How to compute the Laplace transform of a normally distributed density function?

Question

I'm trying to compute the Laplace transform of a random variable $X$ with a normal density function. So $X \sim N(\mu, \sigma^{2})$. This means that $f_{X} (t) = \frac{1}{\sqrt{2 \pi \sigma^{2}}} e^{-\frac{(t-\mu)^{2}}{2 \sigma^{2}}}$. Let $f(t) := f_{X} (t)$ for convenience. Then the Laplace transform $F$ amounts to: $F(s) = \int_{0}^{\infty} f(t) e^{-st} dt$.

Somewhere along the calculation of this integral, I get stuck. Here's what I tried so far:

\begin{align*} \int_{0}^{\infty} \frac{1}{\sqrt{2 \pi \sigma^{2}}} e^{-\frac{(t-\mu)^{2}}{2 \sigma^{2}}} e^{-st} &= \frac{1}{\sqrt{2 \pi \sigma^{2}}} \int_{0}^{\infty} e^{- \frac{t^{2} - 2 \mu t + \mu^{2} + 2 \sigma^{2} s t}{2 \sigma^{2}} } dt\\ &= \frac{1}{\sqrt{2 \pi \sigma^{2}}} \int_{0}^{\infty} e^{- \frac{t^{2} + 2t(\sigma^{2} s - \mu) + \mu^{2}}{}} dt \\ &= \frac{1}{\sqrt{2 \pi \sigma^{2}}} \int_{0}^{\infty} e^{- \frac{(t+(\sigma^{2}s - \mu))^{2} - (\sigma^{2} s - \mu)^{2} + \mu^{2}}{2 \sigma^{2}}} dt \\ &= \frac{1}{\sqrt{2 \pi \sigma^{2}}} e^{ \frac{(\sigma^{2} s - \mu)^{2} - \mu^{2}}{2 \sigma^{2}}} \int_{0}^{\infty} e^{- \frac{(t+(\sigma^{2} s - \mu))^{2}}{2 \sigma^{2}}} dt \\ &= \frac{1}{\sqrt{2 \pi \sigma^{2}}} e^{-s \mu} e^{\frac{\sigma^{2} s^{2}}{2}} \int_{0}^{\infty} e^{- \frac{(t+(\sigma^{2} s - \mu))^{2}}{2 \sigma^{2}}} dt \end{align*}

At this point, I'm stuck. I don't know how to evaluate the last integral. Furthermore, when I put it in wolframalpha, it seems the result as a whole is not correct (the error function suddenly appears, which is not present in the correct result). So maybe I made a mistake along the way.

So I suppose I have two questions:

1. Did I make a mistake along the way? If so, which one, and how do I fix it?
2. If I did not make a mistake yet, how does one calculate the last integral?

Answer 1

Let's just do the case $\mu=0$, $\sigma=1$ and use the standard normal c.d.f. $$\Phi(x)=\frac1{\sqrt{2\pi}}\int_{-\infty}^xe^{-t^2/2}\,dt=\frac12+\frac12\mathrm{erf}(x/\sqrt 2).$$ By completing the square, $$\begin{eqnarray*}F(s)&=&\frac1{\sqrt{2\pi}}\int_0^\infty e^{-t^2/2}e^{-st}\,dt\\&=&\frac1{\sqrt{2\pi}}\int_0^\infty \exp\left(-\frac12(t^2+2st)\right)\,dt\\ &=&\frac1{\sqrt{2\pi}}\int_0^\infty \exp\left(-\frac12\left((t+s)^2-s^2\right)\right)\,dt\\ &=&e^{s^2/2}\frac1{\sqrt{2\pi}}\int_s^\infty e^{-t^2/2}\,dt\\&=&e^{s^2/2}(1-\Phi(s)).\end{eqnarray*}$$ But erf is not an elementary function, so it seems it cannot be simplified further.