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The Euler constant arises in the integrals

$$ \int_{0}^{\infty} \ln x ~e^{-x} dx=-\gamma $$

$$ \int_{0}^{\infty} (\ln x)^{2} e^{-x} dx=\gamma^2+\frac{\pi^2}{6}$$

$$ \int_{0}^{\infty} (\ln x)^{3} e^{-x} dx=-\gamma^3-\frac{\gamma \pi^2}{2}-2\zeta(3)$$

$$ \int_{0}^{\infty} (\ln x)^{4} e^{-x} dx=\gamma^4+\gamma^2 \pi^2+\frac{3 \pi^4}{2}+8 \gamma\zeta(3)$$

where $\zeta(\cdot)$ is Riemann zeta function.

Is there general form for the integral$$ \int_{0}^{\infty} (\ln x)^i e^{-x} dx; i=0,1,2,\dots.?$$

J.G.
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ELAWADY
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  • Note that $\int_0^\infty t^{z-1} e^{-t} dt=\Gamma(z)$. So an approach would be to take $i$ derivatives with respect to $z$ then plugg in $z=1$. – Zacky Aug 29 '19 at 19:05

1 Answers1

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Since $\int_0^\infty x^{s-1}e^{-x}dx=\Gamma(s)$, $\int_0^\infty x^{s-1}e^{-x}\ln^i xdx=\Gamma^{(i)}(s)$, so you're evaluating $\Gamma^{(i)}(1)$. Theorem 2.1 here gives the recursion $$\Gamma^{(i)}(1)=-\gamma\Gamma^{(i-1)}(1)+(i-1)!\sum_{k=0}^{i-2}\frac{(-1)^{i-k}\zeta(i-k)\Gamma^{(k)}(1)}{k!}.$$

J.G.
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