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Motivation. I was thinking in the more simple case of (6), I am saying the case $r=s=1$, that tell us this MathWorld. Then in this case one can check that the identity holds using integration methods. When I did it I've considered from Hata's integral the following different integral in this

Question. What's about $$\int_0^1\int_0^1\frac{\log(xy)xy}{-1+\log(xy)}dxdy?$$

Remark. The context was that I was trying with my imagination different expressions for an hypothetical similar (6) but now for $\sum_{k=r+1}^\infty\frac{\mu(n)}{n^3}$, where $\mu(n)$ is the Möbius function. For example I tried also consider $\int_0^1\left(\int_1^\infty\left(\int_1^\infty\frac{dx}{xy(1-(1-xy)z)}\right)dy\right)dz$, but I understand that it is science fiction to find an expression for $\sum_{k=r+1}^\infty\frac{\mu(n)}{n^3}$ from this random method.

My attempt. I can deduce the following statement $$\int_0^1\int_0^1\frac{\log(xy)xy}{-1+\log(xy)}dxdy=\frac{1}{4}+\int_0^1\int_0^1\frac{dxdy}{-1+\log(xy)},$$ and I know that $$\int\frac{dx}{-1+\log(xy)}=e\frac{\operatorname{Ei}(\log(xy)-1)}{y}+\operatorname{ constant}.$$ Then can we finish the example to get the integral in the Question in terms of particular values of special functions? Thanks in advance.

References:

Hata, A New Irrationality Measure for $\zeta(3)$, Acta Arith. 92, 47-57 (2000).

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    What exactly is (6)? politeness would dictate that you don't make potential repliers go hunt down what exactly you are referring to. – 2017-01-27
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    @nbubis many thanks for your attention. If you do the calculations for $r=s=1$ of (6) you can deduce a similar integral that I've written. Many thanks. – 2017-01-27
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    Again, you haven't said what $(6)$ is. – 2017-01-27
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    I'm sorry then, but it is (6) of the MathWorld article. Then is the identity (6) case $r=s=1$. Many thanks @nbubis – 2017-01-27
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    All users $$\int_0^1\left(\int_1^\infty\left(\int_1^\infty\frac{dx}{xy(1-(1-xy)z)}\right)dy\right)dz$$ corresponds to my *attempt* to relate the case $r=s=1$ with the specialization $r=1$ of this tail $\sum_{k=r+1}^\infty\frac{\mu(n)}{n^3}$. As I said I understand that it isn't a reasoning mathematical, but I am interestng in to know an integral representation for this $\sum_{k=r+1}^\infty\frac{\mu(n)}{n^3}$, if such similar representation is feasible. If you know some idea to exploit it and you want to edit a new question in this site Mathematics Stack Exchange it is the best. Many thanks. – 2017-01-28

2 Answers 2

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Change variables $xy/e=t$ to get $$ \int_0^1\int_0^1\frac{dxdy}{-1+\log(xy)}=\int_0^1 dy \frac{e}{y}\int_0^{y/e}\frac{dt}{\log t}=\int_0^1 dy \frac{e}{y}\mathrm{li}(y/e)\ , $$ where $\mathrm{li}(z)$ is the logarithmic integral. The second integral admits an antiderivative $$ \int dy \frac{1}{y}\mathrm{li}(y/e)=\text{li}\left(\frac{y}{e}\right) \ln \left(\frac{y}{e}\right)-\frac{y}{e}+C\ , $$ therefore your double integral reads $$ \int_0^1\int_0^1\frac{dxdy}{-1+\log(xy)}=-1-e \text{li}\left(\frac{1}{e}\right)\ . $$

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    Many thanks for your early answer, and attention. Now I am studying the answer. – 2017-01-27
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    Many thanks one more time for your trick to get the integratio and refer the antiderivative. – 2017-01-27
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$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\int_{0}^{1}\int_{0}^{1}{\ln\pars{xy}xy \over -1 + \ln\pars{xy}}\,\dd x\,\dd y = \int_{0}^{1}\int_{0}^{1}\ln\pars{xy}xy \bracks{-\int_{0}^{\infty}\pars{xy \over \expo{}}^{t}\,\dd t}\,\dd x\,\dd y \\[5mm] = &\ -\int_{0}^{\infty}\expo{-t}\int_{0}^{1}\int_{0}^{1}\ln\pars{xy}x^{t + 1} y^{t + 1}\,\dd x\,\dd y\,\dd t \\[5mm] = &\ -2\int_{0}^{\infty}\expo{-t} \bracks{\int_{0}^{1}\ln\pars{x}x^{t + 1} \,\dd x}\pars{\int_{0}^{1}y^{t + 1}\,\dd y}\,\dd t = 2\int_{0}^{\infty}{\expo{-t} \over \pars{t + 2}^{3}}\,\dd t \\[5mm] = &\ 2\expo{}^2\int_{2}^{\infty}{\expo{-t} \over t^{3}}\,\dd t = 2\expo{}^{2}\bracks{\mrm{E}_{3}\pars{2} \over 2^{3 - 1}} = \bbx{\ds{{1 \over 2}\,e^{2}\,\mrm{E}_{3}\pars{2}}} \approx 0.1113 \end{align}

$\ds{\mrm{E}_{p}\pars{z}}$ is the Generalized Exponenential Integral Function.

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    Many thanks, you are very generous with this. I am going to study your nice answer. – 2017-01-28
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    @user243301 Thanks. It's nice it was helpful for you. – 2017-01-28
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    I understand the first identity $\int a^tdt=a^t/log(a)+\text{constant}$, and next calculations in second line, but I don't know where is the $\log(y)$ in the third line. Many thanks. – 2017-01-28
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    $\int_{0}^{1}\int_{0}^{1}\ln(x)x^{t + 1}y^{t + 1}\,dx\,dy + \int_{0}^{1}\int_{0}^{1}\ln(y)x^{t + 1}y^{t + 1}\,dx\,dy$. If you exchange $x$ by $y$( and $y$ by $x$ ) in the second integral you get the first one. So, you have twice the first integral. Note the factor $2$ which multiplies the whole double integration. – 2017-01-28
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    Oh I'm sorry, very thanks much for these calculations, sincerely are nice. – 2017-01-28