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Question Number 122045 by mohammad17 last updated on 13/Nov/20

$$provethat\mathrm{\Gamma }s\mathrm{\Gamma }1-s=\frac{\pi }{sins\pi }$$

Commented by Dwaipayan Shikari last updated on 14/Nov/20

$$\mathrm{\Gamma }\left(s\right)=\frac{e^{-\gamma s}}{s}\prod ^{\mathrm{\infty }}{e}^{\frac{s}{n}}{(1+\frac{s}{n})}^{-1}$$$$\mathrm{\Gamma }(1-s)=\frac{e^{-\gamma +\gamma s}}{1-s}\prod ^{\mathrm{\infty }}{e}^{\frac{1-s}{n}}{(1+\frac{1-s}{n})}^{-1}$$$$\mathrm{\Gamma }\left(s\right)\mathrm{\Gamma }(1-s)=\frac{e^{-\gamma }}{s(1-s)}\prod ^{\mathrm{\infty }}{e}^{\frac{1}{n}}{(1+\frac{s}{n})}^{-1}{(1-\frac{s}{n}+\frac{1}{n})}^{-1}$$$$\frac{\pi s}{sin\pi s}=\prod ^{\mathrm{\infty }}{(1-\frac{s}{n})}^{-1}{(1+\frac{s}{n})}^{-1}$$$$\mathrm{\Gamma }\left(s\right)\mathrm{\Gamma }(1-s)=\frac{e^{-\gamma +\sum ^{\mathrm{\infty }}\frac{1}{n}}}{s(1-s)}.\frac{\pi s}{sin\pi s}.\prod ^{\mathrm{\infty }}(1-\frac{s}{n})\prod ^{\mathrm{\infty }}{(1-\frac{s}{n}+\frac{1}{n})}^{-1}$$$$\mathrm{\Gamma }\left(s\right)\mathrm{\Gamma }(1-s)=\frac{e^{log\left(n\right)}}{(1-s)}.\frac{\pi }{sin\pi s}((1-s)\left(\frac{2-s}{2}\right)...)\frac{1.2.3}{(2-s)(3-s)(4-s)}$$$$\mathrm{\Gamma }\left(s\right)\mathrm{\Gamma }(1-s)=\underset{n\to \mathrm{\infty }}{\lim }\frac{\pi n}{sin\pi s}.\frac{1}{n-\mathit{s}+1}$$$$\mathrm{\Gamma }\left(s\right)\mathrm{\Gamma }(1-s)=\frac{\pi }{sin\pi s}$$

Commented by mnjuly1970 last updated on 14/Nov/20

$$bravomrdwaipayan..$$$$veryniceasalways...$$

Commented by Dwaipayan Shikari last updated on 14/Nov/20

With pleasure��

Answered by mnjuly1970 last updated on 14/Nov/20

$$proofbyusingdouble$$$$integral.$$$$\mathrm{\Gamma }\left(s\right)={\int }_0^{\mathrm{\infty }}{x}^{s-1}{e}^{-x}dx...\left(i\right)$$$$\mathrm{\Gamma }(1-s)={\int }_0^{\mathrm{\infty }}{y}^{-s}{e}^{-y}dy...\left(ii\right)$$$$from\left(i\right)and\left(ii\right):$$$$\mathrm{\Gamma }\left(s\right)\mathrm{\Gamma }(1-s)={\int }_0^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}\frac{1}{x}{\left(\frac{x}{y}\right)}^s{e}^{-(x+y)}dxdy$$$$\{{}_{\frac{x}{y}=v}^{x+y=u}\Rightarrow \{{}_{x=\frac{uv}{1+v}}^{y=\frac{u}{1+v}}$$$$dxdy=\mid \frac{\partial (x,y)}{\partial (u,v)}\mid dudvwhere$$$$\frac{\partial (x,y)}{\partial (u,v)}=J(u,v)=\left|\begin{array}{c}\frac{\partial x}{\partial u}\frac{\partial x}{\partial v}\\ \frac{\partial y}{\partial u}\frac{\partial y}{\partial v}\end{array}\right|$$$$\therefore J(u,v)=x_u{y}_v-x_v{y}_u=\left(\frac{v}{1+v}\right)\left(\frac{-u}{{(1+v)}^2}\right)-\left(\frac{u}{{(1+v)}^2}\right)\left(\frac{1}{1+v}\right)$$$$J(u,v)=\frac{-u}{{(1+v)}^2}✓$$$$\mathrm{\Gamma }\left(s\right)\mathrm{\Gamma }(1-s)={\int }_0^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}\frac{1+v}{uv}{v}^s{e}^{-u}\mid \frac{-u}{{(1+v)}^2}\mid dudv$$$$={\int }_0^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}\frac{v^{s-1}}{1+v}{e}^{-u}dudv={\int }_0^{\mathrm{\infty }}\frac{v^{s-1}}{1+v}dv\underset{analysis}{\stackrel{complex}{=}}\frac{\pi }{sin\left(\pi s\right)}✓$$$$...m.n...$$$$$$

Commented by Dwaipayan Shikari last updated on 14/Nov/20

$$Great!$$

Commented by mnjuly1970 last updated on 14/Nov/20

$$thankyou..$$$$$$

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