Indeed it is different. Did I do something wrong? Is this the same you suggested when you said it could be expressed in terms of a sum over $\exp(ipx/2)$ ?

Also the final term looks like it is a divergent sum, am I correct? I quickly did a simulation with respect to $N$ and it looks like it diverges quickly.

Thank you! However, when I look for the geometric series sum $\exp(ixk/2)$ from $k=0$ to $k=N-1$, I see that it is $ \frac{\cos(1/4 (1 + N) x) \sin((N x)/4)}{ \sin(x/4)} + i \frac{sin((N x)/4) sin(1/4 (1 + N) x)} {sin(x/4)}$ in a trigonometric form. It is a complex expression. Did you mean that it can be written in a different way, or did you mean it is this exact sum?

Thank you! I appreciate it.

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?

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.

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.