Faster method to calculate the exact solution of the following integral (based on the ideas of Fedor Petrov: https://mathoverflow.net/users/4312/fedor-petrov, Double integral with logarithms, URL (version: 2019-04-15): https://mathoverflow.net/q/328126):

$$J\equiv \int_{0}^{1}{\int_{0}^{1}{\frac{\ln x-\ln y}{x-y}}}dxdy .$$

Since $$f\left( x,y \right)=\frac{\ln x-\ln y}{x-y}=f\left( y,x \right),$$ the surface $f\left( x,y \right) $ is symmetric with respect to the bisector plane $x = y$; so,

$$\frac{J}{2}=\int_{0}^{1}{dx\int_{0}^{x}{\frac{\ln x-\ln y}{x-y}}}dy.$$

With the change of variable$$y\equiv tx,\ t\in \left( 0,\ 1 \right),$$ the integral $$\int_{0}^{x}{\frac{\ln x-\ln y}{x-y}}dy,$$ is transformed into the following one that does not depend on $x$, $$ I\equiv -\int_{0}^{1}{\frac{\ln t}{1-t}\,}dt.$$

The integration on the unit square $(0, 0), (1, 0), (1, 1), (0,1)$ is reduced to the integration on the triangle $(0, 0), (1, 0), (1, 1).$

To solve the integral $I$ we will carry out the new change of variable, $$s\equiv 1-t,$$ by which $I$ is transformed into the integral that defines the dilogarithm, whose value for $s = 1$ coincides with the Riemann zeta $\zeta \left( 2 \right)$,whose value is well known:

$$I=\int_{1}^{0}{\frac{\ln \left( 1-s \right)}{s}\,}ds=\text{L}{{\text{i}}_{2}}\left( 1 \right)=\zeta \left( 2 \right)=\frac{{{\pi }^{2}}}{6}.$$ Therefore, the solution to the proposed integral is $$J=\frac{{{\pi }^{2}}}{3}.$$

Note. I've tried it by polylogarithmic transformations, but I couldn't get the result $\frac{{{\pi }^{2}}}{3}.$


closed as off-topic by YCor, user44191, Pace Nielsen, Gerald Edgar, Piotr Hajlasz Apr 15 at 20:13

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    $\begingroup$ MSE is a right place for such type questions. Both Maple and Mathematica confirm $ \frac{\pi ^2}{3}$. $\endgroup$ – user64494 Apr 15 at 15:07
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    $\begingroup$ I do not think this is such a bad question. I would not expect even every research mathematician to come up with the answer right away. So I vote not to close. $\endgroup$ – RP_ Apr 15 at 15:52
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    $\begingroup$ @user64494 what is a general strategy: when you have an integral which you do not know how to calculate, how should you decide between MSE and MO? $\endgroup$ – Fedor Petrov Apr 15 at 17:48
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    $\begingroup$ @user64494 being the author of several papers devoted to exact closed-form integration, I feel so old now. $\endgroup$ – Fedor Petrov Apr 15 at 20:50
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    $\begingroup$ Learning techniques for a problem here could be illuminating (depending on ...) for future purposes of integration or summation that they relate to or approximated by (if you are into numerical values). Plus, cute arguments are aesthetically pleasing ... $\endgroup$ – T. Amdeberhan Apr 15 at 22:06

By symmetry we have $J/2=\int_0^1 dx \int_0^x f(x, y) dy$ where $f(x, y) $ is your integrand. Integrating against $y$ for fixed $x$ we denote $y=tx$, $t$ varies from 0 to 1 and the integral against $y$ reads as $-\int_0^1 \frac{\log t} {1-t}dt$. It does not depend on $x$ and is well known to be equal to $\pi^2/6$ (you may use the geometric progression expansion $\frac{1}{1 - t} =\sum_{n>0 } t^{n-1}$ and integrate term-wise to get $\sum 1/n^2$).


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