Double integral $\int_{0}^{1}\int_{0}^{1}\frac{x^{a-1}y^{b-1}}{(1+xy)\ln(xy)}\,dx\,dy$

Question

Double integral $$\int_{0}^{1}\int_{0}^{1}\frac{x^{a-1}y^{b-1}}{(1+xy)\ln(xy)}\,dx\,dy$$ $\displaystyle{a,b>0}$

my work

$$I\left( {a,b} \right) = \int\limits_0^1 {\int\limits_0^1 {\frac{{{x^a}{y^b}}}{{\ln \left( {xy} \right)}}dx\,dy} } = \frac{{\ln \left( {\dfrac{{1 + b}}{{1 + a}}} \right)}}{{a-b}}$$

$$\displaystyle{I\left( {a,b} \right) = \int\limits_0^1 {\int\limits_0^1 {\frac{{{x^a}{y^b}}}{{\ln \left( {xy} \right)}}dx\,dy} } = }$$

$$\displaystyle{ = \int\limits_0^1 {\int\limits_0^1 {{x^a}{y^b}\left( { - \int\limits_0^\infty {{e^{t \cdot \ln \left( {xy} \right)}}dt} } \right)dx\,dy} } = - \int\limits_0^\infty {\left( {\int\limits_0^1 {\int\limits_0^1 {{x^a}{y^b}{e^{t \cdot \ln \left( {xy} \right)}}dx\,dy} } } \right)dt} = - \int\limits_0^\infty {\left( {\int\limits_0^1 {\int\limits_0^1 {{x^a}{y^b}{{\left( {xy} \right)}^t}dx\,dy} } } \right)dt} = }$$

$$\displaystyle{ = - \int\limits_0^\infty {\left( {\left( {\int\limits_0^1 {{x^{a + t}}dx} } \right)\left( {\int\limits_0^1 {{y^{b + t}}dy} } \right)} \right)dt} = - \int\limits_0^\infty {\left( {\frac{1}{{a + 1 + t}} \cdot \frac{1}{{b + 1 + t}}} \right)dt} = \frac{1}{{b - a}}\int\limits_0^\infty {\left( { - \frac{1}{{a + 1 + t}} + \frac{1}{{b + 1 + t}}} \right)dt} = \frac{{\ln \left( {\dfrac{{1 + b}}{{1 + a}}} \right)}}{{a - b}}}$$

$$\displaystyle{I = \int\limits_0^1 {\int\limits_0^1 {\frac{{{x^{a - 1}}{y^{b - 1}}}}{{\left( {1 + xy} \right)\ln \left( {xy} \right)}}dx\,dy} } = \int\limits_0^1 {\int\limits_0^1 {\frac{{{x^{a - 1}}{y^{b - 1}}}}{{\ln \left( {xy} \right)}}\left( {\sum\limits_{k = 0}^\infty {{{\left( { - 1} \right)}^k}{{\left( {xy} \right)}^k}} } \right)dx\,dy} } = \sum\limits_{k = 0}^\infty {\left( {{{\left( { - 1} \right)}^k}\left( {\int\limits_0^1 {\int\limits_0^1 {\frac{{{x^{a + k - 1}}{y^{b + k - 1}}}}{{\ln \left( {xy} \right)}}dx\,dy} } } \right)} \right)} }$$

Answer 1

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Define the function $\mathcal{I}$ by the double integral,

$$\mathcal{I}{\left(a,b\right)}:=\int_{0}^{1}\mathrm{d}x\int_{0}^{1}\mathrm{d}y\,\frac{x^{a-1}y^{b-1}}{\left(1+xy\right)\ln{\left(xy\right)}};~~~\small{(a,b)\in\mathbb{R}_{>0}^{2}}.$$

Recalling the definition of the digamma function by

$$\psi{\left(z\right)}:=\frac{d}{dz}\ln{\left(\Gamma{\left(z\right)}\right)};~~~\small{\Re{(z)}>0}.$$

as well as its integral representation

$$\psi{\left(z\right)}=-\gamma+\int_{0}^{1}\mathrm{d}t\,\frac{1-t^{z-1}}{1-t};~~~\small{\Re{(z)}>0},$$

we obtain the following result for $\mathcal{I}$ in the $a\neq b$ case:

$$\begin{align} \mathcal{I}{\left(a,b\right)} &=\int_{0}^{1}\mathrm{d}x\int_{0}^{1}\mathrm{d}y\,\frac{x^{a-1}y^{b-1}}{\left(1+xy\right)\ln{\left(xy\right)}}\\ &=\int_{0}^{1}\mathrm{d}x\int_{0}^{1}\mathrm{d}y\,\frac{x^{a-b}(xy)^{b-1}}{\left(1+xy\right)\ln{\left(xy\right)}}\\ &=\int_{0}^{1}\mathrm{d}x\int_{0}^{x}\mathrm{d}u\,\frac{x^{a-b}u^{b-1}}{x\left(1+u\right)\ln{\left(u\right)}};~~~\small{\left[y=\frac{u}{x}\right]}\\ &=\int_{0}^{1}\mathrm{d}u\int_{u}^{1}\mathrm{d}x\,\frac{x^{a-b-1}u^{b-1}}{\left(1+u\right)\ln{\left(u\right)}}\\ &=\int_{0}^{1}\mathrm{d}u\,\frac{u^{b-1}}{\left(1+u\right)\ln{\left(u\right)}}\int_{u}^{1}\mathrm{d}x\,x^{a-b-1}\\ &=\int_{0}^{1}\mathrm{d}u\,\frac{u^{b-1}}{\left(1+u\right)\ln{\left(u\right)}}\cdot\frac{1-u^{a-b}}{a-b}\\ &=\frac{1}{a-b}\int_{0}^{1}\mathrm{d}u\,\frac{u^{b-1}-u^{a-1}}{\left(1+u\right)\ln{\left(u\right)}}\\ &=\frac{1}{a-b}\int_{0}^{1}\mathrm{d}u\,\frac{1}{1+u}\int_{a}^{b}\mathrm{d}t\,u^{t-1}\\ &=\frac{1}{a-b}\int_{a}^{b}\mathrm{d}t\int_{0}^{1}\mathrm{d}u\,\frac{u^{t-1}}{1+u}\\ &=\frac{1}{a-b}\int_{a}^{b}\mathrm{d}t\,\frac12\left[\psi{\left(\frac{t+1}{2}\right)}-\psi{\left(\frac{t}{2}\right)}\right]\\ &=\frac{1}{a-b}\int_{a}^{b}\mathrm{d}t\,\frac{d}{dt}\left[\ln{\left(\Gamma{\left(\frac{t+1}{2}\right)}\right)}-\ln{\left(\Gamma{\left(\frac{t}{2}\right)}\right)}\right]\\ &=\frac{1}{a-b}\left[\ln{\left(\Gamma{\left(\frac{b+1}{2}\right)}\right)}-\ln{\left(\Gamma{\left(\frac{b}{2}\right)}\right)}-\ln{\left(\Gamma{\left(\frac{a+1}{2}\right)}\right)}+\ln{\left(\Gamma{\left(\frac{a}{2}\right)}\right)}\right]\\ &=\frac{1}{a-b}\ln{\left(\frac{\Gamma{\left(\frac{a}{2}\right)}\,\Gamma{\left(\frac{b+1}{2}\right)}}{\Gamma{\left(\frac{a+1}{2}\right)}\,\Gamma{\left(\frac{b}{2}\right)}}\right)}.\blacksquare\\ \end{align}$$

The $a=b$ case can be recovered from the above result using L'Hopital's rule to calculate the limit.

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Answer 2

$$\displaystyle{ = \sum\limits_{k = 0}^\infty {\left( {{{\left( { - 1} \right)}^k}\frac{{\ln \left( {\dfrac{{k + b}}{{k + a}}} \right)}}{{a - b}}} \right)} = \frac{1}{{a - b}}\sum\limits_{k = 0}^\infty {\left( {{{\left( { - 1} \right)}^k}\ln \left( {\frac{{k + b}}{{k + a}}} \right)} \right)} }$$ . Nevertheless

$$\displaystyle{\sum\limits_{k = 0}^\infty {\left( {{{\left( { - 1} \right)}^k}\ln \left( {\frac{{k + b}}{{k + a}}} \right)} \right)} = \mathop {\lim }\limits_{N \to \infty } \sum\limits_{k = 0}^{2N} {\left( {{{\left( { - 1} \right)}^k}\ln \left( {\frac{{k + b}}{{k + a}}} \right)} \right)} = \mathop {\lim }\limits_{N \to \infty } \ln \left( {\frac{{b\left( {1 + a} \right)\left( {2 + b} \right)..\left( {2N - 1 + a} \right)\left( {2N + b} \right)}}{{a\left( {1 + b} \right)\left( {2 + a} \right)..\left( {2N - 1 + b} \right)\left( {2N + a} \right)}}} \right) = \mathop {\lim }\limits_{N \to \infty } \ln {P_n}}$$ And $$\displaystyle{{P_N} = \frac{{b\left( {1 + a} \right)\left( {2 + b} \right)..\left( {2N - 1 + a} \right)\left( {2N + b} \right)}}{{a\left( {1 + b} \right)\left( {2 + a} \right)..\left( {2N - 1 + b} \right)\left( {2N + a} \right)}} = \frac{{\left( {{b^2}{{\left( {2 + b} \right)}^2}..{{\left( {2N + b} \right)}^2}} \right)\left( {a\left( {1 + a} \right)\left( {2 + a} \right)..\left( {2N - 1 + a} \right)\left( {2N + a} \right)} \right)}}{{\left( {{a^2}{{\left( {2 + a} \right)}^2}..{{\left( {2N + a} \right)}^2}} \right)\left( {b\left( {1 + b} \right)\left( {2 + b} \right)..\left( {2N - 1 + b} \right)\left( {2N + b} \right)} \right)}} = }$$

$$\displaystyle{ = \frac{{{{\left( {\frac{b}{2}\left( {1 + \frac{b}{2}} \right)..\left( {N + \frac{b}{2}} \right)} \right)}^2}\left( {a\left( {1 + a} \right)\left( {2 + a} \right)..\left( {2N - 1 + a} \right)\left( {2N + a} \right)} \right)}}{{{{\left( {\frac{a}{2}\left( {1 + \frac{a}{2}} \right)..\left( {N + \frac{a}{2}} \right)} \right)}^2}\left( {b\left( {1 + b} \right)\left( {2 + b} \right)..\left( {2N - 1 + b} \right)\left( {2N + b} \right)} \right)}} = }$$

$$\displaystyle{\frac{{{N^b}{{\left( {\dfrac{{\left( {b/2} \right)\left( {1 + b/2} \right)..\left( {N + b/2} \right)}}{{{N^{b/2}}N!}}} \right)}^2}\left( {\dfrac{{a\left( {1 + a} \right)\left( {2 + a} \right)..\left( {2N - 1 + a} \right)\left( {2N + a} \right)}}{{{{\left( {2N} \right)}^a}\left( {2N} \right)!}}} \right){{\left( {2N} \right)}^a}}}{{{N^a}{{\left( {\dfrac{{\left( {a/2} \right)\left( {1 + a/2} \right)..\left( {N + a/2} \right)}}{{{N^{a/2}}N!}}} \right)}^2}\left( {\dfrac{{b\left( {1 + b} \right)\left( {2 + b} \right)..\left( {2N - 1 + b} \right)\left( {2N + b} \right)}}{{{{\left( {2N} \right)}^b}\left( {2N} \right)!}}} \right){{\left( {2N} \right)}^b}}}}$$ Taking the limit as N goes to infinity, we get:

$$\lim_{N\to\infty} P_N = 2^{a-b} \frac{\Gamma(b)}{\Gamma(a)} \cdot \frac{\Gamma^2(\frac{a}{2})}{\Gamma^2(\frac{b}{2})}$$

Then, we have:

$$\int_0^1 \int_0^1 \frac{x^{a-1} y^{b-1}}{(1+xy)\ln (xy)} dx dy = \frac{1}{a-b}\sum_{k=0}^{\infty} (-1)^k \ln \left( \frac{k+b}{k+a} \right) + \ln 2 + \frac{1}{a-b}\ln\left(\frac{\Gamma(b)}{\Gamma(a)} \cdot \frac{\Gamma^2(\frac{a}{2})}{\Gamma^2(\frac{b}{2})}\right)$$

If the numbers $\displaystyle{ a,b}$ are positive integers we would have a more elegant result