Is there a way to calculate the fractional moments of the Cauchy distribution without complex analysis?

Solution 1:

Solution using Fourier series

$$\begin{split} \int_{\mathbb R}\frac {|x|^p}{1+x^2}dx &= 2\int_0^{+\infty}\frac {x^p}{1+x^2}dx\\ &=2\left(\int_0^1\frac {x^p}{1+x^2}dx+\int_1^{+\infty}\frac {x^p}{1+x^2}dx\right)\\ &=\int_0^1\frac {u^{\frac p 2-\frac 1 2}}{1+u}du+\int_0^{1}\frac {y^{-\frac p2-\frac 1 2}}{1+y}dy \,\,\,\,\text{ with }u=x^2 \text{ and }y=\frac 1 {x^2} \end{split}$$ We thus need to compute, for $-1<\alpha<1$ $$\int_0^1\frac{v^{\alpha}}{1+v}dv =\int_0^1\sum_{n\in\mathbb N}(-1)^nv^{\alpha+n}dv=\sum_{n\in\mathbb N}(-1)^n\int_0^1v^{\alpha+n}dv=\sum_{n\in\mathbb N}\frac{(-1)^n}{\alpha+n+1}$$ In other words, $$\begin{split} \int_{\mathbb R}\frac {|x|^p}{1+x^2}dx &= \sum_{n\in\mathbb N}\frac{(-1)^n}{n+\frac p 2+\frac 1 2}+\sum_{n\in\mathbb N}\frac{(-1)^n}{n-\frac p 2+\frac 1 2}\\ &=\sum_{n\in\mathbb N}\frac{(-1)^n}{n-\frac p 2+\frac 1 2} + \sum_{n\in\mathbb N}\frac{(-1)^n}{-n+\frac p 2-\frac 1 2}\\ &= \sum_{n\in\mathbb N}\frac{(-1)^n}{n-\frac p 2+\frac 1 2} + \sum_{n\geq 1}\frac{(-1)^n}{-n-\frac p 2+\frac 1 2}\,\,\,\text{ (second index shifted by 1)}\\ &= \frac{1}{\frac 1 2 - \frac p 2} + (1-p)\sum_{n\geq 1}\frac{(-1)^{n+1}}{n^2-\left(\frac 1 2-\frac p 2\right)^2} \end{split}$$ This classic sum can be computed via Fourier series, among other ways. For $s\notin \mathbb Z$, define $f_s$ as the $2\pi$-periodic function defined by $f_s(t)=\cos(st)$ for $|t|<\pi$. You can verify that its Fourier series expansion is $$\cos{st} = \frac{\sin{\pi s}}{\pi s} \left [1+2 s^2 \sum_{n\geq 1}\frac{(-1)^{n+1} \cos{n t}}{n^2-s^2} \right ]$$ Evaluating at $t=0$ gives $$\frac{\pi}{\sin(\pi s)}=\frac 1 s + 2s\sum_{n\geq 1}\frac{(-1)^{n+1} }{n^2-s^2}$$ Thus, plugging in $s=\frac 1 2 - \frac p 2$, $$\int_{\mathbb R}\frac {|x|^p}{1+x^2}dx=\frac {\pi}{\cos\left(\frac {\pi p} 2\right)}$$