Re: Did you hear about Euler-Mascheroni integrals?

David W. Cantrell wrote:
I(n) = (-1)^n * \int_0^\infty [ln(z)]^n * exp(-z) dz.

Your guess is correct. The most important part of the integral lies near 0,
and so, for large n, a reasonable approximation of I(n) can be obtained by
replacing \infty by 1 and exp(-z) by 1, giving

(-1)^n * \int_0^1 [ln(z)]^n dz

which is indeed n! .

How can I find out of what size the error of this approximation is? From
numerical experiments I guess that n! = I(n) + O( e^{-n} ) holds. But
how to prove this?

Thanks for any hints.

Joachim
.