Normal Distribution Identity I have the following problem. I am reading the paper which uses this identity for a proof, but I can't see why or how to prove its true. Can you help me?
\begin{align}
  \int_{x_{0}}^{\infty} e^{tx} n(x;\mu,\nu^2)dx &= e^{\mu t+\nu^2 t^2 /2} N(\frac{\mu - x_0 }{\nu} +\nu t )   
\end{align}
where $n(\cdot)$ is the normal pdf with mean $\mu$ and variance $\nu^2$. $N(\cdot)$ refers to the normal cdf. 
\begin{align}
 \int_{x_{0}}^{\infty} e^{tx} n(x;\mu,\nu^2)dx &= \int_{x_{0}}^{\infty} e^{tx} \frac{1}{\nu \sqrt{2\pi}} e^{-\frac{(x-\mu)^2}{2 \nu^2} } dx \\
 &\int_{x_{0}}^{\infty} \frac{1}{\nu \sqrt{2\pi}} e^{-\frac{(x-\mu)^2}{2 \nu^2} + tx } dx \\
&\int_{x_{0}}^{\infty} \frac{1}{\nu \sqrt{2\pi}} e^{-\frac{x^2 -2x\mu +\mu^2 +2\nu^2tx}{2 \nu^2}} dx \\
\end{align}
and I'm stuck here
Thank you.

*

*Substitute the expression for the Normal pdf.

*Gather together the powers of $e$.

*Complete squares in the exponent of $e$ to get the square of something plus a constant.

*Take the constant powers of $e$ out of the integral.

*Change variables to turn the integral into a integral of the standard normal pdf from $-\infty$ to some number $a$.

