14
$\begingroup$

UPDATED

Hi I am trying to prove the following$$ I:=\int_1^\infty \log \log \left(x\right)\frac{dx}{1-x+x^2}=\frac{2\pi}{\sqrt 3}\left(\frac{5}{6}\log (2\pi)-\log \Gamma \big(\frac{1}{6}\big)\right). $$ I am not sure at all how to get started on this one. This looks quite intimidating. Something I realized was $$ \int_1^\infty \log \log \left(x\right)\frac{dx}{1-x+x^2}=\int_0^1 \log \log \left(\frac{1}{x}\right)\frac{dx}{1-x+x^2}. $$Thanks.

Note the Gamma function is given by $$ \Gamma(n)=(n-1)!,\quad \Gamma(z)=\int_0^\infty t^{z-1}e^{-t}\, dt. $$ EDIT: THE incorrect integral I first posted was because of a typo. The result of it is given by (notice the denominator sign mistake I made) $$ I_2:=\int_0^1 \log \log \left(\frac{1}{x}\right)\frac{dx}{1+x+x^2}=\frac{\pi}{\sqrt 3}\log\left(\frac{\sqrt[3]{2\pi}\Gamma(2/3)}{\Gamma(1/3)}\right). $$ as you can see the results are different, enjoy both. Obviously, I am ONLY interested in solving I thanks.

$\endgroup$
  • 1
    $\begingroup$ $\displaystyle\large\ln\left(1 \over x\right) < 0$ when $\displaystyle\large x > 1$. $\endgroup$ – Felix Marin May 14 '14 at 4:46
  • 2
    $\begingroup$ @integrals : Can you check the question? Changing the limits to 0..1 also does not match with the answer numerically. $\endgroup$ – gar May 14 '14 at 5:20
  • 2
    $\begingroup$ This is explicitly calculated in a paper of Victor Adamchik. "A class of logarithmic integrals. Proceedings ISSAC, 1–8, 1997". It goes back to Bierens de Haan, "Nouvelles Tables d’Intégrales Définies" (Table 148 (5)-page 208) (1867). $\endgroup$ – Omran Kouba May 14 '14 at 7:56
  • 3
    $\begingroup$ @mrf it was merely a typo, i meant to write $1-x+x^2$ in denominator. What context are you looking for? "The usual lack of context", what does this refer to? Thanks for your rude comment:) $\endgroup$ – Jeff Faraci May 14 '14 at 16:42
  • 1
    $\begingroup$ This integral is also computed in Adamchick's paper, page 8, formula (31). repository.cmu.edu/cgi/… $\endgroup$ – Leucippus May 15 '14 at 4:20
2
$\begingroup$

Here is an answer. Clear, letting $x\to 1/x$, we have \begin{eqnarray*} I=\int_0^1 \log(-\log x)\frac{1}{1-x+x^2}dx. \end{eqnarray*} Then the rest follows from A closed form of $\int_0^1\frac{\ln\ln\left({1}/{x}\right)}{x^2-x+1}\mathrm dx$.

$\endgroup$
  • $\begingroup$ You're cheating Mr. xpaul! +1 ≧◠◡◠≦✌ $\endgroup$ – Anastasiya-Romanova 秀 Sep 2 '14 at 15:23
  • $\begingroup$ @V-Moy, what do you mean? $\endgroup$ – xpaul Sep 3 '14 at 22:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.