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It is known that $\int_{0}^{\infty}\left(\frac{\log x}{e^x}\right)^n dx=-\gamma$ for $n=1$ and for $n=2$ we have :$\frac{1}{12}(\pi^2+6(\gamma+\log 2)^2)$ and for $n=3$ we have this form ,

What I have noted is that for $n$ odd the integrand is negative and for $n$ even the integrand is positive , Now my question here is : How I prove that integral always have a closed form for any fixed integer $n$ ? And can we expect a general formula or any reccurence relation for that integrand for aribitrary integer $n$ ? Also what about its irrationality ?

For attempt: I have tried to use this method for $n=1$ in order to generalise it for any fixed $n$ : \begin{equation} \int\limits_{0}^{\infty} \mathrm{e}^{-x} x^k \mathrm{d} x = \Gamma(k+1) \end{equation} Differentiate with respect to $k$ \begin{equation} \int\limits_{0}^{\infty} \mathrm{e}^{-x} x^k \mathrm{ln}(x) \mathrm{d} x = \frac{d\Gamma(k+1)}{dt} = \Gamma(k+1) \psi^{(0)}(k+1) \end{equation}

Taking the limit $k \to 0$ yields \begin{equation} \int\limits_{0}^{\infty} \mathrm{e}^{-x} \mathrm{ln}(x) \mathrm{d} x = \Gamma(1) \psi^{(0)}(1) = -\gamma \end{equation} Now for $n=2$ I should use the integration by part I can up to the result but by this way it would be two long and complicated , Then I want probably an abreviate path to deduce any general formula for any arbitary $n$ ?

it is clear only it is irrational for $n=1,2$

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  • $\begingroup$ $$\begin{align} \int_0^\infty \left(\frac{\log(x)}{e^x}\right)^n\,dx&=\int_0^\infty e^{-nx}\log^n(x)\,dx\\\\ &=\frac1n\sum_{k=0}^n\binom{n}{k}\log^{n-k}(n)\int_0^\infty e^{-x}\log^n(x)\,dx\\\\ &=\frac1n\sum_{k=0}^n\binom{n}{k}\log^{n-k}(n) \left.\left(\frac{d^n \Gamma(x+1)}{dx^n}\right)\right|_{x=0} \end{align}$$ $\endgroup$
    – Mark Viola
    Jul 14, 2020 at 19:01
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    $\begingroup$ Note that $\gamma$ is not known to be irrational. $\endgroup$
    – GEdgar
    Jul 14, 2020 at 20:39

1 Answer 1

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Not a complete answer, but I'm sure it'll help:

Let $$I(a) = \int_0^\infty e^{-nx}x^a\,dx$$ $$\implies \frac{d^nI(a)}{da^n} = \int_0^\infty e^{-nx}x^a(\ln x)^n\,dx$$ Put $nx \rightarrow v$ in the first integral to get: $$I(a) = \frac1{n^{1+a}}\int_0^\infty e^{-v}v^a\,dv$$ $$\implies I(a) = \frac{\Gamma(1+a)}{n^{1+a}}$$ Now $$\implies \frac{d^nI(a)}{da^n}\bigg|_{a=0} = \frac{d^n}{da^n}\left(\frac{\Gamma(1+a)}{n^{1+a}}\right)\bigg|_{a=0}$$ Which evaluates to: $$\frac1{n}\sum_{k=0}^n(-1)^k\binom{n}{k}\Gamma^{(n-k)}(1+a)\ln^k(n)\bigg|_{a=0}$$ Where $\Gamma^{(n-k)}(1+a)$ is the $(n-k)$th derivative of the Gamma function.

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    $\begingroup$ Alternatively, $$\begin{align} \int_0^\infty \left(\frac{\log(x)}{e^x}\right)^n\,dx&=\int_0^\infty e^{-nx}\log^n(x)\,dx\\\\ &=\frac1n\sum_{k=0}^n\binom{n}{k}\log^{n-k}(n)\int_0^\infty e^{-x}\log^n(x)\,dx\\\\ &=\frac1n\sum_{k=0}^n\binom{n}{k}\log^{n-k}(n) \left.\left(\frac{d^n \Gamma(x+1)}{dx^n}\right)\right|_{x=0} \end{align}$$ $\endgroup$
    – Mark Viola
    Jul 14, 2020 at 19:02
  • $\begingroup$ Thanks for ur hint $\endgroup$ Jul 14, 2020 at 19:07
  • $\begingroup$ @MarkViola , could u add somethings about its irrationalitry ? $\endgroup$ Jul 14, 2020 at 19:13
  • $\begingroup$ @zeraouliarafik I have no idea as to the meaning of "its irrationality" in the context of my comment. $\endgroup$
    – Mark Viola
    Jul 14, 2020 at 19:38
  • $\begingroup$ for every integer n we have a value , Does this value irrational ? ..., or you may deduce it from your finalized closed form $\endgroup$ Jul 14, 2020 at 19:40

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