Calculate $$\int_k^\infty e^{ax}f(x) dx$$ where $f$ is the probability density of the normal distribution, i.e. $f(x) = \frac{1}{\sigma \sqrt{2 \pi}} e^{\frac{-(x-\mu)^2}{2 \sigma^2}}$. I tried calculating the integral for a general integrable $f(x)$ by using integration by parts without any luck. Solving it for the particular $f(x)$ by using same methods only yields very messy expressions. Any ideas?
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Partly using this answer, this is
\begin{align*} &\,\int_{k}^{\infty}\frac{1}{\sqrt{2\pi}\sigma}\exp\left(-\frac{(x-\mu)^2}{2\sigma^2}+ax\right)\mathrm{d}x\\=&\,\int_{k}^{\infty}\frac{1}{\sqrt{2\pi}\sigma}\exp\left(-\frac{(x-\mu-a\sigma^2)^2}{2\sigma^2}+a\mu+\frac{a^2\sigma^2}{2}\right)\mathrm{d}x\\ =&\,\exp\left(\frac{a^2\sigma^2}{2}+a\mu\right)\int_{k}^{\infty}\frac{1}{\sqrt{2\pi}\sigma}\exp\left(-\frac{(x-\mu-a\sigma^2)^2}{2\sigma^2}\right)\mathrm{d}x \\ &= \exp\left(\frac{a^2\sigma^2}{2}+a\mu\right)\left( 1- \Phi\left(\frac{k-(\mu + a \sigma^2)}{\sigma^2} \right) \right) \end{align*} where $\Phi(x)$ is the CDF of the standard normal distribution. This comes from simply using the fact that the integrand in the last integral is the normal variable $X \sim (\mu+a\sigma^2, \sigma)$, and the integral is the equivalent of $P(X \geq k)$. Then one just standardizes the variable.
