# completing the square?

• Aug 13th 2009, 07:03 AM
Katina88
completing the square?

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

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

$
\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..
$
\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:
$e^{t\mu + \frac{t^2\sigma^2}{2}}$

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

• Aug 13th 2009, 08:28 AM
matheagle
• Aug 13th 2009, 11:17 AM
kobylkinks
Quote:

Originally Posted by Katina88

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

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

$
\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..
$
\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:
$e^{t\mu + \frac{t^2\sigma^2}{2}}$

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