In Berger, Casella (2002) I read that the standard normal pdf, given by $$ \int_0^\infty \exp\left(-\frac{z^2}{2}\right)\,dz$$ is equal to the gamma function $\Gamma(\alpha)$ evaluated in $\alpha=\frac{1}{2}$ when $w=\frac{1}{2}z^2$, which results in: $$\Gamma\left(\frac{1}{2}\right) =\int^\infty_0w^{-\frac{1}{2}}e^{-w}dw=\sqrt{\pi}.$$
However, when I do my calculations and substitute $dz=\frac{dw}{\sqrt{2w}}$ I obtain a slightly different result, that is: $$2^{-\frac{1}{2}}\int^\infty_0w^{-\frac{1}{2}}e^{-w}dw.$$
Can somebody tell me what I missed?
The standard normal distribution is $\displaystyle \frac 1 {\sqrt{2\pi}} e^{-z^2/2} \, dz.$
You have $\displaystyle\int_{-\infty}^{+\infty} \frac 1 {\sqrt{2\pi}} e^{-z^2/2}\, dz=1.$
Thereofore $ \displaystyle \int_0^{+\infty} \sqrt{\frac 2 \pi} e^{-z^2/2} \, dz = 1.$
If $u = z^2/2$ then $z = \sqrt{2u\,\,}, $ and $du = z\,dz = \sqrt{2u\,} \,dz,$ and so $\dfrac{du}{\sqrt{2u\,}} = dz.$
So we get $$ 1 = \int_0^{+\infty} \sqrt{\frac2\pi} e^{-z^2/2}\, dz = \int_0^{+\infty} \sqrt{\frac 2\pi} e^{-u} \, \frac{du}{\sqrt{2u}} = \frac {\Gamma(1/2)}{\sqrt\pi}. $$