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What is the antiderivative, and line-by-line derivation, of

$$\int_0^1 \int_0^1 x\cdot y \enspace \ln(x\cdot y) \enspace dx \hspace{1mm} dy = \enspace ?$$

Not sure if this would help, but could we generalize from the following well-known rule

$$\int_{-\infty}^\infty x \ln x \hspace{1mm} dx = \frac{1}{2}x^2 \ln x - \frac{1}{4} x^2 + C$$

develarist
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    The fact that $\ln(xy) = \ln x + \ln y$ should help. (The $\infty$ limits on your indefinite integral make no sense.) – Ethan Bolker Sep 22 '20 at 14:51
  • Wolfram Alpha is giving the following solution that includes complex variables though: (have to copy and paste entire link) https://www.wolframalpha.com/input/?i=integral+integral+xy+ln%28xy%29+dx+dy – develarist Sep 22 '20 at 14:54
  • That link is about integral integral x. Your question is just a real definite integral. Your one dimensional indefinite integral (properly written) and my hint should lead to a solution. – Ethan Bolker Sep 22 '20 at 15:06
  • as i said in the comment, the link if clicked is not the full URL. For some reason stack wont admit the link beyond int int x, so it has to be selected in full manually, copied and then pasted – develarist Sep 22 '20 at 15:12

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Let us compute the given integral, $J$ for short: $$ \begin{aligned} J &=\iint_{[0,1]^2}xy\; \ln(xy)\; dx\; dy\\ &=\iint_{[0,1]^2}xy\; (\ln x+\ln y)\; dx\; dy\\ &= \iint_{[0,1]^2}xy\; \ln x\; dx\; dy + \iint_{[0,1]^2}xy\; \ln y\; dx\; dy \\ &= 2\iint_{[0,1]^2}xy\; \ln x\; dx\; dy\text{ by symmetry} \\ &= \int_{x=0}^1x\; \ln x\; dx \\ &= \int_0^1\frac 12(x^2)'\; \ln x\; dx \\ &=\left[\frac 12(x^2)'\; \ln x\right]_0^1 - \int_0^1\frac 12x^2\; (\ln x)'\; dx \\ &=\frac 12(0-0)-\frac 12\int_0^1x\; dx \\ &=-\frac 14\ . \end{aligned} $$

dan_fulea
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  • thanks, would you be up for another https://math.stackexchange.com/questions/3829062/analytical-closed-form-solution-for-the-entropy-of-a-parametric-copula – develarist Sep 22 '20 at 15:33