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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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