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Question Number 67540 by mathmax by abdo last updated on 28/Aug/19

$$provethat\mid \mathrm{\Gamma }(\frac{1}{2}+it)\mid =\sqrt{\frac{2\pi }{e^{\pi t}+e^{-\pi t}}}$$$$and\mid \mathrm{\Gamma }(1+it)\mid =\sqrt{\frac{2\pi t}{e^{\pi t}-e^{-\pi t}}}$$

Commented by ~ À ® @ 237 ~ last updated on 28/Aug/19

$$$$$$$$$$I{}^{\prime }mnotsureifwealwayshave\stackrel{-}{\mathrm{\Gamma }}\left(z\right)=\mathrm{\Gamma }\left(\stackrel{-}{z}\right)$$$$Butifistrue$$$$\mid \mathrm{\Gamma }(\frac{1}{2}+it){\mid }^2=\mathrm{\Gamma }(\frac{1}{2}+it)\mathrm{\Gamma }(\frac{1}{2}-it)=\mathrm{\Gamma }(\frac{1}{2}+it)\mathrm{\Gamma }(1-(\frac{1}{2}+it))$$$$=\frac{\pi }{sin(\frac{\pi }{2}+i\pi t)}usingcomplementsformulas\mathrm{\Gamma }\left(s\right)\mathrm{\Gamma }(1-s)=\frac{\pi }{sin\left(\pi s\right)}$$$$=\frac{\pi }{cos\left(i\pi t\right)}=\frac{2\pi }{e^{\pi t}+e^{-\pi t}}$$$$\mid \mathrm{\Gamma }(1+it){\mid }^2=\mathrm{\Gamma }(1+it)\mathrm{\Gamma }(1-it)=it\mathrm{\Gamma }\left(it\right)\mathrm{\Gamma }(1-it)=\frac{i\pi t}{sin\left(i\pi t\right)}=\frac{i\pi t}{\frac{e^{-\pi t}-e^{\pi t}}{2i}}=\frac{2\pi t}{e^{\pi t}-e^{-\pi t}}$$$$$$

Commented by mathmax by abdo last updated on 29/Aug/19

$$thankyousir.$$

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