How to derive the density of the square of a standard normal and chi-squared density from the gamma density? My question is two parts. My book defines the gamma density as:
$$g_n(x) = \lambda\frac{(\lambda x)^{n-1}}{(n-1)!}e^{-\lambda x},\space x>0$$
Part 1: In an example my book states that the following is a gamma density with $\lambda = 1/2$ and $n = 1/2$ (density for the square of a standard normal random variable):
$$f_{X^2}(r) = \frac{1}{\sqrt{2\pi r}}e^{-r/2},\space r >0$$
I can't figure out how to get this equation from the above, given that if n = 1/2 then I get a negative factorial in the denominator of the gamma density? 
Part 2: More generally, $f_{X^2}(r)$ above is the density for the square of a standard normal random variable $X$. My book states that the sum of of the square of $m$ independent standard normal random variables ($\mu_X = 0$ and $\sigma_X^2=1$) is a gamma density with $\lambda = 1/2$ and $n = m/2$ (the chi-squared density). For the case of $m=2$ the density given in my book was 
$$f_{R^2}(r)=1/2e^{-r/2},\space r>0$$
I tried computing the convolution between $f_{R^2}$ and $f_{X^2}$ and ended up with:
$$f_{Z^2}(z)=\frac{1}{\sqrt{2\pi}}\sqrt{z}e^{-z/2}, \space z > 0$$
Again I couldn't arrive at this result by plugging in $\lambda = 1/2$ and $n = 3/2$ into the gamma function.
 A: You need to use the general definition of Gamma function. For any $\alpha>0,$ $\Gamma(\alpha)$ is defined as 
$$\Gamma(\alpha)=\int_0^{\infty}x^{\alpha-1}e^{-x}dx.$$
An important property of the Gamma function is that $\Gamma(\alpha+1)=\alpha\Gamma(\alpha).$ This is proved here.
Another fact is that $\Gamma\left(\dfrac{1}{2}\right)=\sqrt{\pi}.$ Here is the proof. 
Moreover, this function reduces to $\Gamma(\alpha)=(\alpha-1)!$ when $\alpha$ is a natural number. With this definition, the Gamma distribution is defined as
$$g_{n,\lambda}(x)=\dfrac{\lambda^nx^{n-1}e^{-\lambda x}}{\Gamma(n)},\,\,x>0.$$
The pdf you have derived for sum of squares of three standard normal random variables, is correct. You can check that this is indeed the $g\left(\dfrac{3}{2},\dfrac{1}{2}\right)$ density, using the fact that $\Gamma\left(\dfrac{3}{2}\right)=\dfrac{1}{2}\times\Gamma\left(\dfrac{1}{2}\right)=\dfrac{\sqrt{\pi}}{2}.$
A: @ArnabAuddy has already answered part 1, so I'll talk about part 2. Let's first consider the case $m=1$, for which we just have the distribution of $X^2$ with $X\sim N(0,\,1)$. The characteristic function of $X^2$ is $$\Bbb E\exp itX^2=\int_{\Bbb R}\frac{1}{\sqrt{2\pi}}\exp\left(-\frac{1-2it}{2}x^2\right)dx=(1-2it)^{-1/2}.$$A general value of $m$ therefore obtains characteristic function $(1-2it)^{-m/2}$, as does the claimed Gamma distribution. Since distinct distributions have different characteristic functions, this completes the proof.
