Laplace transform of normal distribution function? In my notes this was left as an exercise and I am a bit rusty with my calculus.
Starting with the definitions:
$$\mathcal{L}_X(t) = \mathbb{E}[e^{-tX}] = \int_0^\infty e^{-Xt}f(t)dt \;\;\text{ and }\;\;X\sim\mathcal{N}(\mu,\sigma)\;\;\text{ i.e. }\;\;f(t) = \frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{1}{2}\frac{(x-\mu)^2}{\sigma^2}}$$ so
$$\mathcal{L}_X(t) = \int_0^\infty \frac{1}{\sqrt{2\pi}\sigma} e^{-\frac{1}{2}\frac{(x-\mu)^2}{\sigma^2}} e^{-xt}dx$$
$$= \frac{1}{\sqrt{2\pi}\sigma} \int_0^\infty e^{-\frac{1}{2}\big(\frac{(x-\mu)^2}{\sigma^2}\big)-tx  }dx$$
suppose I now say $u = \frac{x-\mu}{\sigma}$ so that $x = u\sigma+\mu$ I get the following but have run out of ideas for how to continue:
$$ = \frac{1}{\sqrt{2\pi}\sigma} \int_{-\frac{\mu}{\sigma}}^\infty e^{-\frac{1}{2}u^2-t(u\sigma+\mu)  }du$$
I considered trying to complete the square but don't think it helped: the exponent becomes $-\frac{1}{2}(u^2-2tu\sigma-2t\mu)$ giving $-\frac{1}{2}((u-t\sigma)^2 -t^2\sigma^2+2t\mu)$, which doesn't seem to be any simpler to integrate. 
EDIT
After thinking about the comments I think this should be a double sided integral, as the expectation would be an integral over the probability distribution's domain (?)
So now I get
$$  \mathcal{L}_X(t) = \frac{1}{\sqrt{2\pi}\sigma} \int_{-\infty}^\infty e^{-\frac{1}{2}u^2-t(u\sigma+\mu)  }du$$
Now I try completing the square as above and get
$$ \frac{1}{\sqrt{2\pi}\sigma} \int_{-\infty}^\infty e^{-\frac{1}{2}((u-t\sigma)^2 -t^2\sigma^2+2t\mu) }du = \frac{1}{\sqrt{2\pi}\sigma} e^{-t^2\sigma^2-2t\mu} \int_{-\infty}^\infty 
e^{-\frac{1}{2}(u-t\sigma)^2}du$$
Now I make a substitution to say $z = \frac{1}{\sqrt{2}}(u-t\sigma)$ and then we get $u = \sqrt{2}z+t\sigma$, and $du = \sqrt{2}dz$ so finally:
$$ \frac{1}{\sqrt{2\pi}\sigma}e^{-t^2\sigma^2-2t\mu} \int_{-\infty}^\infty e^{-z^2} \sqrt{2}dz = 
\frac{1}{\sigma}e^{-t^2\sigma^2-2t\mu}$$
OK, so that is my attempt. I am very not confident about it, so any help/corrections are very welcome.
 A: Instead of doing the $u$-substitution:
$$
\begin{align}
\mathcal{L}_X (t) & = \int_{-\infty}^\infty e^{-tx} \frac{1}{\sqrt{2\pi}\sigma} e^{-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^2} \mathrm{d}x \\
& = \int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}\sigma} e^{-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^2-tx} \mathrm{d}x \\
& = \int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}\sigma} e^{-\frac{1}{2}\left(\left(\frac{x-\mu}{\sigma}\right)^2+2tx\right)} \mathrm{d}x \\
& = \int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}\sigma} e^{-\frac{1}{2}\left(\left(\frac{x-\mu}{\sigma}\right)^2+2t(x-\mu)+2t\mu\right)} \mathrm{d}x \\
& = \int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}\sigma} e^{-\frac{1}{2}\left(\left(\frac{x-\mu}{\sigma}\right)^2+2(t\sigma)\left(\frac{x-\mu}{\sigma}\right)+2t\mu\right)} \mathrm{d}x \\
& = \int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}\sigma} e^{-\frac{1}{2}\left(\left(\frac{x-\mu}{\sigma}\right)^2+2(t\sigma)\left(\frac{x-\mu}{\sigma}\right)+t^2\sigma^2-t^2\sigma^2+2t\mu\right)} \mathrm{d}x \\
& = \int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}\sigma} e^{-\frac{1}{2}\left(\left(\frac{x-\mu}{\sigma}\right)^2+2(t\sigma)\left(\frac{x-\mu}{\sigma}\right)+t^2\sigma^2\right)+\frac{1}{2}t^2\sigma^2-t\mu} \mathrm{d}x \\
& = \int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}\sigma} e^{-\frac{1}{2}\left(\frac{x-\mu}{\sigma}+t\sigma\right)^2+\frac{1}{2}t^2\sigma^2-t\mu} \mathrm{d}x \\
& = \int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}\sigma} e^{-\frac{1}{2}\left(\frac{x-\mu}{\sigma}+t\sigma\right)^2}e^{\frac{1}{2}t^2\sigma^2-t\mu} \mathrm{d}x \\
& = e^{\frac{1}{2}t^2\sigma^2-t\mu} \int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}\sigma} e^{-\frac{1}{2}\left(\frac{x-\mu}{\sigma}+t\sigma\right)^2} \mathrm{d}x \\
& = e^{\frac{1}{2}t^2\sigma^2-t\mu} \int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}\sigma} e^{-\frac{1}{2}\left(\frac{(x+t\sigma^2)-\mu}{\sigma}\right)^2} \mathrm{d}x
\end{align}
$$
Let $y=x+t\sigma^2$. $\mathrm{d}y=\mathrm{d}x$, $\lim_{x\to\infty}y\to\infty$, $\lim_{x\to-\infty}y\to-\infty$.
$$
\begin{align}
\mathcal{L}_X (t) & = e^{\frac{1}{2}t^2\sigma^2-t\mu} \int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}\sigma} e^{-\frac{1}{2}\left(\frac{y-\mu}{\sigma}\right)^2} \mathrm{d}y \\
& = e^{\frac{1}{2}t^2\sigma^2-t\mu}
\end{align}
$$
