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Take an elementary convergent integral like:

$\int^\infty_0 e^{- \lambda x} = \frac{1}{\lambda} $

If you series expand it every term and integrate term-by-term every term integrates to infinity. Is there a systematic way to cut-off the integral if you keep the $n^{th}$ term in the series so that you can reasonably approximate the integral to some quantified error?

EDIT: Clearly the series expansion is of little use in the above integral, but I am interested in a potential case where, for example, I find an integral that converges when I numerically integrate it but the analytical series diverges.

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  • $\begingroup$ This, unfortunately, is one of those times where series expansion of the integrand won't help you in any way... a number of improper integrals are just like that. $\endgroup$ – J. M. is a poor mathematician Jun 27 '12 at 16:20
  • $\begingroup$ Thanks, I slightly changed the question, although I don't think this will change the answer much. $\endgroup$ – DJBunk Jun 27 '12 at 16:25
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    $\begingroup$ I don't think things work that way! In fact $\int_0^\infty f(x)dx$ is the limit, when it exists, of $\int_0^R f(x)dx$ for $R \to \infty$. So first you integrate on a "finite" interaval, then you take the limit as the parameter $R$ tends to infinity. In the example you gave for $\lambda\ne0$ if you first integrate the power series of $f(x)=\exp(-\lambda x)$ over $[0,R]$, what you'll get is the power series of $F(R)=(1-\exp(-\lambda R))/\lambda$. So if you take the limit now you still get $1/\lambda$. $\endgroup$ – Mercy King Jun 27 '12 at 17:24

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