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Question Number 134845 by mnjuly1970 last updated on 07/Mar/21

$$advancedcalculus...$$$${\int }_0^{1}x\psi (1+x)dx=??$$$$...m.n...$$

Answered by Dwaipayan Shikari last updated on 08/Mar/21

$${\int }_0^{1}x\psi (1+x)dx$$$$={\int }_0^1(x\psi \left(x\right)+1)dx$$$$=xlog\left(\mathrm{\Gamma }\left(x\right)\right)-{\int }_0^{1}log\left(\mathrm{\Gamma }\left(x\right)\right)dx+1$$$$=0-\frac{1}{2}{\int }_0^{1}log\left(\frac{\pi }{sin\left(\pi x\right)}\right)+1$$$$=1-\frac{1}{2}log\left(\pi \right)+\frac{1}{2\pi }{\int }_0^{\pi }log\left(sin\left(x\right)\right)dx$$$$=1-\frac{1}{2}log\left(\pi \right)-\frac{\pi }{2\pi }log\left(2\right)=1-log\left(\sqrt{2\pi }\right)$$

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