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.