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nice-calculus-prove-that-i-pi-sinh-pi-Euler-gamma-function-m-n-july-1970-




Question Number 117979 by mnjuly1970 last updated on 14/Oct/20
          ... nice  calculus...       prove  that:            ∣  Γ ( i ) ∣=^?  (√(π/(sinh(π))))            Γ: Euler gamma function                        ...m.n.july.1970...
$$\:\:\:\:\:\:\:\:\:\:…\:{nice}\:\:{calculus}… \\ $$$$\:\:\:\:\:{prove}\:\:{that}: \\ $$$$\:\:\:\:\:\:\:\:\:\:\mid\:\:\Gamma\:\left(\:{i}\:\right)\:\mid\overset{?} {=}\:\sqrt{\frac{\pi}{{sinh}\left(\pi\right)}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\Gamma:\:\mathscr{E}{uler}\:{gamma}\:{function}\:\: \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…{m}.{n}.{july}.\mathrm{1970}… \\ $$
Commented by Lordose last updated on 14/Oct/20
THEY ARE NOT EQUAL
Answered by mindispower last updated on 14/Oct/20
Γ(1−i)Γ(i)=(π/(sin(iπ)))  Γ(1−i)=−iΓ((−i)  ⇒Γ(−i)Γ(i)=(π/(−isin(iπ)))=(π/(sh(π)))  Γ(−i).Γ(i)=(π/(sh(π)))⇒∣Γ(i)∣^2 =(π/(sh(π)))  ∣Γ(i)∣=(√(π/(sh(π))))
$$\Gamma\left(\mathrm{1}−{i}\right)\Gamma\left({i}\right)=\frac{\pi}{{sin}\left({i}\pi\right)} \\ $$$$\Gamma\left(\mathrm{1}−{i}\right)=−{i}\Gamma\left(\left(−{i}\right)\right. \\ $$$$\Rightarrow\Gamma\left(−{i}\right)\Gamma\left({i}\right)=\frac{\pi}{−{isin}\left({i}\pi\right)}=\frac{\pi}{{sh}\left(\pi\right)} \\ $$$$\Gamma\left(−{i}\right).\Gamma\left({i}\right)=\frac{\pi}{{sh}\left(\pi\right)}\Rightarrow\mid\Gamma\left({i}\right)\mid^{\mathrm{2}} =\frac{\pi}{{sh}\left(\pi\right)} \\ $$$$\mid\Gamma\left({i}\right)\mid=\sqrt{\frac{\pi}{{sh}\left(\pi\right)}} \\ $$$$ \\ $$
Commented by mnjuly1970 last updated on 14/Oct/20
thank you
$${thank}\:{you} \\ $$
Commented by mindispower last updated on 14/Oct/20
withe pleasur
$${withe}\:{pleasur} \\ $$

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