Deriving the Chi-squared distribution using characteristic functions I would like to directly derive the probability density function (PDF) for a Chi-squared distribution with $k$ degrees of freedom using characteristic functions. 
If $X_{1}, X_{2}, \dots, X_{k}$ are independent, standard normal random variables, then $$ Y = \sum_{i=1}^{k} X_{i}^{2} $$ and $Y$ is chi-squared distributed with $k$ degrees of freedom. The PDF for $Y$ when $k = 1$ is given by $$ f(x, 1) = \frac{1}{\sqrt{2 \pi x}} e^{-\frac{1}{2} x} $$ and its respective characteristic function is $$ \varphi_{Y_{1}} (\omega) = \int_{-\infty}^{\infty}  f(x, 1) e^{iwx} dx = (1 - i2 \omega)^{-\frac{1}{2}} \text{,}$$ where $i$ is the imaginary number. For a general $k$ degrees of freedom, $Y$'s characteristic function is given by $$\varphi_{Y} (\omega) = (1 - i2 \omega)^{-\frac{k}{2}} \text{.}$$ Is it possible to explicitly derive $f(x, k)$ using the inverse Fourier transform, where $$ f(x, k) = \frac{1}{2 \pi} \int_{-\infty}^{\infty} (1 - i2 \omega)^{-\frac{k}{2}} e^{-iwx} d\omega ?$$
I have had no success with this approach, but I'm probably missing something very obvious.
Failing that, is it possible to derive a formula for $k=2$, where $$f(x, 2) = \frac{1}{2 \pi} \int_{-\infty}^{\infty} (1 - i2 \omega)^{-1} e^{-iwx} d\omega ?$$ This would at least allow me derive the PDF inductively. In addition, I am aware that $f(x, 1)$ can be represented as the Gamma distribution $\text{Gamma}(x, \frac{1}{2}, \frac{1}{2})$ and the sum of independent Gamma random variables is known to be Gamma distributed, therefore, for this example, $$ Y \sim \text{Gamma} ( \cdot, \frac{k}{2}, \frac{1}{2}) \text{.} $$
 A: I have exact the exact motivation of finding the pdf of the chi squared through the characteristic function and got stuck in the same place. But finally I worked it out.
What we want to find is:
$$\int_{-\infty}^{\infty} f(\omega) d\omega=\int_{-\infty}^{\infty} (1 - i2 \omega)^{-k/2} e^{-iwx} d\omega =?$$
For solving the integral, we will use the residue theorem.
The integrand has one pole at -i/2. For calculating the residue of this pole, we will use the Cauchy integral formula for order n and extend it using the concepts of fractional calculus.
The Cauchy integral formula goes as:
$$g^{(n)}(a)=\frac{n!}{2\pi i} \oint_{\gamma} \frac{g(z)}{(z-a)^{n+1}}dz$$
Where $\gamma$ is a closed curve enclosing $a$ and oriented counterclockwise.
Following the principles of fractional calculus, we substitute $n! = \Gamma(n+1)$, rearranging terms and changing the variable n for s to make explicit that we are now in a continuous formula, we have:
$$\oint_{\gamma} \frac{g(z)}{(z-a)^{s}}dz=\frac{2\pi i}{\Gamma(s)}g^{(s-1)}(a)\:,\: s\: \epsilon \: \mathbb{R} $$
In our case, we have $g(z)=e^{-ixz}$, $s=k/2$  and  $a=-i/2$.
From the fractional calculus we know that the fractional derivative of the exponential is:
$$\frac{d^{s}}{dz^{s}}e^{az}=a^se^{az}\: \: \Rightarrow \: \: g^{(k)}(z)=(-ix)^ke^{-ixz}$$
Making the substitutions, we get:
$$\oint_{\gamma}\frac{e^{-ixz}}{(1-2iz)^{\frac{k}{2}}}dz=(-2i)^{-\frac{k}{2}}\oint_{\gamma}\frac{e^{-ixz}}{(z+\frac{i}{2})^{\frac{k}{2}}}dz=(-2i)^{-\frac{k}{2}}\frac{2\pi i}{\Gamma(k/2)}(-ix)^{\frac{k}{2}-1}e^{-ix(-i/2)}$$
Simplifying:
$$\oint_{\gamma}\frac{e^{-ixz}}{(1-2iz)^{k/2}}dz=-\frac{2^{1-\frac{k}{2}}x^{\frac{k}{2}-1}e^{-\frac{x}{2}}\pi}{\Gamma(\frac{k}{2})} = Res(f(z),-\frac{i}{2}) =R$$
Now, we take the integral contour $\gamma$ as in the picture below.
(Note, it seems that I'm not allowed to embed images because I don't have enough reputation)
Integral Contour
[A segment from $r$ to $-r$, with $r\, \epsilon\,  \mathbb{R} > 1/2 $, closed by a semicircle centered at the origin with raidus r extending from $\pi$ to $2\pi$. Closed curve oriented counterclockwise. The semicircle path is named $C_{\gamma}^+$, with the + sign indicating the orientation]
If we make $r\rightarrow \infty$, we have:
$$\oint_{\gamma}f(z)dz=\int_{C_{\gamma}^+}f(z)dz-\int_{-\infty}^{\infty}f(\omega)d\omega=R$$
Where the minus in the right hand comes from the integration orientation.
Now, we have to prove that the limit of the integral in $C_{\gamma}^+$ as $r\rightarrow \infty$ is zero:
$$\lim_{r\rightarrow \infty} f(re^{i\theta})=\lim_{r\rightarrow \infty} \frac{e^{-ixre^{i\theta}}}{(1-2ire^{i\theta})^{\frac{k}{2}}} =\lim_{r\rightarrow \infty} \frac{e^{-ixr(\cos(\theta)+i\sin(\theta))}}{(-2i)^{\frac{k}{2}}r^{\frac{k}{2}}e^{\frac{k}{2}i\theta}}$$
$$=(-2i)^{-\frac{k}2{}}\lim_{r\rightarrow \infty}\frac{e^{xr\sin(\theta)}e^{-i(xr\cos(\theta)+\frac{k}{2}\theta)}}{r^{\frac{k}{2}}}=(-2i)^{-\frac{k}2{}}\lim_{r\rightarrow \infty}\frac{e^{xr\sin(\theta)}e^{-ixr\cos(\theta)}}{r^{\frac{k}{2}}}$$
We have used for $r\rightarrow \infty$ $\left | 2ire^{i\theta} \right |\gg 1$ and $\left | xr\cos(\theta) \right |\gg \frac{k}{2}\theta$  for  $\theta\neq \pm \pi/2$
The limit of $e^{-ixr\cos(\theta)}$ as $r\rightarrow \infty$ does not exist, however its bounded as the modulus is one, therefore the limit will depend on the other two terms. $r^{k/2}$ tends to infinity, and $e^{xr\sin(\theta)}$ tends to zero for $x>0$
and  $\theta \epsilon (\pi,2\pi)$,  which is precisely the range corresponding to $C_{\gamma}^+$.
It is immediate to see that for the cases $x=0$ and $\theta=-\pi/2$ the limit is also zero.
Thus we have:
$$\forall \theta \epsilon (\pi,2\pi), x\geq 0 : \lim_{r\rightarrow \infty}f(re^{i\theta})=0$$
$$\int_{C_{\gamma}^+}f(z)dz=0 \Rightarrow \int_{-\infty}^{\infty}f(\omega)d\omega=-R=\frac{2^{1-\frac{k}{2}}x^{\frac{k}{2}-1}e^{-\frac{x}{2}}\pi}{\Gamma(\frac{k}{2})} $$
Finally, the probability density function of the Chi Squared distribution is:
$$pdf(x,k)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{e^{-ix\omega}}{(1-i2\omega)^{\frac{k}{2}}}d\omega=\frac{1}{2\pi}\frac{2^{1-\frac{k}{2}}x^{\frac{k}{2}-1}e^{-\frac{x}{2}}\pi}{\Gamma(\frac{k}{2})}=\frac{x^{\frac{k}{2}-1}e^{-\frac{x}{2}}}{2^{\frac{k}{2}}\Gamma(\frac{k}2{})}, x\geq0$$
Which is, of course, the value we were looking for.
