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Question Number 216445 by Tawa11 last updated on 08/Feb/25

$$\mathrm{Prove}\mathrm{that}\mathrm{\Gamma }\left(\frac{1}{2}\right)=\sqrt{\pi }$$

Commented by Mathstar last updated on 08/Feb/25

$$\mathrm{\Gamma }\left(\mathrm{x}\right)\mathrm{\Gamma }(1-\mathrm{x})=\frac{\pi }{\sin \left(\pi \mathrm{x}\right)}$$$$\mathrm{Let}\mathrm{x}=\frac{1}{2}$$$${\mathrm{\Gamma }}^2\left(\frac{1}{2}\right)=\pi \Rightarrow \mathrm{\Gamma }\left(\frac{1}{2}\right)=\sqrt{\pi }$$

Commented by Tawa11 last updated on 10/Feb/25

$$\mathrm{Thanks}\mathrm{sir}.$$

Answered by MrGaster last updated on 08/Feb/25

$$={\int }_0^{\mathrm{\infty }}{t}^{-\frac{1}{2}}{e}^{-t}dt$$$$\mathrm{Let}t=x^2\Rightarrow dt=2xdx\Rightarrow {\int }_0^{\mathrm{\infty }}{t}^{-\frac{1}{2}}{e}^{-t}dt={\int }_0^{\mathrm{\infty }}{x}^{-1}{e}^{-x^2}2xdx=2{\int }_0^{\mathrm{\infty }}{e}^{-x^2}dx$$$$\mathrm{Let}I={\int }_0^{\mathrm{\infty }}{e}^{-x^2}dx\Rightarrow I^2=\left({\int }_0^{\mathrm{\infty }}{e}^{-x^2}dx\right)\left({\int }_0^{\mathrm{\infty }}{e}^{-y^2}dy\right)={\int }_0^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}{e}^{-(x^2+y^2)}dxdy$$$$x=r\cos \theta ,y=r\sin \theta \Rightarrow dxdy=rdrd\theta$$$${\int }_0^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}{e}^{-(x^2+y^2)}dxdy={\int }_0^{\frac{\pi }{2}}{\int }_0^{\frac{\pi }{2}}{e}^{-r^2}rdrd\theta =\frac{\pi }{2}{\int }_0^{\mathrm{\infty }}{e}^{-r^2}rdr$$$$\mathrm{Let}u=r^2\Rightarrow du=2rdr\Rightarrow {\int }_0^{\mathrm{\infty }}{e}^{-r^2}rdr=\frac{1}{2}{\int }_0^{\mathrm{\infty }}{e}^{-u}du=\frac{1}{2}$$$$I^2=\frac{\pi }{2}\cdot \frac{1}{2}=\frac{\pi }{4}\Rightarrow I=\frac{\sqrt{\pi }}{2}$$$$\overline{)\begin{array}{c}\mathrm{\Gamma }\left(\frac{1}{2}\right)=2I=2\cdot \frac{\sqrt{\pi }}{2}=‘‘{\sqrt{\pi }}^{″}\end{array}}$$

Commented by issac last updated on 08/Feb/25

$$\mathrm{Oh}...\mathrm{I}\mathrm{think}{\mathrm{you}}^{\prime }\mathrm{re}\mathrm{more}\mathrm{accurate}.$$$$\mathrm{I}\mathrm{solved}\mathrm{using}\mathrm{a}\mathrm{function}$$$$\int t^{\alpha -1}{e}^{-t}dt=-\mathrm{\Gamma }(\alpha ,z)+C$$$$\mathrm{that}\mathrm{was}\mathrm{already}\mathrm{defined}\mathrm{and}$$$$\mathrm{you}\mathrm{approached}\mathrm{it}\mathrm{at}\mathrm{little}\mathrm{closer}\mathrm{to}\mathrm{the}$$$$\mathrm{method}\mathrm{of}\mathrm{Calculus}$$

Commented by Tawa11 last updated on 10/Feb/25

$$\mathrm{Thanks}\mathrm{sir}.$$

Answered by issac last updated on 08/Feb/25

$$\mathrm{\Gamma }\left(z\right)={\int }_0^{\mathrm{\infty }}{t}^{z-1}{e}^{-t}dt,\mathfrak{R}\left(z\right)>0$$$$\mathrm{\Gamma }\left(\frac{1}{2}\right)={\int }_0^{\mathrm{\infty }}\frac{e^{-t}}{\sqrt{t}}dt$$$$\int t^{\alpha }{e}^{-t}dt=-\mathrm{\Gamma }(\alpha +1,t)+C$$$$\left(\mathrm{\Gamma }(\alpha ,t)\mathrm{is}\mathrm{incomplete}\mathrm{gamma}\mathrm{function}\right)$$$$\int t^{-\frac{1}{2}}{e}^{-t}dt=-\mathrm{\Gamma }(\frac{1}{2},t)+C$$$$\therefore {\int }_0^{\mathrm{\infty }}{t}^{-\frac{1}{2}}{e}^{-t}dt={[-\mathrm{\Gamma }(\frac{1}{2},t)]}_{t=0}^{t=\mathrm{\infty }}=\sqrt{\pi }.$$

Commented by Tawa11 last updated on 10/Feb/25

$$\mathrm{Thanks}\mathrm{sir}.$$

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