matematikuvol
Oct14-11, 11:27 AM
\Gamma(x)=\int^{\infty}_0t^{x-1}e^{-t}dt
\Gamma(\frac{1}{2})=\int^{\infty}_0\frac{e^{-t}}{\sqrt{t}}dt=
take t=x^2
dt=2xdx
x=\sqrt{t}
=\int^{\infty}_0\frac{e^{-x^2}}{x}2xdx
Why here we can here reducing integrand by x?
\Gamma(\frac{1}{2})=\int^{\infty}_0\frac{e^{-t}}{\sqrt{t}}dt=
take t=x^2
dt=2xdx
x=\sqrt{t}
=\int^{\infty}_0\frac{e^{-x^2}}{x}2xdx
Why here we can here reducing integrand by x?