Since
\(n!=\Gamma(n+1)=\displaystyle\int_0^\infty t^ne^{-t}\,\mathrm dt\)
you have
\(\left(\dfrac12\right)!=\displaystyle\int_0^\infty t^{1/2}e^{-t}\,\mathrm dt\)
Let \(u=t^{1/2}\), so that \(u^2=t\) and \(2u\,\mathrm du=\mathrm dt\). This means the integral is equivalent to
\(\displaystyle2\int_0^\infty u^2e^{-u^2}\,\mathrm du\)
Integrating by parts, setting \(f=u\), \(\mathrm df=\mathrm du\) and \(\mathrm dg=ue^{-u^2}\,\mathrm du\), \(g=-\dfrac12e^{-u^2}\) yields
\(\displaystyle2\int_0^\infty u^2e^{-u^2}\,\mathrm du=2\left(-\dfrac12ue^{-u^2}\bigg|_{u=0}^{u\to\infty}+\dfrac12\int_0^\infty e^{-u^2}\,\mathrm du\right)=\int_0^\infty e^{-u^2}\,\mathrm du\)
You might already know that the value of this integral is \(\dfrac{\sqrt\pi}2\), so you're done. Or, if you're not familiar, you can derive this result.
If \(\mathcal I(x)=\displaystyle\int_0^\infty e^{-x^2}\,\mathrm dx\), then you have
\(\displaystyle\mathcal I(x)\mathcal I(y)=\left(\int_0^\infty e^{-x^2}\,\mathrm dx\right)\left(\int_0^\infty e^{-y^2}\,\mathrm dy\right)=\int_0^\infty\int_0^\infty e^{-(x^2+y^2)}\,\mathrm dx\,\mathrm dy\)
Converting to polar coordinates yields
\(\displaystyle\int_0^{\pi/2}\int_0^\infty re^{-r^2}\,\mathrm dr\,\mathrm d\theta=\dfrac\pi2\int_0^\infty re^{-r^2}\,\mathrm dr=\dfrac\pi4\)
and so
\(\displaystyle\mathcal I(x)^2=\mathcal I(x)\mathcal I(x)=\dfrac\pi4\implies \mathcal I(x)=\dfrac{\sqrt\pi}2\)
