Question

Express (1/2)! in terms of pi.

Answer 1

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

$\displaystyle\mathcal I(x)^2=\mathcal I(x)\mathcal I(x)=\dfrac\pi4\implies \mathcal I(x)=\dfrac{\sqrt\pi}2$