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
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
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.
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) $