Compute $ \int_{0}^{+\infty} \left( \frac{\ln(x)}{e^x}\right)^2 dx $ [closed]

Question

How can I compute this integral? $$ \int_{0}^{+\infty} \left( \frac{\ln(x)}{e^x}\right)^2 dx $$

Answer 1

Let $$f(a):=\int_0^\infty x^{a-1}e^{-2x}\,dx.$$ Then $$f''(1)=\int_0^\infty \ln^2x\,e^{-2x}\,dx,$$ which is the integral in question.

On the other hand, $f(a)=2^{-a}\,\Gamma(a)$, and hence the integral in question is $$f''(1)=\frac{\ln^2 2}2 - \Gamma'(1)\ln2 + \Gamma''(1)/2 =\frac{\pi^2}{12}+ \frac{(\gamma +\ln2)^2}2=1.6293\dots,$$ where $\gamma=0.57721\dots$ is the Euler gamma constant.

(The second equality in the latter display follows because $\Gamma'(1)=-\gamma$ and $\Gamma''(1)=\gamma^2+\pi^2/6$. In turn, the latter two equalities can be obtained using the last two displays in Section Recurrence relation.)

Answer 2

Hint: you may try $n=2$ and follow in general this :\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}

For more information check answers of my question here

For more clarification according to the answer of @Nikunj, try $n=2$ using this steps in general: 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.