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Question Number 135127 by Dwaipayan Shikari last updated on 10/Mar/21

$${\int }_0^1{log}^2\left(\mathrm{\Gamma }\left(x\right)\right)dx$$

Answered by mathmax by abdo last updated on 11/Mar/21

$$\mathrm{we}\mathrm{have}\mathrm{\Gamma }\left(\mathrm{x}\right).\mathrm{\Gamma }(1-\mathrm{x})=\frac{\pi }{\sin \left(\pi \mathrm{x}\right)}\Rightarrow$$$$\log \left(\mathrm{\Gamma }\left(\mathrm{x}\right)\right)+\log \left(\mathrm{\Gamma }(1-\mathrm{x})\right)=\log \left(\pi \right)-\log \left(\sin \left(\pi \mathrm{x}\right)\right)\Rightarrow$$$$\log {\left(\mathrm{\Gamma }\left(\mathrm{x}\right)\right)}^2+2\log \left(\mathrm{\Gamma }\left(\mathrm{x}\right)\right)\log \left(\mathrm{\Gamma }(1-\mathrm{x})\right)+{\log }^2\left(\mathrm{\Gamma }(1-\mathrm{x})\right)$$$$={\log }^2\left(\pi \right)-2\log \pi \log \left(\sin \left(\pi \mathrm{x}\right)\right)+{\log }^2\left(\sin \left(\pi \mathrm{x}\right)\right)\Rightarrow$$$${\int }_0^1{\log }^2\left(\mathrm{\Gamma }\left(\mathrm{x}\right)\right)\mathrm{dx}+{\int }_0^1{\log }^2\left(\mathrm{\Gamma }(1-\mathrm{x})\right)\mathrm{dx}+2{\int }_0^1\log \left(\mathrm{\Gamma }\left(\mathrm{x}\right)\right)\log \left(\mathrm{\Gamma }(1-\mathrm{x})\right)\mathrm{dx}$$$$={\log }^2\left(\pi \right)-2\log \pi {\int }_0^1\log \left(\sin \left(\pi \mathrm{x}\right)\right)\mathrm{dx}+{\int }_0^1{\log }^2\left(\sin \left(\pi \mathrm{x}\right)\right)\mathrm{dx}$$$${\int }_0^1{\log }^2\left(\mathrm{\Gamma }(1-\mathrm{x})\right)\mathrm{dx}{=}_{1-\mathrm{x}=\mathrm{t}}{\int }_0^1{\log }^2\left(\mathrm{\Gamma }\left(\mathrm{t}\right)\right)\mathrm{dt}$$$${\int }_0^1\log \left(\sin \left(\pi \mathrm{x}\right)\right)\mathrm{dx}{=}_{\pi \mathrm{x}=\mathrm{t}}\frac{1}{\pi }{\int }_0^{\pi }\log \left(\mathrm{sint}\right)\mathrm{dt}$$$$=\frac{1}{\pi }{\int }_0^{\frac{\pi }{2}}\log \left(\mathrm{sint}\right)\mathrm{dt}+\frac{1}{\pi }{\int }_{\frac{\pi }{2}}^{\pi }\log \left(\mathrm{sint}\right)\mathrm{dt}(\to \mathrm{t}=\frac{\pi }{2}+\mathrm{u})$$$$=\frac{1}{\pi }(-\frac{\pi }{2}\log 2)+\frac{1}{\pi }(-\frac{\pi }{2}\log 2)=-\log \left(2\right)$$$$\Rightarrow 2{\int }_0^1{\log }^2\left(\mathrm{\Gamma }\left(\mathrm{x}\right)\right)\mathrm{dx}+2{\int }_0^1\log \left(\mathrm{\Gamma }\left(\mathrm{x}\right)\right).\log \left(\mathrm{\Gamma }(1-\mathrm{x})\right)\mathrm{dx}$$$$={\log }^2\pi +2\log \pi \log 2+{\int }_0^1{\log }^2\left(\sin \left(\pi \mathrm{x}\right)\right)\mathrm{dx}...\mathrm{be}\mathrm{continued}...$$

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