Finding characteristic function of normal distribution for positive numbers only $X$~$N\left(\mu,\sigma ^2\right),E\left[e^{\eta X}\right]=?$

Question

We have:
$X$~$N\left(\mu,\sigma^2\right)$
We know that $\eta\ge0$ and thus I need to find $E\left[e^{\eta X}\right]$.
So what I thought of doing:
$$E\left[e^{\eta X}\right]=E\left[e^{\left(-j\right)j\eta X}\right]=\Phi _{_X}\left(-j\eta \right)=e^{j\mu \omega -\frac{1}{2}\sigma ^2\omega ^2}|_{_{_{\omega =-\eta j}}}$$ Problem is, we need $\eta\ge0$
So what I thought:
Normal distribution is symmetric, thus I can say the characteristic function from $0$ to $\infty$ and from $-\infty$ to $0$ is equal?? so I just find the characteristic function normally and then divide by 2 to find for $\eta\ge0$.
Am I right in doing it? From doing what I said I received:
$$E\left[e^{\omega X}\right]=e^{\mu \omega +\frac{1}{2}\sigma ^2\omega ^2}\rightarrow \:E\left[e^{\eta X}\right]=\frac{e^{\mu \:\omega \:+\frac{1}{2}\sigma \:^2\omega \:^2}}{2}$$ Is this true?