# Express integral in terms of CDF of standard normal variable

I am stuck with the following problem. If we let $Z\sim \mathcal{N}(\mu,\sigma^2)$ and let $F(z)$ denote its CDF, and denote by $\Phi(x)$ the CDF of the standard normal distribution, how can we express integrals of the type $$\int_{-\infty}^a t(z)dF(z)$$ for some function $t(z)$ in terms of $\Phi(x)$?

For example for $t(z)=1$ we get $$\int_{-\infty}^a dF(z)=P(Z\leq a)=\Phi\left(\frac{a-\mu}{\sigma}\right)$$ but what about the more general case? Is there a way to do this? Specifically I am interested in the case $t(z)=e^z$. I hope the question makes sense.

Define: $$U:=\frac{Z-\mu}{\sigma}$$ Then $U$ has standard normal distribution, and for any suitable function $g$ we have: $$\int g(z)dF(z)=\mathbb Eg(Z)=\mathbb Eg(\mu+\sigma U)=\int g(\mu+\sigma u)d\Phi(u)$$

Observe that $$\int_{-\infty}^at(z)dF(z)=\int g(z)dF(z)$$ for $g(z):=t(z)1_{(-\infty,a]}(z)$

• of course, thanks! Feb 23 '17 at 16:15