I'm not getting the standard answer of but I'm not sure what's wrong with my reasoning. Any insight appreciated.
Let $Z \sim N(0,1)$. We want to find the MGF of $\left|Z\right|$. By definition:
$$\psi_{\left|Z\right|}(t) = \int_{-\infty}^{\infty} e^{t\left|z\right|}f_z(z)dz$$
Then since $e^{t\left|z\right|}f_z(z)$ is a product of two even functions, it is also an even function, so it's symmetrical around the y-axis and:
$$\psi_{\left|Z\right|}(t) = 2\int_{0}^{\infty} e^{tz}f_z(z)dz$$
Now we can substitute in the PDF definition:
$$\psi_{\left|Z\right|}(t) = 2\int_{0}^{\infty} e^{tz}\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}z^2}dz$$
Now we can do some plain old algebra:
$$\psi_{\left|Z\right|}(t) = \frac{2}{\sqrt{2\pi}}\int_{0}^{\infty} e^{tz}e^{-\frac{1}{2}z^2}dz$$
$$\psi_{\left|Z\right|}(t) = \frac{2}{\sqrt{2\pi}}\int_{0}^{\infty} e^{-\frac{1}{2}z^2 + tz}dz$$
$$e^{-\frac{1}{2}t^2}\cdot\psi_{\left|Z\right|}(t) = e^{-\frac{1}{2}t^2}\frac{2}{\sqrt{2\pi}}\int_{0}^{\infty} e^{-\frac{1}{2}z^2 + tz}dz$$
$$e^{-\frac{1}{2}t^2}\cdot\psi_{\left|Z\right|}(t) = \frac{2}{\sqrt{2\pi}}\int_{0}^{\infty} e^{-\frac{1}{2}z^2 + tz - \frac{1}{2}t^2}dz$$
$$e^{-\frac{1}{2}t^2}\cdot\psi_{\left|Z\right|}(t) = \frac{2}{\sqrt{2\pi}}\int_{0}^{\infty} e^{-\frac{1}{2}(z^2 - 2tz + t^2)}dz$$
$$e^{-\frac{1}{2}t^2}\cdot\psi_{\left|Z\right|}(t) = \frac{2}{\sqrt{2\pi}}\int_{0}^{\infty} e^{-\frac{1}{2}(z-t)^2}dz$$
Put that aside for a moment. From the fact that PDF formula for the normal distribution sums to 1 it is possible to deduce that:
$$\int_{-\infty}^{\infty} e^{-\frac{1}{2\sigma^2}(x-\mu)^2}dx = \sigma\sqrt{2\pi}$$
Then since the function is even we can infer:
$$\int_{0}^{\infty} e^{-\frac{1}{2\sigma^2}(x-\mu)^2}dx = \frac{\sigma\sqrt{2\pi}}{2}$$
So now we can go back, using $\sigma=1, \mu=t$:
$$e^{-\frac{1}{2}t^2}\cdot\psi_{\left|Z\right|}(t) = \frac{2}{\sqrt{2\pi}}\frac{\sqrt{2\pi}}{2}$$
Cancelling gives us:
$$\psi_{\left|Z\right|}(t) = e^{\frac{1}{2}t^2}$$
Then to double check my answer I take the derivative and plug in 0, expecting to get $\mathbb{E}(\left|Z\right|) = \sqrt{\frac{2}{\pi}}$.
$$\psi'_{\left|Z\right|}(t) = te^{\frac{1}{2}t^2}$$ $$\psi'_{\left|Z\right|}(0) = 0$$
Not what I expected. Where did I go wrong?
Let the standard normal density function be $\phi(u)$ and the cumulative distribution function be $\Phi(u)$. Then the final answer is, as given in a comment, $2\Phi(t) e^{\frac12 t^2} $
To sum it up: \begin{eqnarray} M_{|U|}(t) &= \int_{-\infty}^\infty e^{t|u|} \phi(u)\; du \\ &=2 \int_0^\infty e^{t|u|} \frac1{\sqrt{2\pi}} e^{-\frac12 u^2}\;du \\ &=\frac{2}{\sqrt{2\pi}} e^{\frac12 t^2} \int_0^\infty e^{-\frac12 (u-t)^2} \; du \\ &= \frac{2}{\sqrt{2\pi}} e^{\frac12 t^2} \int_{-t}^\infty e^{-\frac12 z^2}\: dz \\ &= 2\Phi(t) e^{\frac12 t^2} \end{eqnarray}