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How to evaluate $$\int_{0}^{1}\int_{0}^{1}\frac{x\ln x\ln y}{1-xy}\frac{dxdy}{\ln(xy)} ?$$

Any ideas on how to even start with this integral? It seems impossible to me.

There's a similar integral that originates from this site.

Zacky
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  • $\frac{x\ln(x)\ln(y)}{(1-xy)(\ln(xy))}$ has an infinite discontinuity at $x=1/y$. I don't know if this means it is undefined or if it is simply hard to compute. – Varun Vejalla Jul 05 '19 at 23:07
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    @automaticallyGenerated The curve $xy=1$ is outside the integration domain (the square $[0,1]^2$). – A.Γ. Jul 05 '19 at 23:15
  • Could you please give us some context about why you think it has a closed form? I would find this integral interesting if this integral does have. – Kemono Chen Jul 05 '19 at 23:27
  • @KemonoChen Wolfram Alpha seems to think the answer may be 1-2γ as a possible form, although it's not sure if that's exact or just an approximation – Brenton Jul 05 '19 at 23:36
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    This integral looks very similar to those discovered by Jonathan Sondow, http://home.earthlink.net/~jsondow/, there may be a way to get there from the existing integral $$\gamma = \int_{0}^{1}\int_{0}^{1} \frac{x-1}{(1-xy)\ln(xy)} dxdy$$ – Joshua Farrell Jul 05 '19 at 23:38
  • @A.Γ. that is correct. I didn't see that. – Varun Vejalla Jul 05 '19 at 23:39
  • I feel like integration by parts must be applied to get to the definition of $\gamma$ in @JoshuaFarrell's comment. I don't know what to use for $u$ and $v$ though. – Varun Vejalla Jul 06 '19 at 00:16

1 Answers1

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We can use Feynman's trick in two dimensions. Consider the following integral:

$$I(n)=\int_0^1\int_0^1 \frac{(xy)^{n-1} x\ln x\ln y}{\ln(xy)}dxdy.$$ Differentiating with respect to $n$ gives $$I'(n)=\int_0^1\int_0^1 (xy)^{n-1} x \ln x \ln y \,dxdy$$ $$=\int_0^1 x^{n} \ln x \, dx \int_0^1 y^{n-1} \ln y\, dy=\frac{1}{(n+1)^2}\frac{1}{n^2}.$$ Now we have to to get back to $I(n)$. Since $I(\infty)=0$, we have $$I(n)=-(I(\infty)-I(n))=-\int_n^\infty \frac{1}{(x+1)^2 x^2 } dx=-\frac{1}{n}-\frac{1}{n+1}+2\ln\left(1+\frac{1}{n}\right).$$ Finally, notice that $$\int_0^1\int_0^1 \frac{ x\ln x\ln y}{(1-xy)\ln(xy)}\,dxdy=\sum_{n=1}^\infty \int_0^1\int_0^1 \frac{(xy)^{n-1} x\ln x\ln y}{\ln(xy)}dxdy$$ $$=\sum_{n=1}^\infty \left(\underbrace{\frac{1}{n}-\frac{1}{n+1}}_{1}-2\left(\underbrace{\frac{1}{n}-\ln\left(1+\frac{1}{n}\right)}_{\gamma}\right)\right)=1-2\gamma.$$ Where $\gamma$ is the Euler-Mascheroni constant, also see here for the above identity.

Zacky
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    I didn't mention in the answer, but evaluate that integral it's enough to make the following substitutions: $$\int_n^\infty \frac{1}{(x+1)^2 x^2}dx\overset{x=\frac{1}{t}}=\int_0^\frac{1}{n} \frac{1}{\left(\frac{1}{t}+1\right)^2}dt\overset{t+1=y}=\int_1^{\frac{1}{n}+1} \frac{(y-1)^2}{y^2}dy$$ Also: $$\int_0^1 x^k dx=\frac{1}{k+1}\Rightarrow \int_0^1 x^k \ln x dx=\frac{d}{dk} \frac{1}{k+1}=-\frac{1}{(k+1)^2}$$ – Zacky Jul 19 '19 at 15:48
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    (+1) Sorry about the (now deleted) comment, which arose from misreading your original text. I have made the break between $I$ and $I'$ more distinctive in the above edit. – John Bentin Jul 19 '19 at 15:50