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The sum I am looking for is the following sum as $M \to \infty$:

$$ L(\omega) = \sum_{m = 1}^{M} \frac{\sin\left( N \frac{\omega_m - \omega}{2} \right)}{\sin\left( \frac{\omega_m - \omega}{2} \right)} \cos\left(N \frac{\omega_m - \omega}{2} + \beta_m \right) $$

where

  • $\omega_m$ is a random number from a Gaussian distribution having the parameters (mean $\mu$ and variance $\sigma^2$). $$ \omega_m \sim \mathcal{N}(\mu, \sigma^2) $$
  • the $\beta_m$ are random numbers drawn from a uniform distribution from $-\pi$ to $+\pi$,

$$ \beta_m \sim \mathcal{U}[-\pi, +\pi] $$

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    $\begingroup$ since $L$ is a random quantity, in what sense do you wish to evaluate it? The expectation value of $L$ is zero... $\endgroup$
    – Carlo Beenakker
    Commented Dec 2, 2022 at 16:33

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Since the average over $\beta$ will give a vanishing expectation value of $L$, let me omit it for now and set $\beta=0$. I will also simplify the question by setting $\omega=\mu=0$ and $\sigma=1$. Then the expectation value of $L$ has a compact expression $$\mathbb{E}[L(0)]=M\;\int_{-\infty}^\infty \frac{dx}{\sqrt{2\pi}}e^{-x^2/2} \frac{\sin\left( N x/2 \right)}{\sin\left( x/2\right)} \cos\left(N x/2 \right)=Me^{-\frac{1}{8} (2 N-1)^2} \sum _{j=1}^N e^{-\frac{1}{2} (j-1) (j-2 N)}.$$

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  • $\begingroup$ Thank you for the answer. Really appreciate it. I think I did some mistake arriving at this formula from my original physical problem. Indeed, $\beta$ term makes it a zero expectation and that’s not expected of the original physical problem. However, I was also looking for a solution when $\beta$ is not present. Thank you for this. I will again look into the physical problem. $\endgroup$
    – CfourPiO
    Commented Dec 4, 2022 at 1:55
  • $\begingroup$ I again looked at my physical problem and found that I do not need the cosine term in the expression at all. So, the expectation of $L$ is a summation over the Dirictlet Kernel function ($\sin(N x/2)/\sin(x/2)$) when $x$ is Gaussian distributed. I actually tried solving the integral that you suggested in the answer on Mathematica. However, I couldn't get a solution. It takes long to compute and doesn't compute it. How did you arrive at such an expression? I want to follow the same to derive the expression when there is no cosine term. $\endgroup$
    – CfourPiO
    Commented Dec 5, 2022 at 13:29
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    $\begingroup$ which integral are you referring to? the integral in my answer can be readily evaluated for any given $N$; presumably if you want to ask a new question you will want to start a new post (one question per post is the general principle here) $\endgroup$
    – Carlo Beenakker
    Commented Dec 5, 2022 at 13:36
  • $\begingroup$ I was referring to the integral in your answer. Yes, the expression is very nice and can be evaluated for any $N$. I was just curious how this expression is derived from the integral. Is it manually done by parts? $\endgroup$
    – CfourPiO
    Commented Dec 5, 2022 at 13:40
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    $\begingroup$ the integrand can be worked out as a product of $e^{-x^2/2}$ times a sum of exponentials $e^{ipx/2}$, with integer $p$, which can then be integrated term by term. $\endgroup$
    – Carlo Beenakker
    Commented Dec 5, 2022 at 13:43

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