I'm trying to show that $Z(t)=\exp{(\sigma X(t)-(\sigma \mu + \frac{1}{2} \sigma^2)t)}$ is a martingale.
Attempt:
I want to show that $E[Z(t)|\mathcal{F}(s)] = Z(s)$
$E[Z(t)|\mathcal{F}(s)] = E[Z(t)/ Z(s) * Z(s)|\mathcal{F}(s)]$
$=Z(s)E[Z(t)/Z(s)|\mathcal{F}(s)] = Z(s)E[Z(t)/Z(s)]$
So at this point it looks like I need to determine that $E[Z(t)/Z(s)]=1$, but I'm not seeing it:
$E[Z(t)/Z(s)] = E[\exp{(\sigma (X(t)-X(s))-(t-s)(\sigma \mu + \frac{1}{2} \sigma^2))}]$
I'm aware that $X(t)-X(s) \sim N(0,t)$, but I'm still not able to see why the above expectation should be 1.
-----edit 1 (comment from Kurt G.)-------
$X(t)-X(s)~\sim N(0,t-s)$
Let $X(t)-X(s)=u$
$E[\exp{\sigma(X(t)-X(s))}] = E[\exp{(\sigma u)}]$
$=\int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi(t-s)}}e^{\frac{-u^2}{2(t-s)}}*e^{\sigma u}du =\int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi(t-s)}}e^{\sigma u - \frac{u^2}{2(t-s)}}du $
$=e^{\frac{1}{2}\sigma^2(t-s)} * \int_{-\infty}^{\infty} f(u)du = e^{\frac{1}{2}\sigma^2(t-s)}$
$E[Z(t)/Z(s)] = e^{-(t-s)(\sigma \mu + \frac{1}{2}\sigma^2)} * e^{\frac{1}{2}\sigma^2(t-s)}$
$=e^{-(t-s)\sigma \mu}$
$Z(s) E[Z(t)/Z(s)|\mathcal{F}(s)] = \exp{(\sigma X(s) - (\sigma \mu + \frac{1}{2}\sigma^2)s -(t-s)\sigma \mu})$
$=e^{\sigma X(s) - \sigma \mu t + \frac{1}{2} \sigma^2 s} \neq Z(s)$
Not sure where I've gone wrong.