This should be a pretty simply question, but I can't seem to quite wrap my head around it.
Suppose that $x$ is normally distributed with mean zero and variance $\sigma^2$. I need to compute the partial expectations $$\int_{a}^\infty e^{kx}$$ for different combinations of the lower bound $a$ and the multiple $k$. I know that if $k=1$, we can use the usual lognormal partial expectation formula: $$\int_{a}^\infty e^{x}=e^{\sigma^2/2}\Phi\left(\frac{\sigma^2 - a}{\sigma}\right).$$ And it's pretty clear that if $a=\infty$, then for $k\geq 1$ $$\int_{-\infty}^\infty e^{kx}=e^{\sigma^2k^2/2}$$ because $kx$ is normally distributed with mean zero and variance $k^2\sigma^2$.
For the general case, I think it is $$\int_{a}^\infty e^{kx}=e^{\sigma^2k^2/2}\Phi\left(\frac{\sigma^2k^2 - a/k}{\sigma k}\right).$$ Is this correct? Thanks in advance.
Edit: Also, I would be interested to know if there is an easy way to compute these partial expectations if $k$ is negative.
Not quite correct, here are the steps and it will solve that worrying negative $k$ problem that appeared.
Completing the square with the exponent term in the normal distribution we have $$ \begin{align} e^{kx}e^{-\frac{(x-\mu))^2}{2\sigma^2}} &= \exp\left\{-\frac{1}{2\sigma^2}\left[x^2 - 2 \mu x + \mu^2 - 2\sigma^2kx\right] \right\} \\ &= \exp \left\{x^2 - 2(\mu +\sigma^2 k)x +\mu^2 \right\} \\ &= \exp\left\{(x-[\mu + \sigma^2 k])^2 + \mu^2 - (\mu+\sigma^2 k)^2\right\}\\ &= \exp\left\{-\frac{\left(x-[\mu + \sigma^2 k]\right)^2}{2\sigma^2}\right\} \exp\left\{\mu k + \frac{\sigma^2k^2}{2}\right\}, \end{align} $$ So plugging that into the integral expression we have $$ \begin{align} \int_a^{\infty}e^{kx}q(x;\mu,\sigma^2)dx &= e^{\mu k + \frac{\sigma^2k^2}{2}}\int_a^{\infty}\frac{1}{\sqrt{2\sigma^2}}e^{-\frac{1}{2\sigma^2}\left(x - [\mu + \sigma^2 k ]\right)^2}dx \\ &= e^{\mu k + \frac{\sigma^2k^2}{2}}\left[1 - \Phi\left(\frac{a-\mu-\sigma^2k}{\sigma}\right) \right]. \end{align} $$