Integral of a random function How is it possible to evaluate the integral:
$$I(\mu,\sigma)=\int_0^{2\pi}\sin(\omega t)^2dt$$
where $\omega$ is a random variable having a normal distribution $N(\mu,\sigma)$?
What is the $pdf$ of $I(\mu,\sigma)$? Thanks
 A: You should rewrite the integrale in the form
$$
   I(\omega)=\int_0^{2\pi}dt\frac{1-\cos(2\omega t)}{2}=2\pi-\int_0^{2\pi}dt\cos(2\omega t).
$$
So, you have to evaluate the pdf of
$$
  I_0(\omega)=\int_0^{2\pi}dt\cos(2\omega t).
$$
Now, consider the integral
$$
  I_1(\omega)=\int_0^{2\pi}dt e^{2i\omega t}
$$
and you see that $I_0=Re(I_1)$ and so
$$
   E(e^{2i\omega t})=e^{2i\mu t-2\sigma^2 t^2}.
$$
and so
$$
  E(I_0(\omega))=\int_0^{2\pi}dt\cos(2\mu t)e^{-2\sigma^2 t^2}.
$$
Then,
$$
  E(e^{2i\omega (t+t')})=e^{2i\mu (t+t')-2\sigma^2 (t+t')^2}
$$
and so
$$
  E((I_0(\omega))^2)=\int_0^{2\pi}dt\int_0^{2\pi}dt'\cos(2\mu (t+t'))e^{-2\sigma^2 (t+t')^2}.
$$
This procedure can be repeated for the n-th moment to yield
$$
  E((I_0(\omega))^k)=\int_0^{2\pi}dt_1\ldots\int_0^{2\pi}dt_k\cos\left(2\mu \sum_{n=1}^kt^k\right)e^{-2\sigma^2\left(\sum_{n=1}^kt^k\right)^2}.
$$
These integrals involve erf and so, become even more involved with the order $k$. The pdf is not a normal one but the situation could be alleviated if the upper integration bound is allowed to go to infinity. In this case, the original integral does not seem to exist.
A: This is more of an oversized comment.
Let $Z$ we normal random variables with mean $\mu$ and variance $\sigma^2$. You define a new random variable as follows:
$$
     X(\omega) = \int_{0}^{2 \pi} \sin^2(Z(\omega) t) \mathrm{d}t = \pi \left(1-\operatorname{sinc}\left(4 \pi Z(\omega)\right) \right) = \begin{cases} 0 & Z(\omega)=0 \cr
\pi - \frac{\sin(4 \pi Z(\omega))}{4 Z(\omega)} & Z(\omega) \not= 0  \end{cases}
$$
Thus the problem reduces to finding distribution functions of $X = f(Z)$, for $f(z)$ given above.
Obviously $f(z) \geqslant 0$, but it is also bounded from above. Indeed:
$$
    |f(z)| \leqslant \pi + \pi | \operatorname{sinc}(4 \pi z)| \leqslant 2 \pi
$$
the bound above is rather generous:

Here is how the cdf looks for the special case of standard normal $Z$:

The probability density function looks like this:

As already noted by @Jon, computing moments is within immediate reach, at least numerically:
$$
     \mathbb{E}\left(X^r \right) = \pi^r \mathbb{E}\left( \left(1-\operatorname{sinc}(4 \pi Z)\right)^r \right)
$$
