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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

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2 Answers 2

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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.

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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:

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

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The probability density function looks like this:

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

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