# Does this $\int_{0}^{\infty}(\frac{\log x}{e^x})^n dx$ always have a closed form for $n$ being positive integer ? what about its irrationality?

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 $$n$$ : $$$$\int\limits_{0}^{\infty} \mathrm{e}^{-x} x^k \mathrm{d} x = \Gamma(k+1)$$$$ Differentiate with respect to $$k$$ $$$$\int\limits_{0}^{\infty} \mathrm{e}^{-x} x^k \mathrm{ln}(x) \mathrm{d} x = \frac{d\Gamma(k+1)}{dt} = \Gamma(k+1) \psi^{(0)}(k+1)$$$$

Taking the limit $$k \to 0$$ yields $$$$\int\limits_{0}^{\infty} \mathrm{e}^{-x} \mathrm{ln}(x) \mathrm{d} x = \Gamma(1) \psi^{(0)}(1) = -\gamma$$$$ Now for $$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 $$n=1,2$$

• \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} Jul 14, 2020 at 19:01
• Note that $\gamma$ is not known to be irrational. Jul 14, 2020 at 20:39

## 1 Answer

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.

• 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} Jul 14, 2020 at 19:02
• Thanks for ur hint Jul 14, 2020 at 19:07
• @MarkViola , could u add somethings about its irrationalitry ? Jul 14, 2020 at 19:13
• @zeraouliarafik I have no idea as to the meaning of "its irrationality" in the context of my comment. Jul 14, 2020 at 19:38
• for every integer n we have a value , Does this value irrational ? ..., or you may deduce it from your finalized closed form Jul 14, 2020 at 19:40