A drowning number theorist

(Adopted from Matteo De Zorzi's method.)

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We substitute $u = \ln x$ to get

$\int_1^\infty \frac{\ln \ln x}{x^2} \ dx = \int_0^\infty \frac{\ln u}{e^{2u}} \ \frac{du}{1/e^u} = \int_0^\infty e^{-u}\ln u \ du$

Either by seeing the similarities with the integral form of the gamma function $\displaystyle \Gamma(z) = \int_0^\infty e^{-t}t^{z-1} \ dt$ , or by noting with experience using differentiation through the integral (which often produces logarithms), we introduce a $u^{z-1}$ factor into the integral to get

$\begin{aligned} \int_0^\infty e^{-u}\ln u \ du &= \left. \int_0^\infty e^{-u}u^{z-1}\ln u \ du \ \right|_{z=1} \\ &= \left. \frac{d}{dz} \int_0^\infty e^{-u}u^{z-1} \ du \ \right|_{z=1} \\ &= \left. \frac{d}{dz} \Gamma(z) \ \right|_{z=1} \\ &= \psi(1)\Gamma(1) \\ &= \boxed{-\gamma} \end{aligned}$

where $\psi$ is the digamma function, defined as $\psi = (\ln \Gamma)'$ , and $-\psi(1) = \gamma \approx 0.5772$ is the Euler-Mascheroni constant.

Sorry not able to use latex... Anyway take x=e^t then integral becomes the integral of (e^(-t)lnt) from 0 to infinity.. This is the derivative of "gamma integral" that gives the digamma calculated in 1 that gives eulero mascheroni costant hence result