# A integral from “irresistible integrals”

Evaluate : $$\int_0^1\frac{x}{(1-x+x^2)^2}\ln \ln \frac{1}{x}\text{d}x$$ The answer on the book is $$-\frac{\gamma}{3}-\frac13\ln\frac{6\sqrt{3}}{\pi}+\frac{\pi\sqrt{3}}{27}\left(5\ln 2\pi-6\ln \Gamma\left(\frac16\right)\right)$$ Could anyone show a proof? Refers to chapter 12 of the book pg237.

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And your question is...? – Gerry Myerson Feb 11 '13 at 6:13
@GerryMyerson : I dont know how to get the answer. – Ryan Feb 11 '13 at 6:22
Then maybe you should edit that into the body. While you're there, a page reference to the book would not be amiss. – Gerry Myerson Feb 11 '13 at 6:29
That book is a lot of fun, but many of the results (such as the one here) are completely beyond me. – marty cohen Feb 11 '13 at 7:17
Boros and Moll give a reference to "Adamchik (1997)". Then in the references, they list three 1997 papers by Adamchik. My guess is they mean A class of logarithmic integrals, Proc ISSAC '97. Have you looked for that paper? Maybe a better reference is MR2546186 (2010h:33031) Medina, Luis A.; Moll, Victor H.; A class of logarithmic integrals, Ramanujan J. 20 (2009), no. 1, 91–126. – Gerry Myerson Feb 11 '13 at 23:44

If you're still interested, Ryan, here is a link to Adamchiks paper on the topic.

You can rewrite $\displaystyle \frac{x}{(x^{2}-x+1)^{2}}=\frac{x^{3}}{(x^{3}+1)^{2}}+\frac{2x^{2}}{(x^{3}+1)^{2}}+\frac{x}{(x^{3}+1)^{2}}$ and use the formulae he derives.

One of which is $\displaystyle \int_{0}^{1}\frac{x^{n-1}}{(1+x^{n})^{2}}\log\left(\log(1/x)\right)dx=\frac{-1}{2n}\left(\gamma+\log(\frac{2n}{\pi})\right)$

Your integral is down near (43).

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I don't have the full result yet, but I believe this is a way to get there. Perform a substitution $x = e^{-v}$ and get the integral as

$$\int_0^{\infty} dv \frac{e^{-2 v}}{(1-e^{-v}+e^{-2 v})^2} \log{v}$$

We can attack this by the Residue Theorem by considering the following integral about a keyhole contour $C$:

$$\oint_C dz \frac{e^{-2 z}}{(1-e^{-z}+e^{-2 z})^2} \log^2{z}$$

There are poles in the integrand at $z=\pm i\pi/3$. This integrand is $i 2 \pi$ times the sum of the residues at these poles, the calculation of which I will spare you. The contour integral then has value

$$\oint_C dz \frac{e^{-2 z}}{(1-e^{-z}+e^{-2 z})^2} \log^2{z} = i 2 \pi \frac{1}{9 \pi} \Re{\left \{ \log{\left ( -i \frac{\pi}{3} \right )} \left [ 9 (\sqrt{3}+i) - (3 - i \sqrt{3}) \pi \log{\left ( -i \frac{\pi}{3} \right )} \right ] \right \}}$$

But we also know that the contour integral vanishes except on the paths along the real axis:

\begin{align}\oint_C dz \frac{e^{-2 z}}{(1-e^{-z}+e^{-2 z})^2} \log^2{z} &= \int_0^{\infty} dv \frac{e^{-2 v}}{(1-e^{-v}+e^{-2 v})^2} [\log^2{v} - (\log{v} + i 2 \pi)^2]\\ &= -i 4 \pi \int_0^{\infty} dv \frac{e^{-2 v}}{(1-e^{-v}+e^{-2 v})^2} \log{v} + 4 \pi^2 \int_0^{\infty} dv \frac{e^{-2 v}}{(1-e^{-v}+e^{-2 v})^2}\\ \end{align}

This last integral on the RHS has a nonzero, real value:

$$4 \pi^2 \int_0^{\infty} dv \frac{e^{-2 v}}{(1-e^{-v}+e^{-2 v})^2} = \frac{4 \pi^2}{27} (9 + 2 \sqrt{3} \pi)$$

so I am a little at a loss of what to do next, or where I may have erred.

EDIT

OK, I found my error, and it was assuming that one of the poles was at $z=-i \pi/3$, which is not allowable here, as $z \in [0, 2 \pi)$. Thus the poles are at $z = i \pi/3$ and $z=i 5 \pi/3$. More later, as I am still not sure where the $\gamma$ and $\Gamma{(1/6)}$ come from.

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Looking forward for that :) – Ryan Feb 13 '13 at 11:36