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Question-142301




Question Number 142301 by 777316 last updated on 29/May/21
Answered by Dwaipayan Shikari last updated on 29/May/21
sin (((πs)/2))Γ(1−s)=sin (((πs)/2))(π/(sin(πs)Γ(s)))  lim_(s→2n) =(π/(2cos ((π/2)s)Γ(s)))=(π/(2Γ(2n)(−1)^n ))
$$\mathrm{sin}\:\left(\frac{\pi{s}}{\mathrm{2}}\right)\Gamma\left(\mathrm{1}−{s}\right)=\mathrm{sin}\:\left(\frac{\pi{s}}{\mathrm{2}}\right)\frac{\pi}{{sin}\left(\pi{s}\right)\Gamma\left({s}\right)} \\ $$$$\underset{{s}\rightarrow\mathrm{2}{n}} {\mathrm{lim}}=\frac{\pi}{\mathrm{2cos}\:\left(\frac{\pi}{\mathrm{2}}{s}\right)\Gamma\left({s}\right)}=\frac{\pi}{\mathrm{2}\Gamma\left(\mathrm{2}{n}\right)\left(−\mathrm{1}\right)^{{n}} } \\ $$

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