4

While messing around trying to calculate a power series for the Gamma function, I ran across this integral:

$$\int_0^\infty(\log t)^n e^{-t}\ dt,\ n \in \mathbb{N}$$

I've looked at it for a while and tried a couple of things, but I'm stumped. Is there a way to calculate this?

Javier
  • 7,298
  • Where do you get it? Try it here http://www.wolframalpha.com/ – NECing Mar 17 '13 at 00:15
  • Mathematica gives an answer in term of Euler's $\gamma$ and values of the $\zeta$ functions. For example, for $n=5$ the result is $-20 \gamma ^2 \zeta (3)-\frac{10 \pi ^2 \zeta (3)}{3}-24 \zeta (5)-\gamma ^5-\frac{5 \gamma ^3 \pi ^2}{3}-\frac{3 \gamma \pi ^4}{4}$. – Hans Engler Mar 17 '13 at 00:16
  • Just wondering - does this look to anyone else like the laplace transform of the function $(\log t)^n$ evaluated at $s=1$, or is that irrelevant? – Vincent Tjeng Mar 17 '13 at 04:15

2 Answers2

4

It is the $n$th derivative of the gamma function evaluated at the point $s=1$, where gamma function is given by

$$ \Gamma(s)=\int_{0}^{\infty} x^{s-1}e^{-x} dx \implies \Gamma^{(n)}(s)|_{s=1}=\int_{0}^{\infty} (\ln(x))^{n}e^{-x} dx .$$

Added: You can start from the point

$$ \psi(x) = \frac{d}{dx}\ln \Gamma(x) =\frac{\Gamma'(x)}{\Gamma(x)} \implies \Gamma'(x)=\Gamma(x)\psi(x),$$

where $\psi(x)$ is the digamma function.

Mhenni Benghorbal
  • 47,431
0

See the pattern? $$ \int_{0}^{\infty} \operatorname{ln} (t)^{7} \operatorname{e} ^{-t} d t = -\gamma^{7} - \frac{61 \pi^{6} \gamma}{24} - 84 \zeta (5) \pi^{2} - \frac{21 \pi^{4} \zeta (3)}{2} - 280 \zeta (3)^{2} \gamma - 70 \zeta (3) \pi^{2} \gamma^{2} - 504 \zeta (5) \gamma^{2} - \frac{21 \pi^{4} \gamma^{3}}{4} - 70 \zeta (3) \gamma^{4} - \frac{7 \pi^{2} \gamma^{5}}{2} - 720 \zeta (7) $$ (Computed by Maple)

GEdgar
  • 111,679