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

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:

• 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. Oct 19, 2018 at 8:20
• Approximation: $23+\tan \left(\frac{e^{6/7}}{16 \ln ^{\frac{5}{4}}(2)}\right)$ for 10 digits correct. Oct 19, 2018 at 21:04
• @MariuszIwaniuk. Just out of curiosity : how did you get this approximation ? Oct 20, 2018 at 4:38
• @ClaudeLeibovici. Maple have a command identify give exact value from numeric. Oct 20, 2018 at 13:27
• @MariuszIwaniuk. Thanks for the information ! I was not aware of that. It is great ! Oct 20, 2018 at 14:53

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 $$0.05$$%.