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I need to compute a time-average which is of the form

$$ \lim_\limits{T \to \infty} \frac{1}{T} \int_0^T \frac{\mathrm{d}t}{1+\sum_k c_k e^{i x_k t}} $$

where $x_k \in \mathbb{R}$, $c_k \in \mathbb{C}$ and the sum in the denominator is finite, $k=1,\ldots,K$.

I tried substituting $z = e^{it}$ and using the residue theorem but in the end I didn't get far. Is there some way of getting a closed form expression for the limit? Alternatively, I don't mind if I have to use numerical techniques, but preferably I'd do it on a limit where the convergence is more controlled.

user3237992
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    Do we have any guarantees the sum in the denominator is always nonzero in the region of integration? – eyeballfrog Jul 05 '22 at 14:41
  • Adding the zero term in the sum with $c_0=1,x_0=0$ might make the argument easier – Snoop Jul 05 '22 at 14:48
  • Actually that can be done in the case I'm looking at. I'll modify the question. The original integrand is $e^{i(x_k-x_{k'})t}/\sum_l|\sum_{k,k'} c_{k,k',l} e^{i(x_k-x_{k'})t}|^2$. One of the exponentials in the nominator is the same as another exponential in the denominator. – user3237992 Jul 05 '22 at 21:49

1 Answers1

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If the $c_k$ are small enough (say, $\sum_k |c_k| < 1$), then expanding the inverse in a Taylor series will work: $$ \frac{1}{1+\sum_k c_k e^{ix_k t}}=1-\sum_kc_ke^{ix_k t}+\sum_{k,l}c_kc_le^{i(x_k+x_l)t}-\sum_{k,l,m}c_kc_lc_me^{i(x_k+x_l+x_m)t}+\ldots $$ This can be regrouped by combining terms with the same number of copies of each $x_i$: $$ =\sum_{n_1=0}^{\infty}\sum_{n_2=0}^{\infty}\cdots\sum_{n_K=0}^{\infty}(n_1+n_2+\ldots+n_K)!\exp\left(i(n_1 x_1+n_2 x_2 +\ldots + n_K x_K)t\right)\prod_{i=1}^{K}\frac{(-c_i)^{n_i}}{n_i!} $$ Each exponential term averages to $0$ as $T\rightarrow\infty$, unless the weighted sum of the $x$'s in the exponent is zero, in which case the term averages to (indeed, uniformly equals) $1$. So the form of the result will depend on exactly which sums of the $x$'s vanish. If none do (except for the trivial sum), the result is $1$. If $x_1=-x_2$ (and all the other $x_i$ are algebraically independent), then the non-vanishing terms will be those where $n_1=n_2$ (and all the other $n_i=0$), leading to $$ \sum_{n=0}^{\infty}{{2n}\choose{n}}(c_1 c_2)^n=-1-2c_1c_2+\frac{1}{\sqrt{1-4c_1c_2}}. $$ And so on, for arbitrarily complicated rational relationships between the $x_i$.

mjqxxxx
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  • Nice solution. I'll leave the question open though in case a more general solution is possible, without having to resort to a convergent expansion.

    I realised that in the specific case I need, the integral kernel in its original form is something like $\exp(-i x_{k,k'} t) / \sum_{k,k'} c_{k,k'} \exp(-i x_{k,k'} t)$. Here $x_{k,k'} = -x_{k',k}$ holds and also $c_{k,k'} = c_{k',k}^*$ where the asterisk denotes complex conjugation. Further, it can be safely assumed that $\sum_k |c_{k,k}|^2 > 0$.

    Would this be enough for a solution that does not need a Taylor series?

     – user3237992 Jul 23 '22 at 15:45