Expected value of normally distributed variable at $e^{\theta x}$ I need help in working out some equations regarding the expected value.
The question is:

Let $X$ be a normally distributed random variable with mean $\mu$ and variance $\sigma^2$. Show that:
$E[e^{\theta X } ]= e^{\theta \mu + \sigma^2 \theta^2 /2}$

So far, I've tried to solve this by working with the definition of the expected value (the integral), and I've tried to work backwards from the solution.
At the end, I end up with the following equation i am simply not able to show:
$$\int_{-\infty}^\infty e^{\theta x} \frac 1 { \sqrt{2\sigma^2 \pi}} e^{-0,5(x-\mu)^2/\sigma^2} \, dx = \int_{-\infty}^\infty  \frac 1 {\sqrt{2\sigma^2 \pi}}e^{-0,5(x-\mu-\theta \sigma^2)^2/\sigma^2  }{e^{\mu \theta +0.5 \theta^2 \sigma^2}} \, dx $$
 A: If $X\sim N(\mu,\sigma^2)$ then $Z= \dfrac{X-\mu}\sigma\sim N(0,1)$ and $X=\mu+\sigma Z$, so we can say
$$
\operatorname{E}(e^{\theta X}) = \operatorname{E}(e^{\theta(\mu+\sigma Z)}) = \operatorname{E}(e^{\theta\mu} e^{\theta\sigma Z}).
$$
In the last expression, the factor $e^{\theta\mu}$ is not random; therefore it can be pulled out, getting
$$
e^{\theta\mu} \operatorname{E} (e^{\theta\sigma Z}).
$$
Let $\chi(\theta) = \operatorname{E}(e^{\theta Z}).$ Then we have
$$
e^{\theta\mu} \operatorname{E}(e^{\theta\sigma Z}) = e^{\theta\mu} \chi(\theta\sigma).
$$
So if you can find $\chi(\theta) = \operatorname{E}(e^{\theta Z}),$ then you can just put $\theta\sigma$ where $\theta$ was, getting $\chi(\theta\sigma),$ and then multiply it by $e^{\theta\mu}.$
Now let's find $\chi(\theta) = \operatorname{E}(e^{\theta Z}).$ Let $\varphi(z) = \dfrac 1 {\sqrt{2\pi}} e^{-z^2/2}.$
\begin{align}
\chi(\theta) = \operatorname{E}(e^{\theta Z}) & = \int_{-\infty}^\infty e^{\theta z} \varphi(z)\,dz = \int_{-\infty}^\infty e^{\theta z} \frac 1 {\sqrt{2\pi}} e^{-z^2/2} \, dz \\[10pt]
& = \frac 1 {\sqrt{2\pi}} \int_{-\infty}^\infty \exp \left( -\frac{z^2} 2 + \theta z \right) \, dz \\[10pt]
& = \frac 1 {\sqrt{2\pi}} \int_{-\infty}^\infty \exp \left( -\frac 1 2 \left( z^2 + 2\theta z \right) \right) \, dz \\[4pt]
& \qquad \text{This begs for completing the square:} \\[10pt]
& = \frac 1 {\sqrt{2\pi}} \int_{-\infty}^\infty \exp \left( -\frac 1 2 \left( z^2 + 2\theta z + \theta^2 \right) \right) \cdot \exp\left( +\frac 1 2 \theta^2 \right) \, dz \\[6pt]
& \qquad \text{The factor } \exp\left(+\frac 1 2 \theta^2\right) \text{ does not depend on } z, \\
& \qquad \text{so it can be pulled out of the integral}: \\[10pt]
& = \exp \left( \frac 1 2 \theta^2\right) \frac 1 {\sqrt{2\pi}} \int_{-\infty}^\infty \exp\left( -\frac 1 2 (z+\theta)^2 \right) \,dz \\[10pt]
& = \exp\left( \frac 1 2 \theta^2\right) \cdot 1.
\end{align}
We get $1$ because, via the substitution that says $u = z+\theta$ and $du = dz$, we see that we have
$$
\frac 1 {\sqrt{2\pi}} \int_{-\infty}^\infty e^{-u^2/2} \,du = 1.
$$
