How does one verify that $\int_{0}^{1}x\ln[{\ln{x}}\ln(1-x)]\mathrm dx=-\gamma?$

Question

Consider $(1)$

How does one verify that

$$\int_{0}^{1}x\ln[\color{red}{\ln{x}}\ln(1-x)]\mathrm dx=-\gamma\tag1$$

An attempt:

$$\int_{0}^{1}x\ln{(\ln{x})}\mathrm dx+\int_{0}^{1}x\ln{(\ln{(1-x))}}\mathrm dx\tag2$$

$u=\ln{x}$ then $xdu=dx$

$$\int_{0}^{\infty}e^{-2u}\ln{(2u)}\mathrm du+\int_{0}^{\infty}e^{-2u}\ln{(1-e^u)}\mathrm du=I_1+I_2\tag3$$

Applying integration by parts to $I_1$

$$I_1={-e^{-2u}\ln{(2u)}\over 2}|_{0}^{\infty}+{1\over 2}\int_{0}^{\infty}{1\over u}\cdot e^{-2u}\mathrm du\tag4$$

$(4)$ seem to be divergent

How else can we tackle $(1)?$

Answer 1

There's a subtle problem with your split: the terms you've defined, e.g. $x\ln(\ln(x))$, aren't actually defined (over the reals) on the domain $(0,1)$ because $\ln(x)$ is negative on this range! Instead, you have to split things as $\displaystyle\int_0^1x\ln(-\ln(x))dx+\int_0^1x\ln(-\ln(1-x))dx$; this works because the minus signs internally cancel when the terms are multiplied!

But now we can substitute $u=1-x$ into the second term: minus signs from $du=-dx$ and from switching the order of integration from $\int_1^0$ back to $\int_0^1$ cancel, leaving the second integral as $\displaystyle\int_0^1(1-u)\ln(-\ln(u))du$. And then we can add this to the first term of the split to get $\displaystyle\int_0^1\ln(-\ln(x))dx$. But this formula (more often written in the form $\displaystyle\int_0^1\ln(\ln(\frac1x))dx$) is one of the classical expressions for $-\gamma$.

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\int_{0}^{1}x\ln\pars{\ln\pars{x}\ln\pars{1 - x}}\,\dd x = \int_{0}^{1}x\ln\pars{-\ln\pars{x}}\,\dd x + \int_{0}^{1}x\ln\pars{-\ln\pars{1 - x}}\,\dd x \\[5mm] = &\ \int_{0}^{1}x\ln\pars{-\ln\pars{x}}\,\dd x + \int_{0}^{1}\pars{1 - x}\ln\pars{-\ln\pars{x}}\,\dd x = \int_{0}^{1}\ln\pars{-\ln\pars{x}}\,\dd x \\[5mm] \stackrel{x\ =\ \exp\pars{-t}}{=} &\ \int_{0}^{\infty}\ln\pars{t}\expo{-t}\,\dd t = \bbx{\ds{-\,\gamma}} \end{align}

Answer 3

Letting $x \mapsto 1-x$, $$ I = \int_{0}^{1} \ln [\ln (1-x) \ln x] d x-I $$ we have $$ \begin{aligned} I &=\frac{1}{2} \int_{0}^{1} \ln [\ln (1-x) \ln x] d x \\ &=\frac{1}{2}\left[\int_{0}^{1} \ln [-\ln (1-x)]+\ln (-\ln x)d x\right] \\ &=\frac{1}{2}\left[\int_{0}^{1} \ln (-\ln x)d x+\int_{0}^{1} \ln (-\ln x) d x\right] \\ &=\int_{0}^{1} \ln (-\ln x) d x \\&= \int_{0}^{\infty} e^{-x} \ln x d x\\ &=-\gamma \end{aligned} $$