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In my work I came across the following integral which stems from basically computing the statistics of an output signal of a nonlinear system given white noise at the input, so the integral is (if I did not make a mistake) represent a probability.

$$\int_{0}^{\infty} \frac{\lambda_i^{N_i} x^{N_i-1} e^{-\lambda_i x}}{(N_i-1)!} \prod_{j=1,j\neq i}^{N} (e^{-\lambda_j x} \sum_{m=0}^{N_j-1}\frac{(\lambda_j x)^m}{m!} ) \prod_{l=N+1}^M (1-e^{-\lambda_l x} \sum_{n=0}^{N_l - 1} \frac{(\lambda_n x)^n}{n!}) dx $$

$N_i$ are natural numbers in the range of 1 to 8, $M$ is a natural in the range of 20, $N$ equals 8. $\lambda_i$ are positive real values. There is actually another sum across $i$, however, due to linearity of the integral it is of no interest here. That is where the i index comes from.

I had a look at the residuum theorem, but nothing seems to exist for this complicated case which could help solving this integral. One can formally compute the products involving the polynomials and then integrate each sum, however, this gets very difficult to tract afterwards if I want to evaluate this as the number of summands, without any relation between the resulting coefficients, is rather large.

Anyone got an idea how to solve this one?

StrictlyStationaryPoster
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    Let $p_\lambda(a,,b)$ denote $P(a\le X\le b)$ for $X\sim\operatorname{Poisson}(\lambda)$ so your integral is$$\lambda_i^{2-N_i}\int_0^\infty p_{\lambda x}(N_i-1,,N_i-1)\prod_{j\ne i}p_{\lambda_j x}(0,,N_i-1)\prod_{l=N+1}^Mp_{\lambda_nx}(N_i,,\infty)dx.$$There's probably some clever way to restate that as a statistics problem that doesn't require much integration. But before we go further, is that $\lambda_i^{2-N_i}$ factor definitely what you intended? Or was $\lambda_ix^{N_i-1}$ meant to be $(\lambda_ix)^{N_i-1}$? – J.G. Jun 24 '22 at 12:31
  • Thank you, it was suppsed to be $\lambda_i^{N_i}$ and not $\lambda_i$, I edited the original post. Now it should be correct. – StrictlyStationaryPoster Jun 24 '22 at 12:43
  • Ah, so we want the prefactor in my formalism to be neither $\lambda_i^{2-N_i}$ nor $1$, but $\lambda$. – J.G. Jun 24 '22 at 12:46

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