Let's say you have a cost function $C(x)$ and you want to understand the expected cost if the input follows the normal distribution
$$X \sim \mathcal{N}(\mu,\sigma ^2)\\ $$
If I want to find my expected cost, I would need to integrate over the normal distribution
$$ E(C(x)) = \frac{1}{\sigma \sqrt{2 \pi }}\int_{-\infty}^\infty C(x)e^{-\frac{(x-\mu)^2}{2\sigma^2}}dx $$
I have found ways to approximate the integral of the normal distribution (e.g. taylor series), but nothing on how to integrate another function over it.
I can numerically estimate the integral given an arbitrary cost function, but I was wondering if there is any way to simplify it analytically and see how the expected cost changes as a function of $\mu$ and $\sigma$ and my cost function?
If the arbitrary cost function is too hard, would it be possible to solve for a simplier case such as $C(x) = x^2$? Even in this case I'm struggling to simplify it, although it just might not be possible.
HINT:
If $C(x)=x^2$, then we can find a closed form for its expected value. Recall that
$$I(a)=\int_{-\infty}^{\infty}e^{-ax^2}dx=\sqrt{\frac{\pi}{a}}$$
and
$$-I'(a)=\int_{-\infty}^{\infty}x^2e^{-ax^2}dx=\frac{\pi^{1/2}}{2a^{3/2}}$$
Then, change variables from $(x-\mu)/\sqrt{2\sigma}$ to $x$ and expand the quadratic inside the integral.
Can you finish from here?