I'm trying to find a series representation for a integral, but I think there's something I'm missing, as even though the algebraic manipulations I'm doing are valid (I think!), the series representation of the integral (which I know to converge) ends up diverging. Here's what I'm doing (specifics omitted for brevity, p and q are large polynomials with non-integer exponents, deg(q)>deg(p), q(x) has no roots in the positive reals): $$\mathcal{I}=\int_0^\infty \frac{p(\lambda)}{q(\lambda)}e^{-k\lambda^2}d\lambda$$ $$\mathcal{I}=\int_0^\infty \frac{p(\lambda)}{q(\lambda)}\sum_{n=0}^\infty\frac{(-1)^nk^n\lambda^{2n}}{n!}d\lambda $$ $$\mathcal{I}=\sum_{n=0}^\infty\frac{(-1)^nk^n}{n!}\int_0^\infty \frac{\lambda^{2n}p(\lambda)}{q(\lambda)}d\lambda$$ $$\mathcal{I}=\sum_{n=0}^\infty\frac{(-1)^nk^n}{n!}C_i$$ , where $C_i$ is a constant involving the ratios of Gamma functions. Only problem is, the last expression for $\mathcal{I}$ does not converge (or, at least, oscillates enormously beyond the ability of my computer to calculate- it pegs the 200th partial sum at around $10^{2500}$), when in fact the first expression gives a accurate value of around 0.02.

My question is, what error (of concept or execution) have I wound up inadvertently committing and, if possible, what means is there to correct it so I wind up with a convergent series that can be used?



1 Answer 1


The exponential series converges uniformly on compact intervals but not on the whole of $[0,+\infty)$. You cannot interchange the integral with the sum of that series.

  • $\begingroup$ Alright, thanks. Does this mean that I could replace the upper limit of infinity with, say, 10 or 50 or some point where the integrand is essentially zero and have the series give an approximation to the integral? $\endgroup$ Jul 14, 2013 at 3:54
  • $\begingroup$ Yes. ${}{}{}{}$ $\endgroup$ Jul 14, 2013 at 3:56
  • $\begingroup$ One more question: Is there a uniformly convergent representation of exp(-x) on the positive reals? $\endgroup$ Jul 14, 2013 at 4:05

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