How do we estimate the following kind of integrals - $\int_0^{\infty} e^{-x^r} dx $ For r > 1. I basically need this to find normalization constants for probability measures with densities proportional to the integrand(had to write the last line to meet quality standards).
Asked
Active
Viewed 62 times
1 Answers
0
$$\int_0^\infty \mathrm{e}^{-x^r}\, \mathrm{d}x = \Gamma\left(1+\frac{1}{r}\right) \quad , r>1$$
Where $\Gamma$ is the gamma function.
This is 3.326 in Gradshteyn & Ryzhik (1979/1980), using $\Gamma\left(1+\frac{1}{r}\right) = \frac{1}{r}\Gamma\left(\frac{1}{r}\right)$. Turns out the above is valid for $\Re\ r > 0$.
Eric Towers
- 67,037
-
-
@ClaudeLeibovici: It is valid for any $r$ with positive real part, as I noted in the last line. – Eric Towers Jun 15 '14 at 04:23