How to prove the Cauchy distribution has no moments?

Question

How can I prove the Cauchy distribution has no moments?;

$$E(X^n)=\int_{-\infty}^\infty\frac{x^n}{\pi(1+x^2)}\,\mathrm dx.$$

I can prove myself, letting $n=1$ or $n=2$ that it does not have any moment. However, I would I prove for ALL $n$, that the Cauchy distribution has no moments?

Answer 1

Hint

For example, if $x>1$, then $$\int_1^\infty \frac{x^n}{1+x^2}\,\mathrm d x\geq \int_1^\infty \frac{x}{1+x^2}\,\mathrm d x.$$

Answer 2

For convenience, let us write $f_n(x)$ for the expression in the integrand.

Of course for positive $n$, $\int_{-1}^{1}f_n(x)\;dx$ converges, so the problem is with the tails of the integral.

Now of course $x^{m}>x^n$ for $x>1$ for $m>n \geq 1$, so $$I_m(R) \equiv\int_1^{R}f_m(x)\;dx \geq \int_1^{R}f_n(x)\;dx\equiv I_n(R)$$ You have shown that $I_1(R)$ diverges as $R\to\infty$, so necessarily so does $I_m(R)$ for $m > 1$.

A similar statement is true on $(-\infty,0]$, but you will have to take the sign into account.

Putting these together, you get your desired result.