- #1
matematikuvol
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\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?
