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


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.

  • $\begingroup$ Feel free to refer the literature if my integral is easily deduced from other. $\endgroup$
    – user243301
    Commented Jun 24, 2018 at 18:12
  • 1
    $\begingroup$ The actual numerical result of your integral is negative. You have incorrectly shown a positive result. $\endgroup$ Commented Jun 24, 2018 at 21:39
  • 2
    $\begingroup$ 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
  • 3
    $\begingroup$ 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


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}

  • $\begingroup$ Many thanks I need to read and study your great answer. $\endgroup$
    – user243301
    Commented Jun 25, 2018 at 12:07

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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