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Evaluate :

$$\int_{0}^{1} \log^2\left(\frac{\Gamma(x)}{\Gamma(1-x)}\right)\log^2\left(\frac{\Gamma(x)}{\Gamma(1+x)}\right) dx$$

This is my attempt in below ... but what I want is simplify more to answer....thanks.

My own attempt: enter image description here

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    $\begingroup$ Welcome to the site ! You must understand that a lot of people here are ready to help you but that no one will do your homework. Tell us what you already tried and explain where you are stuck. Cheers. $\endgroup$ Oct 19, 2018 at 8:20
  • $\begingroup$ Approximation: $23+\tan \left(\frac{e^{6/7}}{16 \ln ^{\frac{5}{4}}(2)}\right)$ for 10 digits correct. $\endgroup$ Oct 19, 2018 at 21:04
  • $\begingroup$ @MariuszIwaniuk. Just out of curiosity : how did you get this approximation ? $\endgroup$ Oct 20, 2018 at 4:38
  • $\begingroup$ @ClaudeLeibovici. Maple have a command identify give exact value from numeric. $\endgroup$ Oct 20, 2018 at 13:27
  • $\begingroup$ @MariuszIwaniuk. Thanks for the information ! I was not aware of that. It is great ! $\endgroup$ Oct 20, 2018 at 14:53

1 Answer 1

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I really do not see how you could simplify the monster. I really wonder if a closed form expression could exist. I would try numerical integration.

We can get an approximate solution using the following series expansions at $x=0$. $$\log (\Gamma (x))=-\log (x)-\gamma x+\frac{\pi ^2 x^2}{12}+\frac{x^3 \psi ^{(2)}(1)}{6}+\frac{\pi ^4 x^4}{360}+O\left(x^5\right)$$ $$\log (\Gamma (1-x))=\gamma x+\frac{\pi ^2 x^2}{12}-\frac{x^3 \psi ^{(2)}(1)}{6}+\frac{\pi ^4 x^4}{360}+O\left(x^5\right)$$ $$\log (\Gamma (1+x))=-\gamma x+\frac{\pi ^2 x^2}{12}+\frac{x^3 \psi ^{(2)}(1)}{6}+\frac{\pi ^4 x^4}{360}+O\left(x^5\right)$$ which would make the integrand to be $$\log ^4(x)+4 \gamma x \log ^3(x)+4 \gamma ^2 x^2 \log ^2(x)-\frac{2}{3} x^3 \left(\psi ^{(2)}(1) \log ^3(x)\right)-\frac{4}{3} x^4 \left(\gamma \psi ^{(2)}(1) \log ^2(x)\right)+O\left(x^5\right)$$ and the antiderivative $$x \left(\log ^4(x)-4 \log ^3(x)+12 \log ^2(x)-24 \log (x)+24\right)+\frac{1}{2} \gamma x^2 \left(4 \log ^3(x)-6 \log ^2(x)+6 \log (x)-3\right)+\frac{4}{27} \gamma ^2 x^3 \left(9 \log ^2(x)-6 \log (x)+2\right)+\frac{1}{192} x^4 \psi ^{(2)}(1) \left(-32 \log ^3(x)+24 \log ^2(x)-12 \log (x)+3\right)-\frac{4}{375} x^5 \left(\gamma \psi ^{(2)}(1) \left(25 \log ^2(x)-10 \log (x)+2\right)\right)+O\left(x^6\right)$$ Then, for the definite integral $$\gamma \left(\frac{16 \zeta (3)}{375}-\frac{3}{2}\right)-\frac{\zeta (3)}{32}+24+\frac{8 \gamma ^2}{27}\approx 23.2249$$ while the numerical integration leads to $23.2372$ that is to say in error by $0.05$%.

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