Evaluate $\int_{0}^{1}{\int_{0}^{1}{\frac{x^{\alpha -1}y^{\beta -1}}{\left( 1+xy \right)\ln xy}dxdy}}$

On AoPS (Art of Problem Solving), the following integral was posted by fellow user pprime, but no one has been able to come close to solving it. Many attempts have been made, but none have succeeded. I have come here as a last resort so that some kind user would help me evaluate this integral.

http://www.artofproblemsolving.com/community/c7h1288609_extreme_integration_marathon

Evaluate $$\int_{0}^{1}{\int_{0}^{1}{\frac{x^{\alpha -1}y^{\beta -1}}{\left( 1+xy \right)\ln xy}dxdy}}$$

Hint: $\int_{0}^{+\infty }{\exp \left( -u\ln \left( xy \right) \right)du}=\frac{1}{\ln x+\ln y}$

• What is AoPS ?... – Jean Marie Aug 14 '16 at 7:03
• @JeanMarie Art of Problem Solving – Why Do You Care Aug 14 '16 at 7:10
• Thanks for the information. I should maybe have guessed... – Jean Marie Aug 14 '16 at 7:15
• $$\frac{1}{\alpha-\beta} \log{\left( \frac{\Gamma(\frac{\alpha}{2}) \Gamma(\frac{\beta +1}{2})}{\Gamma(\frac{\alpha +1}{2}) \Gamma(\frac{\beta}{2})} \right)} \text{ for } \alpha < \beta$$ Can someone confirm my answer? – Jack Lam Aug 14 '16 at 7:57
• @JackLam Looks like your answer ended up being correct! – Why Do You Care Aug 14 '16 at 18:06

3 Answers

Just to note that with Frullani's theorem we can arrive at the infinite product rather quickly. We have that $$I(\alpha,\beta)=\int_{0}^{1}\int_{0}^{1}\frac{x^{\alpha-1}y^{\beta-1}}{\left(1+xy\right)\log\left(xy\right)}dxdy\stackrel{xy=u}{=}\int_{0}^{1}y^{\beta-\alpha-1}\int_{0}^{y}\frac{u^{\alpha-1}}{\left(1+u\right)\log\left(u\right)}dudy$$ $$\stackrel{IBP}{=}\frac{1}{\beta-\alpha}\int_{0}^{1}\frac{y^{\alpha-1}-y^{\beta-1}}{\left(1+y\right)\log\left(y\right)}dy=\frac{1}{\beta-\alpha}\sum_{k\geq0}\left(-1\right)^{k}\int_{0}^{1}\frac{y^{\alpha-1+k}-y^{\beta-1+k}}{\log\left(y\right)}dy$$ $$\stackrel{y=e^{-v}}{=}\frac{1}{\beta-\alpha}\sum_{k\geq0}\left(-1\right)^{k+1}\int_{0}^{\infty}\frac{e^{-v(\alpha+k)}-e^{-v(\beta+k)}}{v}dv$$ and now we can use the Frullani's theorem and get $$I(\alpha,\beta)=\frac{1}{\beta-\alpha}\sum_{k\geq0}\left(-1\right)^{k+1}\log\left(\frac{\beta+k}{\alpha+k}\right)=\frac{1}{\beta-\alpha}\log\left(\prod_{k\geq0}\left(\frac{\beta+2k+1}{\alpha+2k+1}\right)\left(\frac{\beta+2k}{\alpha+2k}\right)^{-1}\right)$$ now note that $$\prod_{k=0}^{N}\left(a+2k\right)=2^{N}\prod_{k=0}^{N}\left(\frac{a}{2}+k\right)=2^{N}a\left(\frac{a}{2}+1\right)_{N}$$ and $$\prod_{k=0}^{N}\left(a+2k+1\right)=2^{N}\prod_{k=0}^{N}\left(k+\frac{a+1}{2}\right)=\left(a+1\right)2^{N}\left(\frac{a+1}{2}+1\right)_{N}$$ where $\left(x\right)_{N}$ is the Pochhammer symbol. So $$\prod_{k=0}^{N}\left(\frac{\beta+2k+1}{\alpha+2k+1}\right)\left(\frac{\beta+2k}{\alpha+2k}\right)^{-1}=\frac{\beta+1}{\alpha+1}\frac{\alpha}{\beta}\frac{\left(\frac{\beta+1}{2}+1\right)_{N}}{\left(\frac{\alpha+1}{2}+1\right)_{N}}\frac{\left(\frac{\alpha}{2}+1\right)_{N}}{\left(\frac{\beta}{2}+1\right)_{N}}$$ and now using the asymptotic for the Pochhammer symbol we get

$$I(\alpha,\beta)=\color{red}{\frac{1}{\beta-\alpha}\log\left(\frac{\Gamma\left(\frac{\alpha+1}{2}\right)\Gamma\left(\frac{\beta}{2}\right)}{\Gamma\left(\frac{\alpha}{2}\right)\Gamma\left(\frac{\beta+1}{2}\right)}\right)}$$ where $$\, \beta,\alpha>0,\,\beta\neq \alpha$$

as wanted. If $\alpha=\beta=\gamma$ we have $$I(\gamma)=\int_{0}^{1}\int_{0}^{1}\frac{\left(xy\right)^{\gamma-1}}{\left(1+xy\right)\log\left(xy\right)}dxdy\stackrel{xy=u}{=}\int_{0}^{1}y^{-1}\int_{0}^{y}\frac{u^{\alpha-1}}{\left(1+u\right)\log\left(u\right)}dudy$$ $$\stackrel{IBP}{=}-\int_{0}^{1}\frac{y^{\gamma-1}}{1+y}dy=\sum_{k\geq0}\left(-1\right)^{k+1}\frac{1}{\gamma+k}=\color{red}{-\Phi\left(-1,1,\gamma\right)}$$ where $\gamma>0$ and $\Phi\left(x,y,z\right)$ is the Lerch Transcendent. Obviously for some special values of $\gamma$ we have nice closed forms. For example for $\gamma=1$ we have $$I\left(1\right)=\sum_{k\geq0}\left(-1\right)^{k+1}\frac{1}{1+k}=-\log\left(2\right).$$

$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\color{#f00}{\int_{0}^{1}\int_{0}^{1}{x^{\alpha - 1}\,\,y^{\beta - 1} \over \pars{1 + xy}\ln\pars{xy}}\,\dd x\,\dd y} = -\int_{0}^{1}\int_{0}^{1}{x^{\alpha - 1}\,\,y^{\beta - 1} \over 1 + xy}\,\,\, \overbrace{\int_{0}^{\infty}\pars{xy}^{z}\,\dd z} ^{\ds{-\,{1 \over \ln\pars{xy}}}}\ \,\dd x\,\dd y \\[5mm] = &\ -\int_{0}^{\infty}\int_{0}^{1}\int_{0}^{1} {x^{\alpha -1 + z}\,\,\,y^{\beta - 1 + z} \over 1 + xy} \,\dd x\,\dd y\,\dd z = -\int_{0}^{\infty}\int_{0}^{1}\int_{0}^{1} {\pars{xy}^{\alpha -1 + z}\,\,\,y^{\beta - \alpha - 1} \over 1 + xy} \,\dd\pars{xy}\,\dd y\,\dd z \\[5mm] = &\ -\int_{0}^{\infty}\int_{0}^{1}\int_{0}^{y} {x^{\alpha -1 + z}\,\,\,y^{\beta - \alpha - 1} \over 1 + x} \,\dd x\,\dd y\,\dd z = -\int_{0}^{\infty}\int_{0}^{1}{x^{\alpha -1 + z}\,\,\, \over 1 + x}\int_{x}^{1} y^{\beta - \alpha - 1}\,\,\,\dd y\,\dd x\,\dd z \\[5mm] = &\ {1 \over \beta - \alpha}\int_{0}^{\infty}\int_{0}^{1} {x^{\beta + z - 1}\,\,\, -\,\,\, x^{\alpha + z - 1}\,\,\, \over 1 + x} \,\dd x\,\dd z \\[5mm] = &\ {1 \over \beta - \alpha}\int_{0}^{\infty}\int_{0}^{1} {x^{\beta + z - 1}\,\,\, -\,\,\, x^{\alpha + z - 1}\,\,\, -\,\,\, x^{\beta + z}\,\,\, +\,\,\, x^{\alpha + z} \over 1 - x^{2}} \,\dd x\,\dd z \\[5mm] \stackrel{x^{2}\ \mapsto\ x}{=}\ &\ {1 \over 2\pars{\beta - \alpha}} \int_{0}^{\infty}\int_{0}^{1} {x^{\beta/2 + z/2 - 1}\,\,\, -\,\,\, x^{\alpha/2 + z/2 - 1}\,\,\, -\,\,\, x^{\beta/2 + z/2 - 1/2}\,\,\, +\,\,\, x^{\alpha/2 + z/2 - 1/2} \over 1 - x} \,\,\,\,\dd x\,\dd z \end{align}

With the Digamma Function identity $\pars{~\gamma\ \mbox{is the Euler-Mascheroni Constant}~}$ $$\left.\Psi\pars{\xi} = -\gamma + \int_{0}^{1}{1 - t^{\xi - 1} \over 1 - t}\,\dd t\, \right\vert_{\ \Re\pars{\xi}\ >\ 0\,\,\,\,\,}$$ the integration is reduced to: \begin{align} &\color{#f00}{\int_{0}^{1}\int_{0}^{1}{x^{\alpha - 1}\,\,y^{\beta - 1} \over \pars{1 + xy}\ln\pars{xy}}\,\dd x\,\dd y} \\[5mm] = &\ {1 \over 2\pars{\beta - \alpha}}\int_{0}^{\infty}\bracks{% \Psi\pars{z + \alpha \over 2} + \Psi\pars{z + \beta + 1 \over 2} - \Psi\pars{z + \beta \over 2} - \Psi\pars{z + \alpha + 1 \over 2}}\,\dd z \end{align}
Since $\ds{\Psi\pars{\xi}\ \stackrel{\mrm{def.}}{=}\ \totald{\ln\pars{\Gamma\pars{\xi}}}{\xi}}$ \begin{align} &\color{#f00}{\int_{0}^{1}\int_{0}^{1}{x^{\alpha - 1}\,\,y^{\beta - 1} \over \pars{1 + xy}\ln\pars{xy}}\,\dd x\,\dd y} = \left.{1 \over \beta - \alpha} \ln\pars{\Gamma\pars{z/2 + \alpha/2}\Gamma\pars{z/2 + \beta/2 + 1/2} \over \Gamma\pars{z/2 + \beta/2}\Gamma\pars{z/2 + \alpha/2 + 1/2}} \right\vert_{\ 0}^{\ \infty} \\[5mm] = &\ \color{#f00}{{1 \over \beta - \alpha}\, \ln\pars{\Gamma\pars{\bracks{\alpha + 1}/2}\Gamma\pars{\beta/2} \over \Gamma\pars{\alpha/2}\Gamma\pars{\bracks{\beta + 1}/2}}}\,;\qquad \Re\pars{\alpha} > 0\,,\quad\Re\pars{\beta} > 0 \end{align}

When $\ds{\ul{\beta \to \alpha}}$, the solution becomes: $$\half\bracks{\Psi\pars{\alpha \over 2} - \Psi\pars{\alpha + 1 \over 2}}$$

$I\left( \alpha ,\beta \right)=\int\limits_{0}^{1}{\int\limits_{0}^{1}{\frac{x^{\alpha -1}y^{\beta -1}}{\left( 1+xy \right)\ln xy}dxdy}}=-\int\limits_{0}^{1}{\int\limits_{0}^{1}{\int\limits_{0}^{+\infty }{\frac{x^{\alpha -1}y^{\beta -1}}{1+xy}\left( xy \right)^{j}djdxdy}}}=-\int\limits_{0}^{+\infty }{\int\limits_{0}^{1}{\int\limits_{0}^{1}{\frac{x^{\alpha -1}y^{\beta -1}}{1+xy}\left( xy \right)^{j}dxdydj}}}$

$=-\int\limits_{0}^{+\infty }{\int\limits_{0}^{1}{\int\limits_{0}^{1}{\sum\limits_{k=0}^{+\infty }{\left( -xy \right)^{k}\cdot }x^{\alpha +j-1}y^{\beta +j-1}dxdydj}}}=-\sum\limits_{k=0}^{+\infty }{\left( -1 \right)^{k}\int\limits_{0}^{+\infty }{\int\limits_{0}^{1}{\int\limits_{0}^{1}{x^{\alpha +k+j-1}y^{\beta +k+j-1}dxdydj}}}}$

$=-\sum\limits_{k=0}^{+\infty }{\left( -1 \right)^{k}\int\limits_{0}^{+\infty }{\left( \int\limits_{0}^{1}{x^{\alpha +k+j-1}dx} \right)\left( \int\limits_{0}^{1}{y^{\beta +k+j-1}dy} \right)dj}}=-\sum\limits_{k=0}^{+\infty }{\left( -1 \right)^{k}\int\limits_{0}^{+\infty }{\left( \frac{1}{\left( \alpha +k+j \right)\left( \beta +k+j \right)} \right)dj}}$

$=-\frac{1}{\alpha -\beta }\sum\limits_{k=0}^{+\infty }{\left( -1 \right)^{k}\int\limits_{0}^{+\infty }{\left( \frac{\alpha +k+j-\left( \beta +k+j \right)}{\left( \alpha +k+j \right)\left( \beta +k+j \right)} \right)dj}}$

$=-\frac{1}{\alpha -\beta }\sum\limits_{k=0}^{+\infty }{\left( -1 \right)^{k}\int\limits_{0}^{+\infty }{\left( \frac{1}{\beta +k+j}-\frac{1}{\alpha +k+j} \right)dj}}=\frac{1}{\alpha -\beta }\sum\limits_{k=0}^{+\infty }{\left( -1 \right)^{k}\ln \left( \frac{\beta +k}{\alpha +k} \right)}$

$=\frac{1}{\alpha -\beta }\left( \left( -1 \right)^{0}\ln \left( \frac{\beta +0}{\alpha +0} \right)+\left( -1 \right)^{1}\ln \left( \frac{\beta +1}{\alpha +1} \right)+\left( -1 \right)^{2}\ln \left( \frac{\beta +2}{\alpha +2} \right)+\left( -1 \right)^{3}\ln \left( \frac{\beta +3}{\alpha +3} \right)+... \right)$

$=\frac{1}{\alpha -\beta }\left( \ln \left( \frac{\beta +0}{\alpha +0} \right)-\ln \left( \frac{\beta +1}{\alpha +1} \right)+\ln \left( \frac{\beta +2}{\alpha +2} \right)-\ln \left( \frac{\beta +3}{\alpha +3} \right)+... \right)$

$=\frac{1}{\alpha -\beta }\left( \ln \left( \frac{\beta +0}{\alpha +0} \right)+\ln \left( \frac{\alpha +1}{\beta +1} \right)+\ln \left( \frac{\beta +2}{\alpha +2} \right)+\ln \left( \frac{\alpha +3}{\beta +3} \right)+... \right)$

$=\frac{1}{\alpha -\beta }\ln \left( \left( \frac{\beta +0}{\alpha +0} \right)\cdot \left( \frac{\alpha +1}{\beta +1} \right)\cdot \left( \frac{\beta +2}{\alpha +2} \right)\cdot \left( \frac{\alpha +3}{\beta +3} \right)\cdot ... \right)=\frac{1}{\alpha -\beta }\ln \prod\limits_{k=0}^{+\infty }{\left( \frac{\beta +2k}{\alpha +2k} \right)\left( \frac{\alpha +1+2k}{\beta +1+2k} \right)}$

$=\frac{1}{\alpha -\beta }\ln \frac{\beta }{\alpha }\cdot \frac{\alpha +1}{\beta +1}\cdot \prod\limits_{k=1}^{+\infty }{\left( \frac{1+\frac{\frac{\beta }{2}}{k}}{1+\frac{\frac{\alpha }{2}}{k}}\cdot \frac{1+\frac{\frac{\alpha +1}{2}}{k}}{1+\frac{\frac{\beta +1}{2}}{k}} \right)}$

$=\frac{1}{\alpha -\beta }\ln \frac{\beta }{\alpha }\cdot \frac{\alpha +1}{\beta +1}\cdot \prod\limits_{k=1}^{+\infty }{\left( \frac{e^{\frac{\frac{\alpha }{2}}{k}}\left( 1+\frac{\frac{\alpha }{2}}{k} \right)^{-1}\cdot e^{\frac{\frac{\beta +1}{2}}{k}}\left( 1+\frac{\frac{\beta +1}{2}}{k} \right)^{-1}}{e^{\frac{\frac{\beta }{2}}{k}}\left( 1+\frac{\frac{\beta }{2}}{k} \right)^{-1}\cdot e^{\frac{\frac{\alpha +1}{2}}{k}}\left( 1+\frac{\frac{\alpha +1}{2}}{k} \right)^{-1}} \right)}$

$=\frac{1}{\alpha -\beta }\ln \frac{\beta }{\alpha }\cdot \frac{\alpha +1}{\beta +1}\cdot \prod\limits_{k=1}^{+\infty }{\left( \frac{e^{\frac{\frac{\alpha }{2}}{k}}\left( 1+\frac{\frac{\alpha }{2}}{k} \right)^{-1}\cdot e^{\frac{\frac{\beta +1}{2}}{k}}\left( 1+\frac{\frac{\beta +1}{2}}{k} \right)^{-1}}{e^{\frac{\frac{\beta }{2}}{k}}\left( 1+\frac{\frac{\beta }{2}}{k} \right)^{-1}\cdot e^{\frac{\frac{\alpha +1}{2}}{k}}\left( 1+\frac{\frac{\alpha +1}{2}}{k} \right)^{-1}} \right)}$

$=\frac{1}{\alpha -\beta }\ln \left( \frac{\frac{e^{-\gamma \frac{\alpha }{2}}}{\frac{\alpha }{2}}\prod\limits_{k=1}^{+\infty }{\left( e^{\frac{\frac{\alpha }{2}}{k}}\left( 1+\frac{\frac{\alpha }{2}}{k} \right)^{-1} \right)}\cdot \frac{e^{-\gamma \frac{\beta +1}{2}}}{\frac{\beta +1}{2}}\prod\limits_{k=1}^{+\infty }{\left( e^{\frac{\frac{\beta +1}{2}}{k}}\left( 1+\frac{\frac{\beta +1}{2}}{k} \right)^{-1} \right)}}{\frac{e^{-\gamma \frac{\beta }{2}}}{\frac{\beta }{2}}\prod\limits_{k=1}^{+\infty }{\left( e^{\frac{\frac{\beta }{2}}{k}}\left( 1+\frac{\frac{\beta }{2}}{k} \right)^{-1} \right)}\cdot \frac{e^{-\gamma \frac{\alpha +1}{2}}}{\frac{\alpha +1}{2}}\prod\limits_{k=1}^{+\infty }{\left( e^{\frac{\frac{\alpha +1}{2}}{k}}\left( 1+\frac{\frac{\alpha +1}{2}}{k} \right)^{-1} \right)}} \right)$

$=\frac{1}{\alpha -\beta }\ln \left( \frac{\Gamma \left( \frac{\alpha }{2} \right)\cdot \Gamma \left( \frac{\beta +1}{2} \right)}{\Gamma \left( \frac{\beta }{2} \right)\cdot \Gamma \left( \frac{\alpha +1}{2} \right)} \right)$

• In your first step you make a replace the $ln(xy)$ with an integral over $j$ of $(xy)^j$. According to your hint shouldn't this be $(xy)^{-j}$? – Spencer Aug 15 '16 at 5:15
• Nope, I'm correct. – Why Do You Care Aug 15 '16 at 5:24
• Can you explain why? I was going to upvote your answer until I saw that. – Spencer Aug 15 '16 at 5:27
• If you look carefully at the exponentiation, you'll see a double negative on the exponent as per my hint. And that results in the exponent being a positive. – Why Do You Care Aug 15 '16 at 5:28
• Ok I think I see it. The negative in front of the integral on the second step converts $\ln(xy)=-\ln(1/xy)$ then you apply the hint to that and get $(xy)^j$ for positive $j$. – Spencer Aug 15 '16 at 5:33