}}$

I know that eventually I'm meant to have
$\displaystyle \int_{-\infty}^{\infty} \frac{1}{\sqrt{2\Pi}} e^{\frac{-(x-\mu)^2}{2 \sigma{^2}}} dx = 1$

2. Originally Posted by Katina88

$\displaystyle \int_{-\infty}^{\infty} \frac{e^{tx}}{\sigma\sqrt{2\Pi}} e^{\frac{-(x-\mu)^2}{2 \sigma{^2}}} dx$

Then i let, $\displaystyle z = \frac{x - \mu}{\sigma}; dz = \frac{1}{\sigma}; x = z\sigma + \mu$

$\displaystyle \int_{-\infty}^{\infty} \frac{e^{tz\sigma + \mu}}{\sqrt{2\Pi}} e^{\frac{z^2}{2}} dz$

hmm! i really don't know what I'm doing but I'll just keep going..
$\displaystyle \int_{-\infty}^{\infty} \frac{1}{\sqrt{2\Pi}} e^{\frac{z^2+ 2t(z\sigma + \mu)}{2}}e^{\frac{t^2}{2}} dz$

yeahh i really don't know what to do but the answer is meant to be:
$\displaystyle e^{t\mu + \frac{t^2\sigma^2}{2}}$

I know that eventually I'm meant to have
$\displaystyle \int_{-\infty}^{\infty} \frac{1}{\sqrt{2\Pi}} e^{\frac{-(x-\mu)^2}{2 \sigma{^2}}} dx = 1$