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Question Number 134845 by mnjuly1970 last updated on 07/Mar/21
           advanced  calculus...           ∫_0 ^( 1) xψ(1+x)dx=??          ...m.n...
$$\:\:\:\:\:\:\:\:\:\:\:{advanced}\:\:{calculus}… \\ $$$$\:\:\:\:\:\:\:\:\:\int_{\mathrm{0}} ^{\:\mathrm{1}} {x}\psi\left(\mathrm{1}+{x}\right){dx}=?? \\ $$$$\:\:\:\:\:\:\:\:…{m}.{n}… \\ $$
Answered by Dwaipayan Shikari last updated on 08/Mar/21
∫_0 ^1 xψ(1+x)dx  =∫_0 ^1 (xψ(x)+1)dx  =xlog(Γ(x))−∫_0 ^1 log(Γ(x))dx+1  =0−(1/2)∫_0 ^1 log((π/(sin(πx))))+1  =1−(1/2)log(π)+(1/(2π))∫_0 ^π log(sin(x))dx  =1−(1/2)log(π)−(π/(2π))log(2)=1−log((√(2π)))
$$\int_{\mathrm{0}} ^{\mathrm{1}} {x}\psi\left(\mathrm{1}+{x}\right){dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \left({x}\psi\left({x}\right)+\mathrm{1}\right){dx} \\ $$$$={xlog}\left(\Gamma\left({x}\right)\right)−\int_{\mathrm{0}} ^{\mathrm{1}} {log}\left(\Gamma\left({x}\right)\right){dx}+\mathrm{1} \\ $$$$=\mathrm{0}−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} {log}\left(\frac{\pi}{{sin}\left(\pi{x}\right)}\right)+\mathrm{1} \\ $$$$=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}{log}\left(\pi\right)+\frac{\mathrm{1}}{\mathrm{2}\pi}\int_{\mathrm{0}} ^{\pi} {log}\left({sin}\left({x}\right)\right){dx} \\ $$$$=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}{log}\left(\pi\right)−\frac{\pi}{\mathrm{2}\pi}{log}\left(\mathrm{2}\right)=\mathrm{1}−{log}\left(\sqrt{\mathrm{2}\pi}\right) \\ $$

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