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This can be done without invoking the Beta function explicitly. Directly from the integral definition of $\Gamma(x)$,the product in question is a double integral: $$ \Gamma(x)\Gamma(1-x) =\int^\infty_0\int^\infty_0 (u/v)^x u^{-1} e^{-(u+v)} du\, dv $$ Switching to polar coordinates via $u=x^2=r^2\cos^2\phi$, $v=y^2=r^2\sin^2\phi$, we have $$ \Gamma(x)\Gamma(1-x)= 4\int_0^{\pi/2} (\tan\phi)^{2x-1} \, d\phi \int_0^\infty r e^{-r^2}\, dr = 2\int_0^{\pi/2} (\tan\phi)^{2x-1} d\phi $$ Finally, $\phi= \tan^{-1}\sqrt{s}$, brings us to $$ \Gamma(x)\Gamma(1-x)=\int^\infty_0 {s^{x-1} ds\over 1+s} $$ which can be evaluated straightforwardly by contour integration as shown elsewhere on this page.

This can be done without invoking the Beta function explicitly. Directly from the integral definition of $\Gamma(x)$,the product in question is a double integral: $$ \Gamma(x)\Gamma(1-x) =\int^\infty_0\int^\infty_0 (u/v)^x u^{-1} e^{-(u+v)} du\, dv $$ Switching to polar coordinates via $v=x^2=r^2\cos^2\phi$, $u=y^2=r^2\sin^2\phi$, we have $$ \Gamma(x)\Gamma(1-x)= 4\int_0^{\pi/2} (\tan\phi)^{2x-1} \, d\phi \int_0^\infty r e^{-r^2}\, dr = 2\int_0^{\pi/2} (\tan\phi)^{2x-1} d\phi $$ Finally, $\phi= \tan^{-1}\sqrt{s}$, brings us to $$ \Gamma(x)\Gamma(1-x)=\int^\infty_0 {s^{x-1} ds\over 1+s} $$ which can be evaluated straightforwardly by contour integration as shown elsewhere on this page.

This can be done without invoking the Beta function explicitly. Directly from the integral definition of $\Gamma(x)$,the product in question is a double integral: $$ \Gamma(x)\Gamma(1-x) =\int^\infty_0\int^\infty_0 (u/v)^x u^{-1} e^{-(u+v)} du\, dv $$ Switching to polar coordinates via $u=X^2=r^2\cos^2\phi$, $v=Y^2=r^2\sin^2\phi$, the $r$ and $\phi$ integrals decouple and we have $$ \Gamma(x)\Gamma(1-x)= 4\int_0^{\pi/2} (\tan\phi)^{2x-1} \, d\phi \int_0^\infty r e^{-r^2}\, dr = 2\int_0^{\pi/2} (\tan\phi)^{2x-1} d\phi $$ Finally, $\phi= \tan^{-1}\sqrt{s}$, brings us to $$ \Gamma(x)\Gamma(1-x)=\int^\infty_0 {s^{x-1} ds\over 1+s} $$ which can be evaluated straightforwardly by contour integration as above.

This can be done without invoking the Beta function explicitly. Directly from the integral definition of $\Gamma(x)$,the product in question is a double integral: $$ \Gamma(x)\Gamma(1-x) =\int^\infty_0\int^\infty_0 (u/v)^x u^{-1} e^{-(u+v)} du\, dv $$ Switching to polar coordinates via $u=x^2=r^2\cos^2\phi$, $v=y^2=r^2\sin^2\phi$, we have $$ \Gamma(x)\Gamma(1-x)= 4\int_0^{\pi/2} (\tan\phi)^{2x-1} \, d\phi \int_0^\infty r e^{-r^2}\, dr = 2\int_0^{\pi/2} (\tan\phi)^{2x-1} d\phi $$ Finally, $\phi= \tan^{-1}\sqrt{s}$, brings us to $$ \Gamma(x)\Gamma(1-x)=\int^\infty_0 {s^{x-1} ds\over 1+s} $$ which can be evaluated straightforwardly by contour integration as shown elsewhere on this page.

This can be done without invoking the Beta function explicitly. Directly from the integral definition of $\Gamma(x)$,the product in question is a double integral: $$ \Gamma(x)\Gamma(1-x) =\int^\infty_0\int^\infty_0 (u/v)^x u^{-1} e^{-(u+v)} du\, dv $$ Switching to polar coordinates via $u=x^2=r^2\cos^2\phi$, $v=y^2=r^2\sin^2\phi$, the $r$ and $\phi$ integrals decouple and we have $$ \Gamma(x)\Gamma(1-x)= 4\int_0^{\pi/2} (\tan\phi)^{2x-1} \, d\phi \int_0^\infty r e^{-r^2}\, dr = 2\int_0^{\pi/2} (\tan\phi)^{2x-1} d\phi $$ Finally, $\phi= \tan^{-1}\sqrt{s}$, brings us to $$ \Gamma(x)\Gamma(1-x)=\int^\infty_0 {s^{x-1} ds\over 1+s} $$ which can be evaluated straightforwardly by contour integration as above.

This can be done without invoking the Beta function explicitly. Directly from the integral definition of $\Gamma(x)$,the product in question is a double integral: $$ \Gamma(x)\Gamma(1-x) =\int^\infty_0\int^\infty_0 (u/v)^x u^{-1} e^{-(u+v)} du\, dv $$ Switching to polar coordinates via $u=X^2=r^2\cos^2\phi$, $v=Y^2=r^2\sin^2\phi$, the $r$ and $\phi$ integrals decouple and we have $$ \Gamma(x)\Gamma(1-x)= 4\int_0^{\pi/2} (\tan\phi)^{2x-1} \, d\phi \int_0^\infty r e^{-r^2}\, dr = 2\int_0^{\pi/2} (\tan\phi)^{2x-1} d\phi $$ Finally, $\phi= \tan^{-1}\sqrt{s}$, brings us to $$ \Gamma(x)\Gamma(1-x)=\int^\infty_0 {s^{x-1} ds\over 1+s} $$ which can be evaluated straightforwardly by contour integration as above.