# Show a detail prove of : $\int_{0}^{1}\int_{0}^{1}\left({x\over 1-xy}\cdot{\ln{x}-\ln{y}\over \ln{x}+\ln{y}}\right)\mathrm dx\mathrm dy=1-2\gamma$

Variation of my recent post. Strangely it leads to the result in term of Euler's constant;$\gamma$

Prove that

$$\int_{0}^{1}\int_{0}^{1}\left({x\over 1-xy}\cdot{\ln{x}-\ln{y}\over \ln{x}+\ln{y}}\right)\mathrm dx\mathrm dy=1-2\gamma$$

Where $\gamma=0.577215...$;Euler's constant.

My try:

$u=xy$ then $du=ydx$

$$\int_{0}^{1}\int_{0}^{y}{{u\over y^2}\over 1-u}\cdot{\ln{u\over y^2}\over \ln{u}}\mathrm du\mathrm dy$$

I can't understand how to or even glimsely to solve double integrals.

Any chance can someone show me a detail of the answer. Thank you.

• the proposed approach in the other question doesn't work? – tired Jan 3 '17 at 11:19
• Changing the order of integration we get $$\int_0^1du\frac{u}{1-u}\int_u^1dv\frac{1}{v^2}\left(1-2\frac{\log(v)}{\log(u)}\right)$$ the inner integral is straightforward and you only need integral representation 5 from here en.wikipedia.org/wiki/… – tired Jan 3 '17 at 11:27
• For a proof of different integral reps of Euler's constant see here: math.stackexchange.com/questions/980593/… – tired Jan 3 '17 at 11:41

Folllowing @tired Comment(s): \begin{align} \text{Let} &\colon\space x\,y=u,\,\,y=v \space\Rightarrow\space x/y=u/v^2,\,\,y\,dx=du,\,\,dy=dv \\[3mm] \color{red}{I} &= \int_{0}^{1}\int_{0}^{1}\left(\frac{x}{1-x\,y}\,\frac{\log{x}-\log{y}}{\log{x}+\log{y}}\right)\,dx\,dy = \int_{0}^{1}\int_{0}^{1}\left(\frac{(x/y)}{1-(x\,y)}\,\frac{\log{(x/y)}}{\log{(x\,y)}}\right)\,(y\,dx)\,dy \\[3mm] &= \int_{0}^{1}\int_{0}^{v}\left(\frac{(u/v^2)}{1-u}\,\frac{\log{(u/v^2)}}{\log{u}}\right)\,du\,dv = \int_{0}^{1}\int_{\color{red}{0}}^{\color{red}{v}}\left[\frac{u}{1-u}\,\frac{1}{v^2}\left(1-2\frac{\log{v}}{\log{u}}\right)\right]\color{red}{\,du\,dv} \\[3mm] &= \int_{0}^{1}\int_{\color{red}{u}}^{\color{red}{1}}\left[\frac{u}{1-u}\,\frac{1}{v^2}\left(1-2\frac{\log{v}}{\log{u}}\right)\right]\color{red}{\,dv\,du} \quad\qquad\qquad\qquad\qquad\qquad\color{blue}{\{1\}} \\[3mm] &= \int_{0}^{1}\frac{u}{1-u}\,du\,\int_{u}^{1}\left(\frac{1}{v^2}-\frac{2}{\log{u}}\frac{\log{v}}{v^2}\right)\,dv \\[3mm] &= \int_{0}^{1}\frac{u}{1-u}\,du\,\left[-\frac{1}{v}+\frac{2}{\log{u}}\frac{1+\log{v}}{v}\right]_{u}^{1} \qquad\qquad\qquad\qquad\qquad\qquad\color{blue}{\{2\}} \\[3mm] &= \int_{0}^{1}\frac{u}{1-u}\left(\frac{-2}{\log{u}}\frac{1-u}{u}-\frac{1+u}{u}\right)\,du = \int_{0}^{1}\left(\frac{-2}{\log{u}}-\frac{1+u}{1-u}\right)\,du \\[3mm] &= \int_{0}^{1}\left(\frac{-2}{\log{u}}-\frac{1+u\color{red}{-1+1}}{1-u}\right)\,du = \int_{0}^{1}\left(\frac{-2}{\log{u}}+\frac{-2}{1-u}+1\right)\,du \\[3mm] &= \int_{0}^{1}\,du\,-2\int_{0}^{1}\left(\frac{1}{\log{u}}+\frac{1}{1-u}\right)\,du = \color{red}{1 - 2\,\gamma} \quad\qquad\qquad\qquad\qquad\color{blue}{\{3\}} \\[3mm] \end{align}

$\,\color{blue}{\{1\}}\colon\,$ Change order integration.
$\,\color{blue}{\{2\}}\colon\,$ IBP: $\,\int\log{x}\,\frac{dx}{x^2},\,\{u=\log{x}\,{\small\Rightarrow}\,du=\frac{dx}{x},\, dv=\frac{dx}{x^2}\,{\small\Rightarrow}\,v=\frac{-1}{x}\},\,=-\frac{\log{x}}{x}+\int\frac{dx}{x^2}=-\frac{1+\log{x}}{x}$.
$\,\color{blue}{\{3\}}\colon\,$ Integral Representation of Euler–Mascheroni Constant: $\quad\gamma = \int_{0}^{1}\left(\frac{1}{\log{x}}+\frac{1}{1-x}\right)\,dx$.

• @Algebra: Thanks. – Hazem Orabi Jan 21 '17 at 8:31
• No, thanks you for answering the question @Hazem Orabi – gymbvghjkgkjkhgfkl Jan 21 '17 at 17:55


$\ds{\int_{0}^{1}\int_{0}^{1}{x \over 1 - xy}\, {\ln\pars{x} - \ln\pars{y} \over \ln\pars{x} + \ln\pars{y}}\,\dd x\,\dd y = 1 - 2\gamma\,,\qquad\gamma:\ Euler\mbox{-}Mascheroni\ Constant}$.

\begin{align} &\int_{0}^{1}\int_{0}^{1}{x \over 1 - xy}\, {\ln\pars{x} - \ln\pars{y} \over \ln\pars{x} + \ln\pars{y}}\,\dd x\,\dd y \\[5mm] = &\ \left.\partiald{}{\mu}\int_{0}^{1}\int_{0}^{1}{x \over 1 - xy} \pars{x^{\mu} - y^{\mu}} \bracks{-\int_{0}^{\infty}\pars{xy}^{\,t}\,\dd t}\,\dd x\,\dd y\, \right\vert_{\ \mu\ =\ 0} \\[5mm] = &\ \left.\partiald{}{\mu}\int_{0}^{\infty}\int_{0}^{1}\int_{0}^{1} {x^{t + 1}y^{t + \mu} - x^{t + \mu + 1}y^{t} \over 1 - xy} \,\dd x\,\dd y\,\dd t\,\right\vert_{\ \mu\ =\ 0} \\[5mm] = &\ \left.\partiald{}{\mu}\int_{0}^{\infty}\int_{0}^{1}\int_{0}^{y} {x^{t + 1}y^{\mu - 2} - x^{t + \mu + 1}y^{-\mu - 2} \over 1 - x} \,\dd x\,\dd y\,\dd t\,\right\vert_{\ \mu\ =\ 0} \\[5mm] = &\ \left.\partiald{}{\mu}\int_{0}^{\infty}\int_{0}^{1}\int_{x}^{1} {x^{t + 1}y^{\mu - 2} - x^{t + \mu + 1}y^{-\mu - 2} \over 1 - x} \,\dd y\,\dd x\,\dd t\,\right\vert_{\ \mu\ =\ 0} \\[5mm] = &\ \left.\partiald{}{\mu}\int_{0}^{\infty}\int_{0}^{1} \pars{x^{t + 1}\,{1 - x^{\mu - 1} \over \mu - 1} + x^{t + \mu + 1}\,{1 - x^{-\mu - 1}\over \mu + 1}} {\dd x\,\dd t \over 1 - x}\,\right\vert_{\ \mu\ =\ 0} \\[5mm] = &\ \left.\partiald{}{\mu}\int_{0}^{\infty}\int_{0}^{1} \pars{{1 \over \mu - 1}\,{x^{t + 1} - x^{t + \mu} \over 1 - x} + {1 \over \mu + 1}\,{x^{t + \mu + 1} - x^{t} \over 1 - x}} \dd x\,\dd t\,\right\vert_{\ \mu\ =\ 0} \\[5mm] = &\ \left.\partiald{}{\mu}\int_{0}^{\infty} \bracks{{\Psi\pars{t + \mu + 1} - \Psi\pars{t + 2} \over \mu - 1} + {\Psi\pars{t + 1} - \Psi\pars{t + \mu + 2} \over \mu + 1}} \dd t\,\right\vert_{\ \mu\ =\ 0} \\[5mm] = &\ \left.\partiald{}{\mu}\bracks{% {1 \over \mu - 1}\,\ln\pars{\Gamma\pars{t + \mu + 1} \over \Gamma\pars{t + 2}} + {1 \over \mu + 1}\,\ln\pars{\Gamma\pars{t + 1} \over \Gamma\pars{t + \mu + 2}}} _{\ t\ =\ 0}^{\ t\ \to\ \infty}\, \right\vert_{\ \mu\ =\ 0} \\[5mm] = &\ \partiald{}{\mu}\bracks{% -\,{\ln\pars{\Gamma\pars{\mu + 1}} \over \mu - 1} + {\ln\pars{\Gamma\pars{\mu + 2}} \over \mu + 1}}_{\ \mu\ =\ 0} = \Psi\pars{1} + \Psi\pars{2} = 2\Psi\pars{1} + 1 = \bbx{\ds{1 - 2\gamma}} \end{align}