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

Prove that Γ((1/2)) = (√π)

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

Commented by Mathstar last updated on 08/Feb/25

Γ(x)Γ(1−x)=(π/(sin(πx))) Let x=(1/2) Γ^2 ((1/2))=π ⇒ Γ((1/2)) = (√π)

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

Commented by Tawa11 last updated on 10/Feb/25

Thanks sir.

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

Answered by MrGaster last updated on 08/Feb/25

=∫_0 ^∞ t^(−(1/2)) e^(−t) dt Let t=x^2 ⇒dt=2xdx⇒∫_0 ^∞ t^(−(1/2)) e^(−t) dt=∫_0 ^∞ x^(−1) e^(−x^2 ) 2xdx=2∫_0 ^∞ e^(−x^2 ) dx Let I=∫_0 ^∞ e^(−x^2 ) dx⇒I^2 =(∫_0 ^∞ e^(−x^2 ) dx)(∫_0 ^∞ e^(−y^2 ) dy)=∫_0 ^∞ ∫_0 ^∞ e^(−(x^2 +y^2 )) dxdy x=r cos θ,y=r sin θ⇒dxdy=r dr dθ ∫_0 ^∞ ∫_0 ^∞ e^(−(x^2 +y^2 )) dxdy=∫_0 ^(π/2) ∫_0 ^(π/2) e^(−r^2 ) rdrdθ=(π/2)∫_0 ^∞ e^(−r^2 ) r dr Let u=r^2 ⇒du=2rdr⇒∫_0 ^∞ e^(−r^2 ) rdr=(1/2)∫_0 ^∞ e^(−u) du=(1/2) I^2 =(π/2)∙(1/2)=(π/4)⇒I=((√π)/2) determinant (((Γ((1/2))=2I=2∙((√π)/2)=“(√π)”)))

$$=\int_{\mathrm{0}} ^{\infty} {t}^{−\frac{\mathrm{1}}{\mathrm{2}}} {e}^{−{t}} {dt} \\ $$$$\mathrm{Let}\:{t}={x}^{\mathrm{2}} \Rightarrow{dt}=\mathrm{2}{xdx}\Rightarrow\int_{\mathrm{0}} ^{\infty} {t}^{−\frac{\mathrm{1}}{\mathrm{2}}} {e}^{−{t}} {dt}=\int_{\mathrm{0}} ^{\infty} {x}^{−\mathrm{1}} {e}^{−{x}^{\mathrm{2}} } \mathrm{2}{xdx}=\mathrm{2}\int_{\mathrm{0}} ^{\infty} {e}^{−{x}^{\mathrm{2}} } {dx} \\ $$$$\mathrm{Let}\:{I}=\int_{\mathrm{0}} ^{\infty} {e}^{−{x}^{\mathrm{2}} } {dx}\Rightarrow{I}^{\mathrm{2}} =\left(\int_{\mathrm{0}} ^{\infty} {e}^{−{x}^{\mathrm{2}} } {dx}\right)\left(\int_{\mathrm{0}} ^{\infty} {e}^{−{y}^{\mathrm{2}} } {dy}\right)=\int_{\mathrm{0}} ^{\infty} \int_{\mathrm{0}} ^{\infty} {e}^{−\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)} {dxdy} \\ $$$${x}={r}\:\mathrm{cos}\:\theta,{y}={r}\:\mathrm{sin}\:\theta\Rightarrow{dxdy}={r}\:{dr}\:{d}\theta \\ $$$$\int_{\mathrm{0}} ^{\infty} \int_{\mathrm{0}} ^{\infty} {e}^{−\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)} {dxdy}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {e}^{−{r}^{\mathrm{2}} } {rdrd}\theta=\frac{\pi}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} {e}^{−{r}^{\mathrm{2}} } {r}\:{dr} \\ $$$$\mathrm{Let}\:{u}={r}^{\mathrm{2}} \Rightarrow{du}=\mathrm{2}{rdr}\Rightarrow\int_{\mathrm{0}} ^{\infty} {e}^{−{r}^{\mathrm{2}} } {rdr}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} {e}^{−{u}} {du}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${I}^{\mathrm{2}} =\frac{\pi}{\mathrm{2}}\centerdot\frac{\mathrm{1}}{\mathrm{2}}=\frac{\pi}{\mathrm{4}}\Rightarrow{I}=\frac{\sqrt{\pi}}{\mathrm{2}} \\ $$$$\begin{array}{|c|}{\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\mathrm{2}{I}=\mathrm{2}\centerdot\frac{\sqrt{\pi}}{\mathrm{2}}=``\sqrt{\pi}''}\\\hline\end{array} \\ $$

Commented by issac last updated on 08/Feb/25

Oh... I think you′re more accurate. I solved using a function ∫ t^(α−1) e^(−t) dt=−Γ(α,z)+C that was already defined and you approached it at little closer to the method of Calculus

$$\mathrm{Oh}...\:\mathrm{I}\:\mathrm{think}\:\mathrm{you}'\mathrm{re}\:\mathrm{more}\:\mathrm{accurate}. \\ $$$$\mathrm{I}\:\mathrm{solved}\:\mathrm{using}\:\mathrm{a}\:\mathrm{function} \\ $$$$\int\:\:{t}^{\alpha−\mathrm{1}} {e}^{−{t}} {dt}=−\Gamma\left(\alpha,{z}\right)+{C} \\ $$$$\mathrm{that}\:\mathrm{was}\:\mathrm{already}\:\mathrm{defined}\:\mathrm{and} \\ $$$$\mathrm{you}\:\mathrm{approached}\:\mathrm{it}\:\mathrm{at}\:\mathrm{little}\:\mathrm{closer}\:\mathrm{to}\:\mathrm{the}\: \\ $$$$\mathrm{method}\:\mathrm{of}\:\mathrm{Calculus} \\ $$

Commented by Tawa11 last updated on 10/Feb/25

Thanks sir.

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

Answered by issac last updated on 08/Feb/25

Γ(z)=∫_0 ^( ∞) t^(z−1) e^(−t) dt , R(z)>0 Γ((1/2))=∫_0 ^( ∞) (e^(−t) /( (√t))) dt ∫ t^α e^(−t) dt=−Γ(α+1,t)+C (Γ(α,t) is incomplete gamma function) ∫ t^(−(1/2)) e^(−t) dt=−Γ((1/2),t)+C ∴∫_0 ^( ∞) t^(−(1/2)) e^(−t) dt=[−Γ((1/2),t)]_(t=0) ^(t=∞) =(√π).

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

Commented by Tawa11 last updated on 10/Feb/25

Thanks sir.

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

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