Evaluate the double integral $\int _0^1\int _0^1\frac{x+i}{(1-ix y) \ln (x y)} \,dx\,dy$

Question

We know that

$$\int _0^1\int _0^1\frac{x-1}{(1+x y) \ln (x y)} \, dx\,dy=\gamma$$

$$\int _0^1\int _0^1\frac{x+1}{(1-x y) \ln (x y)}\,dx\,dy=\ln \frac4\pi$$

I wonder what would be

$$\int _0^1\int _0^1\frac{x+i}{(1-ix y) \ln (x y)}\,dx\,dy$$

Mathematica fails.

Answer 1

It is quite easy to prove that: $$\iint_{(0,1)^2}\frac{(xy)^n}{\log(xy)}\,dx\,dy = -\frac{1}{n+1},\tag{1}$$ $$\iint_{(0,1)^2}\frac{x^{n+1}y^n}{\log(xy)}\,dx\,dy = -\log\frac{n+2}{n+1},\tag{2}$$ hence: $$\begin{eqnarray*} I &=& \iint_{(0,1)^2}\frac{(x+i)}{(1-i xy)\log(xy)}\,dx\,dy = \sum_{n\geq 0}\iint_{(0,1)^2}\frac{i^n(x+i)(xy)^n}{\log(xy)}\,dx\,dy\\&=&-\sum_{n\geq 0}i^n \left(\frac{i}{n+1}+\log\frac{n+2}{n+1}\right)=\frac{1}{2}\log 2-\frac{\pi}{4}i+\sum_{n\geq 0}i^n\left(\log(n+1)-\log(n+2)\right)\\&=&\frac{1}{2}\log 2-\frac{\pi}{4}i+\color{red}{S_1}+i\color{blue}{S_2}\tag{3}\end{eqnarray*} $$ where: $$\color{red}{S_1} = \sum_{n\geq 0}(-1)^n\left(\log(2n+1)-\log(2n+2)\right),$$ $$\color{blue}{S_2} = \sum_{n\geq 0}(-1)^n\left(\log(2n+2)-\log(2n+3)\right),\tag{4}$$ from which: $$ \color{red}{S_1} = \log\prod_{n=0}^{+\infty}\frac{4n+1}{4n+2}\cdot\frac{4n+4}{4n+3},$$ $$ \color{blue}{S_2} = \log\prod_{n=0}^{+\infty}\frac{4n+2}{4n+3}\cdot\frac{4n+5}{4n+4}.\tag{5}$$ Using now the Euler product for the $\Gamma$ function it is not difficult to check that: $$ \color{red}{S_1}=\log\frac{\sqrt{2\pi^3}}{\Gamma\left(\frac{1}{4}\right)^2},\qquad \color{blue}{S_2}=\log\frac{4\sqrt{2\pi}}{\Gamma\left(\frac{1}{4}\right)^2}.\tag{6}$$

Answer 2

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\norm}[1]{\left\vert\left\vert\, #1\,\right\vert\right\vert} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{\int_{0}^{1}\int_{0}^{1}{x + \ic \over \pars{1 - \ic xy}\ln\pars{xy}} \,\dd x\,\dd y:\ {\large ?}}$.

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\begin{align}&\color{#66f}{\large \int_{0}^{1}\int_{0}^{1}{x + \ic \over \pars{1 - \ic xy}\ln\pars{xy}} \,\dd x\,\dd y} =\int_{0}^{1}\pars{1 + {\ic \over x}} \int_{0}^{x}{\dd y \over \pars{1 - \ic y}\ln\pars{y}}\,\dd x \\[5mm]&=\int_{0}^{1}{\dd y \over \pars{1 - \ic y}\ln\pars{y}} \int_{y}^{1}\pars{1 + {\ic \over x}}\,\dd x =\int_{0}^{1}{1 - y -\ic\ln\pars{y} \over \pars{1 - \ic y}\ln\pars{y}}\,\dd y \\[5mm]&=\int_{0}^{1}{1 - y \over \pars{1 - \ic y}\ln\pars{y}}\,\dd y -\ic\int_{0}^{1}{\dd y \over 1 - \ic y}\,\dd y \\[5mm]&=\dsc{\int_{0}^{1}{1 - y \over \pars{1 - \ic y}\ln\pars{y}}\,\dd y} +\half\,\ln\pars{2} - {1 \over 4}\,\pi\ic\tag{1} \end{align}

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Lets evaluate the '$\dsc{\mbox{red integral}}$': \begin{align}&\overbrace{ \dsc{\int_{0}^{1}{1 - y \over \pars{1 - \ic y}\ln\pars{y}}\,\dd y}} ^{\ds{\dsc{y}=\dsc{\expo{-t}}\ \imp\ \dsc{t}=\dsc{-\ln\pars{y}}}} =\int_{\infty}^{0} {1 - \expo{-t} \over \pars{1 - \ic\expo{-t}}\pars{-t}}\,\pars{-\expo{-t}\,\dd t} =\int_{0}^{\infty} {\expo{-2t} - \expo{-t} \over 1 - \ic\expo{-t}}\,{\dd t \over t} \\[5mm]&=\sum_{n\ =\ 0}^{\infty}\ \ic^{n} \int_{0}^{\infty}\bracks{\expo{-\pars{n + 2}t} - \expo{-\pars{n + 1}t}} \,{\dd t \over t} \\[5mm]&=\sum_{n\ =\ 0}^{\infty}\ \ic^{n} \bracks{\pars{n + 2}\int_{0}^{\infty}\ln\pars{t}\expo{-\pars{n + 2}t}\,\dd t -\pars{n + 1}\int_{0}^{\infty}\ln\pars{t}\expo{-\pars{n + 1}t}\,\dd t} \end{align}

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Since $\ds{\left.\int_{0}^{\infty}\ln\pars{t}\expo{-a t}\,\dd t\, \right\vert_{\Re\pars{a}\ >\ 0} = -\,{\gamma + \ln\pars{a} \over a}}$: \begin{align}&\dsc{\int_{0}^{1}{1 - y \over \pars{1 - \ic y}\ln\pars{y}}\,\dd y} =-\sum_{n\ =\ 0}^{\infty}\ \ic^{n}\ln\pars{n + 2 \over n + 1} \\[5mm]&=-\sum_{n\ =\ 0}^{\infty}\ \pars{-1}^{n}\ln\pars{n + 1 \over n + 1/2} -\ic\sum_{n\ =\ 0}^{\infty}\ \pars{-1}^{n}\ln\pars{n + 3/2 \over n + 1} \\[5mm]&=\sum_{n\ =\ 0}^{\infty}\ \bracks{ \ln\pars{n + 1 \over n + 3/4} -\ln\pars{n + 1/2 \over n + 1/4}} +\ic\sum_{n\ =\ 0}^{\infty}\ \bracks{ \ln\pars{n + 5/4 \over n + 1} -\ln\pars{n + 3/4 \over n + 1/2}} \\[5mm]&=\fermi\pars{1 \over 4} + \fermi\pars{\half}\ic\quad\mbox{where}\quad \fermi\pars{a}\equiv \sum_{n\ =\ 0}^{\infty}\ \bracks{ \ln\pars{n + a + 3/4 \over n + a + 1/2} -\ln\pars{n + a + 1/4 \over n + a}} \end{align} $\ds{\fermi\pars{a}}$ is easily evaluated by expressing the $\ds{\ln}$'s functions in integral form: $\ds{\ln\pars{n + \beta \over n + \alpha} =\pars{\beta - \alpha}\int_{0}^{1}{\dd t \over n + \alpha + \pars{\beta - \alpha}t}}$. Sum over $\ds{n}$ becomes somehow Digamma Functions which are integrated to yield $\ds{\ln\Gamma}$'s functions. The final result is: $$ \fermi\pars{a}=\ln\pars{\Gamma\pars{a + 1/4}\Gamma\pars{a + 1/2}\over \Gamma\pars{a}\Gamma\pars{a + 3/4}} $$ which yields: \begin{align} \fermi\pars{1 \over 4}&= \ln\pars{\root{\pi}\Gamma\pars{3/4} \over \Gamma\pars{1/4}} \\[5mm] \fermi\pars{\half}&= \ln\pars{\Gamma\pars{3/4} \over \root{\pi}\Gamma\pars{5/4}} \end{align}

The evaluation is completed.