How to attack the gamma function manually Thoughts before Friday.
If $\operatorname{\Gamma(1.5)}=\displaystyle \int_0^\infty e^{-x}\sqrt{x}=\dfrac{\sqrt{\pi}}{2}$, how was this calculated? I proved the relevance of the formula as an analytic continuation of the factorial function but I've been stuck on calculating the value of $0.5!$ from the integral itself.
I'm not exactly sure which pattern to produce and I also have no idea where the π came from. Was it guessed or is there a step-by-step solution? Any help would be appreciated.

This definite integral is difficult, and you need "tricks".
For example, you start from
$$\int_0^\infty\frac{e^{-x}}{2\sqrt x}dx=\int_0^\infty e^{-x^2}dx$$
and switch to $2D$ using Fubini's theorem:
$$I^2=\int_0^\infty e^{-x^2}dx\int_0^\infty e^{-y^2}dy=\int_0^\infty\int_0^\infty e^{-(x^2+y^2)}dx\,dy.$$
Next, you convert to polar coordinates, using the element of area $dx\,dy=r\,dr\,d\theta$, and
$$\int_0^\infty\int_0^\infty e^{-(x^2+y^2)}dx\,dy=\int_0^{\pi/2}\int_0^\infty re^{-r^2}\,dr\,d\theta=\int_0^{\pi/2}d\theta\int_0^\infty re^{-r^2}\,dr=\frac\pi4.$$
This shows you how $\pi$ can appear.
