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I would like to know if it is possible to justify the calculation of next definite integral

$$\int_0^1\int_0^1\frac{(\operatorname{arctanh}(xy))(\arctan(xy))}{\log(xy)}\,dx\,dy.$$

Question.(Corrected, see the comments) My reasonings and expriments with Wolfram Alpha online calculator, suggest me that the following identity holds $$\int_0^1\int_0^1\frac{(\operatorname{arctanh}(xy))(\arctan(xy))}{\log(xy)}\,dx\,dy=-\frac{1}{32}\left(16C-\pi^2+4\pi\log 2\right),$$ where $C$ denotes the Catalan's constant. Am I right? Do you know it or is it possible to justify? Many thanks.

My motivation was to compute an example of a reduction formula for integrals that I've known from a preprint by M.L. Glasser (Universidad de Valladolid). I've deduced the conjeture in the Question from my subsequent calculations for the limits of integration to deduce the closed-form of our doble integral (to me seem that were difficult calculations and the justification should be tedious, this is why I am asking here) with the mentioned CAS.

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  • $\begingroup$ Feel free to refer the literature if my integral is easily deduced from other. $\endgroup$
    – user243301
    Commented Jun 24, 2018 at 18:12
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    $\begingroup$ The actual numerical result of your integral is negative. You have incorrectly shown a positive result. $\endgroup$ Commented Jun 24, 2018 at 21:39
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    $\begingroup$ Might be easier to investigate the following conjecture first $$\int_0^1 \tan ^{-1}(x) \tanh ^{-1}(x) \, dx = \frac{C}{2}-\frac{1}{32} \pi (\pi -4 \log (2))$$ $\endgroup$ Commented Jun 24, 2018 at 21:40
  • $\begingroup$ Many thanks @JamesArathoon $\endgroup$
    – user243301
    Commented Jun 24, 2018 at 22:10
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    $\begingroup$ Using the substitution $(t, u) = (yx, y/x)$, you can check that the integral in question is equal to $$2\int_{0}^{1}\left(\int_{t}^{1}\frac{\operatorname{arctanh}(t)\arctan(t)}{\log t}\,\frac{du}{2u}\right)\,dt=-\int_{0}^{1}\operatorname{arctanh}(t)\arctan(t)\,dt.$$ $\endgroup$ Commented Jun 25, 2018 at 2:48

2 Answers 2

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As already pointed out in the comments, we have $$ I \equiv \int \limits_0^1 \int \limits_0^1 \frac{\arctan(x y) \operatorname{artanh} (x y)}{\log(x y)} \, \mathrm{d} x \, \mathrm{d} y = - \int \limits_0^1 \arctan(t) \operatorname{artanh}(t) \, \mathrm{d} t \, . $$ Using $$ \int \limits_0^x \operatorname{artanh} (t) \, \mathrm{d} t = x \operatorname{artanh}(x) + \frac{1}{2} \log (1 - x^2) = \frac{1}{2} \left[x \log \left(\frac{1+x}{1-x}\right) + \log (1 - x^2) \right] $$ for $x \in (-1,1)$, we can integrate by parts to get \begin{align} I &= - \frac{\pi}{4} \log(2) + \int \limits_0^1 \frac{t \operatorname{artanh}(t) + \frac{1}{2} \log(1-t^2)}{1+t^2} \, \mathrm{d} t \\ &= - \frac{\pi}{4} \log(2) + \frac{1}{2} \int \limits_0^1 \frac{2 t \log(1+t) - t \log(1-t^2) + \log(1-t^2)}{1+t^2} \, \mathrm{d} t \\ &\equiv - \frac{\pi}{4} \log(2) + I_1 + I_2 + I_3 \, . \end{align} We obtain $$ I_1 = \int \limits_0^1 \frac{t \log(1+t)}{1+t^2} \, \mathrm{d} t = \frac{\pi^2}{96} + \frac{\log^2 (2)}{8} $$ from this question and we find $$ I_2 = \frac{1}{2} \int \limits_0^1 \frac{- t \log(1-t^2)}{1+t^2} \, \mathrm{d} t = \frac{1}{4} \int \limits_0^1 \frac{- \log(1-s)}{1+s} \, \mathrm{d} s = \frac{\pi^2}{48} - \frac{\log^2 (2)}{8}$$ using the substitution $t^2 = s$ and this question. A well-known representation of Catalan's constant and the change of variables $t = \tan(s)$ yield \begin{align} I_3 &= \frac{1}{2} \int \limits_0^1 \frac{\log(1-t^2)}{1+t^2} \, \mathrm{d} t = \frac{1}{2} \int \limits_0^1 \frac{\log(t)}{1+t^2} \, \mathrm{d} t + \frac{1}{2} \int \limits_0^1 \frac{\log \left(\frac{1}{t} - t\right)}{1+t^2} \, \mathrm{d} t \\ &= - \frac{\mathrm{C}}{2} + \frac{1}{2} \int \limits_0^{\pi/4} \log \left(\frac{\cos(s)}{\sin(s)} - \frac{\sin(s)}{\cos(s)}\right) \, \mathrm{d} s \\ &= - \frac{\mathrm{C}}{2} + \frac{1}{2} \int \limits_0^{\pi/4} \log \left(\frac{2 \cos(2s)}{\sin(2 s)}\right) \, \mathrm{d} s \\ &= - \frac{\mathrm{C}}{2} + \frac{\pi}{8} \log(2) - \frac{1}{4} \int \limits_0^{\pi/2} \log(\tan(r)) \, \mathrm{d} r \\ &= - \frac{\mathrm{C}}{2} + \frac{\pi}{8} \log(2) \, , \end{align} where the last integral vanishes by symmetry (use $r \rightarrow \frac{\pi}{2} - r$).

Putting everything together, we confirm the conjectured result: \begin{align} I &= - \frac{\pi}{4} \log(2) + \frac{\pi^2}{96} + \frac{\log^2 (2)}{8} + \frac{\pi^2}{48} - \frac{\log^2 (2)}{8} - \frac{\mathrm{C}}{2} + \frac{\pi}{8} \log(2) \\ &= - \frac{\mathrm{C}}{2} - \frac{\pi}{8} \log(2) + \frac{\pi^2}{32} \, . \end{align}

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  • $\begingroup$ Many thanks I need to read and study your great answer. $\endgroup$
    – user243301
    Commented Jun 25, 2018 at 12:07
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Let’s start with $$ I \equiv \int \limits_0^1 \int \limits_0^1 \frac{\arctan(x y) \operatorname{artanh} (x y)}{\log(x y)} \, \mathrm{d} x \, \mathrm{d} y = - \int \limits_0^1 \arctan(t) \operatorname{artanh}(t) \, \mathrm{d} t \, . $$ We are going to handle the integral by first integrating $\arctan x.$ $\displaystyle \begin{aligned}\int \operatorname{arctanh} x dx & =x \operatorname{arctanh} x-\int \frac{x}{1-x^2} \\& =x \operatorname{arctanh} x-\frac{1}{2} \ln \left(1-x^2\right) \\& =\frac{1}{2}\left[x \ln \left(\frac{1+x}{1-x}\right)-\ln \left(1-x^2\right)\right]\end{aligned}\tag*{} $ Then performing integration by parts yields $\displaystyle \begin{aligned}&2 \int \limits_0^1 \arctan(t) \operatorname{artanh}(t) \, \mathrm{d} t \\= & \int_0^1 \arctan x d\left(x \ln \left(\frac{1+x}{1-x}\right)-\ln \left(1-x^2\right)\right) \\= & \underbrace{ \int_0^1 \frac{x}{1+x^2} \ln \left(\frac{1-x}{1+x}\right) d x}_{J} - \underbrace{ \int_0^1 \frac{\ln \left(1-x^2\right)}{1+x^2} d x}_{K} \end{aligned}$

For both integrals $J$ and $K$, we use the substitution $t=\frac{1-x}{1+x}$, then $\displaystyle \begin{aligned}& J=\int_0^1 \frac{\frac{1-t}{1+t}}{\frac{2\left(1+t^2\right)}{(1+t)^2}} \ln t \cdot \frac{2 d t}{(1+t)^2} \\& =\int_0^1 \frac{(1-t) \ln t}{(1+t)\left(1+t^2\right)} d t \\& =\int_0^1 \frac{\ln t}{t+1} d t-\int_0^1 \frac{t \ln t}{t^2+1} d t \\& =\int_0^1 \frac{\ln t}{t+1} d t-\frac{1}{4} \int_0^1 \frac{\ln t}{t+1} d t \\& =\frac{3}{4} \int_0^1 \frac{\ln t}{t+1} d t \\& =\frac{3}{4}\left(-\frac{\pi^2}{12}\right) \\& =-\frac{\pi^2}{16} \\\end{aligned}\tag*{} $


Similarly for $K$, we have $\displaystyle \begin{aligned}K& =\int_0^1 \frac{\ln \frac{4 t}{(1+t)^2}}{\frac{2\left(1+t^2\right)}{(1+t)^2}} \cdot \frac{2 d t}{(1+t)^2} \\& =\int_0^1 \frac{\ln (4 t)-2 \ln (1+t)}{1+t^2} d t \\& =\ln 4 \int_0^1 \frac{d t}{1+t^2}+\int_0^1 \frac{\ln t}{1+t^2} d t-2 \int_0^1 \frac{\ln (1+t)}{1+t^2}dt \\& =\frac{\pi \ln 4}{4}-G-\frac{\pi}{4} \ln 2 (\textrm{ For details, please refer to } \cdots (*))\\& =\frac{\pi}{4} \ln 2-G\end{aligned}\tag*{} $

Now we can conclude that

$\displaystyle \boxed{\int \limits_0^1 \int \limits_0^1 \frac{\arctan(x y) \operatorname{artanh} (x y)}{\log(x y)} \, \mathrm{d} x \, \mathrm{d} y =\frac{\pi^2}{32}+\frac{\pi}{8} \ln 2-\frac{G}{2}}\tag*{} $


Footnote (*)

We first express the integrand in terms of $\tan \theta$. $\displaystyle \int_0^1 \frac{\ln (1+t)}{1+t^2} d t=\int_0^{\frac{\pi}{4}} \ln (\tan \theta+1) d \theta\tag*{} $ Then using the substitution $\theta \mapsto \frac{\pi}{4}-\theta$, we have $$\displaystyle \begin{aligned}\int_0^1 \frac{\ln (1+t)}{1+t^2} d t& =\int_0^{\frac{\pi}{4}} \ln \left(1+\tan \left(\frac{\pi}{4}-\theta\right)\right) d \theta \\& =\int_0^{\frac{\pi}{4}} \ln \left(1+\frac{1-\tan \theta}{1+\tan \theta}\right) d \theta \\& =\int_0^{\frac{\pi}{4}} \ln 2 d \theta-\int_0^1 \frac{\ln (1+t)}{1+t^2} d t \\\int_0^1 \frac{\ln (1+t)}{1+t^2} d t & =\frac{\pi \ln 2}{8 }\end{aligned}\tag*{} $$

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