Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?


share|cite|improve this question
up vote 2 down vote accepted

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.

share|cite|improve this answer
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? – logosintegralis Jul 14 '13 at 3:54
Yes. ${}{}{}{}$ – Mariano Suárez-Alvarez Jul 14 '13 at 3:56
One more question: Is there a uniformly convergent representation of exp(-x) on the positive reals? – logosintegralis Jul 14 '13 at 4:05

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.