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

prove that Γs Γ1−s=(π/(sinsπ))

$${prove}\:{that}\:\Gamma{s}\:\Gamma\mathrm{1}−{s}=\frac{\pi}{{sins}\pi} \\ $$

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

Γ(s)=(e^(−γs) /s)Π^∞ e^(s/n) (1+(s/n))^(−1) Γ(1−s)=(e^(−γ+γs) /(1−s))Π^∞ e^((1−s)/n) (1+((1−s)/n))^(−1) Γ(s)Γ(1−s)=(e^(−γ) /(s(1−s)))Π^∞ e^(1/n) (1+(s/n))^(−1) (1−(s/n)+(1/n))^(−1) ((πs)/(sinπs))=Π^∞ (1−(s/n))^(−1) (1+(s/n))^(−1) Γ(s)Γ(1−s)=(e^(−γ+Σ^∞ (1/n)) /(s(1−s))).((πs)/(sinπs)).Π^∞ (1−(s/n))Π^∞ (1−(s/n)+(1/n))^(−1) Γ(s)Γ(1−s)=(e^(log(n)) /((1−s))).(π/(sinπs))((1−s)(((2−s)/2))...)((1.2.3)/((2−s)(3−s)(4−s))) Γ(s)Γ(1−s)=lim_(n→∞) ((πn)/(sinπs)).(1/(n−s+1)) Γ(s)Γ(1−s)=(π/(sinπs))

$$\Gamma\left({s}\right)=\frac{{e}^{−\gamma{s}} }{{s}}\overset{\infty} {\prod}{e}^{\frac{{s}}{{n}}} \left(\mathrm{1}+\frac{{s}}{{n}}\right)^{−\mathrm{1}} \\ $$$$\Gamma\left(\mathrm{1}−{s}\right)=\frac{{e}^{−\gamma+\gamma{s}} }{\mathrm{1}−{s}}\overset{\infty} {\prod}{e}^{\frac{\mathrm{1}−{s}}{{n}}} \left(\mathrm{1}+\frac{\mathrm{1}−{s}}{{n}}\right)^{−\mathrm{1}} \\ $$$$\Gamma\left({s}\right)\Gamma\left(\mathrm{1}−{s}\right)=\frac{{e}^{−\gamma} }{{s}\left(\mathrm{1}−{s}\right)}\overset{\infty} {\prod}{e}^{\frac{\mathrm{1}}{{n}}} \left(\mathrm{1}+\frac{{s}}{{n}}\right)^{−\mathrm{1}} \left(\mathrm{1}−\frac{{s}}{{n}}+\frac{\mathrm{1}}{{n}}\right)^{−\mathrm{1}} \\ $$$$\frac{\pi{s}}{{sin}\pi{s}}=\overset{\infty} {\prod}\left(\mathrm{1}−\frac{{s}}{{n}}\right)^{−\mathrm{1}} \left(\mathrm{1}+\frac{{s}}{{n}}\right)^{−\mathrm{1}} \\ $$$$\Gamma\left({s}\right)\Gamma\left(\mathrm{1}−{s}\right)=\frac{{e}^{−\gamma+\overset{\infty} {\sum}\frac{\mathrm{1}}{{n}}} }{{s}\left(\mathrm{1}−{s}\right)}.\frac{\pi{s}}{{sin}\pi{s}}.\overset{\infty} {\prod}\left(\mathrm{1}−\frac{{s}}{{n}}\right)\overset{\infty} {\prod}\left(\mathrm{1}−\frac{{s}}{{n}}+\frac{\mathrm{1}}{{n}}\right)^{−\mathrm{1}} \\ $$$$\Gamma\left({s}\right)\Gamma\left(\mathrm{1}−{s}\right)=\frac{{e}^{{log}\left({n}\right)} }{\left(\mathrm{1}−{s}\right)}.\frac{\pi}{{sin}\pi{s}}\left(\left(\mathrm{1}−{s}\right)\left(\frac{\mathrm{2}−{s}}{\mathrm{2}}\right)...\right)\frac{\mathrm{1}.\mathrm{2}.\mathrm{3}}{\left(\mathrm{2}−{s}\right)\left(\mathrm{3}−{s}\right)\left(\mathrm{4}−{s}\right)} \\ $$$$\Gamma\left({s}\right)\Gamma\left(\mathrm{1}−{s}\right)=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{\pi{n}}{{sin}\pi{s}}.\frac{\mathrm{1}}{{n}−\boldsymbol{{s}}+\mathrm{1}} \\ $$$$\Gamma\left({s}\right)\Gamma\left(\mathrm{1}−{s}\right)=\frac{\pi}{{sin}\pi{s}} \\ $$

Commented by mnjuly1970 last updated on 14/Nov/20

bravo mr dwaipayan.. very nice as always...

$${bravo}\:{mr}\:{dwaipayan}.. \\ $$$${very}\:{nice}\:{as}\:{always}... \\ $$

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

With pleasure��

Answered by mnjuly1970 last updated on 14/Nov/20

proof by using double integral. Γ(s)=∫_0 ^( ∞) x^(s−1) e^(−x) dx ...(i) Γ(1−s)=∫_0 ^( ∞) y^(−s) e^(−y) dy...(ii) from (i) and (ii): Γ(s)Γ(1−s)=∫_0 ^( ∞) ∫_0 ^( ∞) (1/x)((x/y))^s e^(−(x+y)) dx dy {_((x/y)=v) ^(x+y=u) ⇒{_(x=((uv)/(1+v))) ^(y=(u/(1+v))) dxdy=∣((∂(x,y))/(∂(u,v)))∣dudv where ((∂(x,y))/(∂(u,v)))=J(u,v)= determinant ((((∂x/∂u) (∂x/∂v))),(((∂y/∂u) (∂y/∂v) ))) ∴ J(u,v)=x_u y_v −x_v y_u =((v/(1+v)))(((−u)/((1+v)^2 )))−((u/((1+v)^2 )))((1/(1+v))) J(u,v)=((−u)/((1+v)^2 )) ✓ Γ(s)Γ(1−s)=∫_0 ^( ∞) ∫_0 ^( ∞) ((1+v)/(uv))v^s e^(−u) ∣((−u)/((1+v)^2 ))∣dudv =∫_0 ^( ∞) ∫_0 ^( ∞) (v^(s−1) /(1+v))e^(−u) dudv=∫_0 ^( ∞) (v^(s−1) /(1+v))dv=_(analysis) ^(complex) (π/(sin(πs))) ✓ ...m.n...

$$\:{proof}\:{by}\:{using}\:{double} \\ $$$$\:\:{integral}. \\ $$$$\:\:\Gamma\left({s}\right)=\int_{\mathrm{0}} ^{\:\infty} {x}^{{s}−\mathrm{1}} {e}^{−{x}} {dx}\:...\left({i}\right) \\ $$$$\:\:\:\Gamma\left(\mathrm{1}−{s}\right)=\int_{\mathrm{0}} ^{\:\infty} {y}^{−{s}} {e}^{−{y}} {dy}...\left({ii}\right) \\ $$$$\:{from}\:\:\left({i}\right)\:{and}\:\left({ii}\right): \\ $$$$\:\:\Gamma\left({s}\right)\Gamma\left(\mathrm{1}−{s}\right)=\int_{\mathrm{0}} ^{\:\infty} \int_{\mathrm{0}} ^{\:\infty} \frac{\mathrm{1}}{{x}}\left(\frac{{x}}{{y}}\right)^{{s}} {e}^{−\left({x}+{y}\right)} {dx}\:{dy} \\ $$$$\:\:\:\left\{_{\frac{{x}}{{y}}={v}} ^{{x}+{y}={u}} \:\Rightarrow\left\{_{{x}=\frac{{uv}}{\mathrm{1}+{v}}} ^{{y}=\frac{{u}}{\mathrm{1}+{v}}} \right.\right. \\ $$$$\:\:\:\:{dxdy}=\mid\frac{\partial\left({x},{y}\right)}{\partial\left({u},{v}\right)}\mid{dudv}\:\:{where} \\ $$$$\:\:\:\:\:\:\frac{\partial\left({x},{y}\right)}{\partial\left({u},{v}\right)}={J}\left({u},{v}\right)=\begin{vmatrix}{\frac{\partial{x}}{\partial{u}}\:\:\:\:\:\:\:\:\:\:\frac{\partial{x}}{\partial{v}}}\\{\frac{\partial{y}}{\partial{u}}\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\partial{y}}{\partial{v}}\:}\end{vmatrix} \\ $$$$\:\:\:\:\therefore\:\:{J}\left({u},{v}\right)={x}_{{u}} {y}_{{v}} −{x}_{{v}} {y}_{{u}} =\left(\frac{{v}}{\mathrm{1}+{v}}\right)\left(\frac{−{u}}{\left(\mathrm{1}+{v}\right)^{\mathrm{2}} }\right)−\left(\frac{{u}}{\left(\mathrm{1}+{v}\right)^{\mathrm{2}} }\right)\left(\frac{\mathrm{1}}{\mathrm{1}+{v}}\right) \\ $$$$\:\:\:\:\:\:\:{J}\left({u},{v}\right)=\frac{−{u}}{\left(\mathrm{1}+{v}\right)^{\mathrm{2}} }\:\checkmark \\ $$$$\:\:\:\:\:\Gamma\left({s}\right)\Gamma\left(\mathrm{1}−{s}\right)=\int_{\mathrm{0}} ^{\:\infty} \int_{\mathrm{0}} ^{\:\infty} \frac{\mathrm{1}+{v}}{{uv}}{v}^{{s}} {e}^{−{u}} \mid\frac{−{u}}{\left(\mathrm{1}+{v}\right)^{\mathrm{2}} }\mid{dudv} \\ $$$$=\int_{\mathrm{0}} ^{\:\infty} \int_{\mathrm{0}} ^{\:\infty} \frac{{v}^{{s}−\mathrm{1}} }{\mathrm{1}+{v}}{e}^{−{u}} {dudv}=\int_{\mathrm{0}} ^{\:\infty} \frac{{v}^{{s}−\mathrm{1}} }{\mathrm{1}+{v}}{dv}\underset{{analysis}} {\overset{{complex}} {=}}\frac{\pi}{{sin}\left(\pi{s}\right)}\:\checkmark \\ $$$$\:\:\:\:\:\:\:...{m}.{n}... \\ $$$$ \\ $$

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

Great!

$${Great}! \\ $$

Commented by mnjuly1970 last updated on 14/Nov/20

thank you..

$${thank}\:{you}.. \\ $$$$ \\ $$

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