How to attack the gamma function manually

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.

This is equivalent to $\Gamma(\tfrac12)=\int_0^\infty\frac{1}{\sqrt{x}}e^{-x}dx=\sqrt{\pi}$. With $y=\sqrt{x}$, this is equivalent to normalizing a Gaussian integral, usually done this way. Alternatively, you may work with the Beta function:$$\Gamma^2(\tfrac12)=\operatorname{B}(\tfrac12,\,\tfrac12)=2\int_0^{\pi/2}d\theta=\pi.$$

Here is a possible solution: $$ \Gamma ^2 \left( {\frac{3}{2}} \right) = \frac{1}{4}\Gamma ^2 \left( {\frac{1}{2}} \right) = \frac{1}{4}\left( {\int_0^{ + \infty } {e^{ - t} \frac{1}{{\sqrt t }}dt} } \right)^2 \\ = \left( {\int_0^{ + \infty } {e^{ - x^2 } dx} } \right)^2 = \int_0^{ + \infty } {\int_0^{ + \infty } {e^{ - x^2 - y^2 } dy} dx} \\ = \int_0^{ + \infty } {\int_0^{ + \infty } {xe^{ - x^2 (1 + s^2 )} ds} dx} = \int_0^{ + \infty } {\int_0^{ + \infty } {xe^{ - x^2 (1 + s^2 )} dx} ds} \\ = \int_0^{ + \infty } {\left[ {\frac{1}{{ - 2(1 + s^2 )}}e^{ - x^2 (1 + s^2 )} } \right]_{x = 0}^{x = + \infty } ds} \\ = \frac{1}{2}\int_0^{ + \infty } {\frac{{ds}}{{1 + s^2 }}} = \frac{1}{2}\left[ {\arctan s} \right]_{s=0}^{s= + \infty } = \frac{\pi }{4} . $$