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?

  • $\begingroup$ Where do you get it? Try it here wolframalpha.com $\endgroup$
    – NECing
    Commented Mar 17, 2013 at 0:15
  • $\begingroup$ Mathematica gives an answer in term of Euler's $\gamma$ and values of the $\zeta$ functions. For example, for $n=5$ the result is $-20 \gamma ^2 \zeta (3)-\frac{10 \pi ^2 \zeta (3)}{3}-24 \zeta (5)-\gamma ^5-\frac{5 \gamma ^3 \pi ^2}{3}-\frac{3 \gamma \pi ^4}{4}$. $\endgroup$ Commented Mar 17, 2013 at 0:16
  • $\begingroup$ 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? $\endgroup$ Commented Mar 17, 2013 at 4:15

2 Answers 2


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.

  • 3
    $\begingroup$ Well, yes, that's where I got it from. $\endgroup$
    – Javier
    Commented Mar 17, 2013 at 0:24
  • $\begingroup$ @JavierBadia: Well, you should have mentioned in your post that you knew it is the nth derivative of the gamma function! $\endgroup$ Commented Mar 17, 2013 at 1:17
  • $\begingroup$ Sorry, I should have been clearer. Expanding $t^{s-1}e^{-t}$ as a power series in $s-1$, you get this integral where the nth derivative goes. But in my question I asked how to calculate it, so this is besides the point. $\endgroup$
    – Javier
    Commented Mar 17, 2013 at 1:26
  • $\begingroup$ @JavierBadia: No problem. $\endgroup$ Commented Mar 17, 2013 at 1:34

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)

  • $\begingroup$ Actually, sorry, but I don't. $\endgroup$
    – Javier
    Commented Mar 17, 2013 at 1:27

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