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Question Number 153105 by alisiao last updated on 04/Sep/21

find ln πšͺ(x) ?

$${find}\:\:\boldsymbol{{ln}}\:\boldsymbol{\Gamma}\left(\boldsymbol{{x}}\right)\:? \\ $$

Answered by puissant last updated on 04/Sep/21

ln(Ξ“(x))=βˆ’Ξ³xβˆ’lnx+Ξ£_(k=1) ^∞ {(x/k)βˆ’ln(1+(x/k))} =(xβˆ’(1/2))lnxβˆ’x+(1/2)ln(2Ο€)+(1/2)Ξ£_(n=2) ^∞ (((nβˆ’1))/(n(n+1)))ΞΆ(n,x+1) =(xβˆ’(1/2))lnxβˆ’x+(1/2)ln(2Ο€)+2∫_0 ^∞ ((arctan((x/a)))/(e^(2Ο€x) βˆ’1))dx.

$${ln}\left(\Gamma\left({x}\right)\right)=βˆ’\gamma{x}βˆ’{lnx}+\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\left\{\frac{{x}}{{k}}βˆ’{ln}\left(\mathrm{1}+\frac{{x}}{{k}}\right)\right\} \\ $$$$=\left({x}βˆ’\frac{\mathrm{1}}{\mathrm{2}}\right){lnx}βˆ’{x}+\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{2}\pi\right)+\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{\left({n}βˆ’\mathrm{1}\right)}{{n}\left({n}+\mathrm{1}\right)}\zeta\left({n},{x}+\mathrm{1}\right) \\ $$$$=\left({x}βˆ’\frac{\mathrm{1}}{\mathrm{2}}\right){lnx}βˆ’{x}+\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{2}\pi\right)+\mathrm{2}\int_{\mathrm{0}} ^{\infty} \frac{{arctan}\left(\frac{{x}}{{a}}\right)}{{e}^{\mathrm{2}\pi{x}} βˆ’\mathrm{1}}{dx}. \\ $$

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