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Question Number 53950 by maxmathsup by imad last updated on 27/Jan/19

$$calculate{\int }_{\frac{1}{3}}^{\frac{1}{2}}\mathrm{\Gamma }\left(x\right)\mathrm{\Gamma }(1-x)dxwith\mathrm{\Gamma }\left(x\right)={\int }_0^{\mathrm{\infty }}{t}^{x-1}{e}^{-t}dtwithx>0.$$

Commented bymaxmathsup by imad last updated on 30/Jan/19

$$wehavefor0<x<1\mathrm{\Gamma }\left(x\right).\mathrm{\Gamma }(1-x)=\frac{\pi }{sin\left(\pi x\right)}\left(complmentsformula\right)\Rightarrow$$ $${\int }_{\frac{1}{3}}^{\frac{1}{2}}\mathrm{\Gamma }\left(x\right).\mathrm{\Gamma }(1-x)dx=\pi {\int }_{\frac{1}{3}}^{\frac{1}{2}}\frac{dx}{sin\left(\pi x\right)}{=}_{\pi x=t}\pi {\int }_{\frac{\pi }{3}}^{\frac{\pi }{2}}\frac{dt}{\pi sin\left(t\right)}$$ $$={\int }_{\frac{\pi }{3}}^{\frac{\pi }{2}}\frac{dt}{sint}{=}_{tan\left(\frac{t}{2}\right)=u}{\int }_{\frac{1}{\sqrt{3}}}^1\frac{1}{\frac{2u}{1+u^2}}\frac{2du}{1+u^2}={\int }_{\frac{1}{\sqrt{3}}}^1\frac{du}{u}={[ln\mid u\mid ]}_{\frac{1}{\sqrt{3}}}^1=-ln\left(\frac{1}{\sqrt{3}}\right)$$ $$=ln\left(\sqrt{3}\right).$$

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