1

For exam preparation I try to solve the following question:

Does the expected value exist for the density function $\frac{1}{\pi} \int_{-\infty}^{\infty} \frac{1}{x^2+1} \; \mathrm{d}x$

As I understand it I need to show that $\int_{-\infty}^{\infty}xf(x)\;dx$ converges

$$ \text{finding antiderivative}\\ \\ \frac{1}{\pi} \int_{-\infty}^{\infty} \frac{x}{x^2+1} \; \mathrm{d}x\\ \\ \text{substitute}\;u = x^2+1, du = 2x\\ \\ \frac{1}{2\pi} \int_{-\infty}^{\infty} \frac{1}{u} \; \mathrm{d}u = \frac{ln(\lvert u \rvert)}{2\pi} \Big|_{-\infty}^{\infty}\\ \\ \text{back substitution I get}\\ \\ \frac{ln(x^2+1)}{2\pi} \Big|_{-\infty}^{\infty} = 0 $$

So the integral exist and converges. What I am doing wrong here?

Any comment is appreciated. Thank you.

wejaeger
  • 21
  • The integral for the expected value exists as a Cauchy principal value. The integrand being an odd function about $x=0$ gives you the zero you're getting. – A rural reader Jun 15 '22 at 16:33
  • For more detail, you can also check out [https://math.stackexchange.com/questions/2610315/how-to-prove-int-infty-inftyx-dx-is-divergent] – Doug Jun 15 '22 at 16:35
  • Regarding the first comment. Though, it's correct that one can give meaning to the integral by means of P.V., I don't think that it's too helpful here, or instructive. If you want to see, why this integral is not converging, read the wiki article on 'improper integrals'. – Tobsn Jun 15 '22 at 16:47

1 Answers1

2

Your assertion $$ \frac{1}{2\pi} \int_{-\infty}^{\infty} \frac{1}{u} \; \mathrm{d}u = \frac{ln(\lvert u \rvert)}{2\pi} \Big|_{-\infty}^{\infty}\\ $$ is not correct. In fact, these four integrals all diverge: $$ \frac{1}{2\pi} \int_0^1 \frac{1}{u}\;du = +\infty \\ \frac{1}{2\pi} \int_1^{+\infty} \frac{1}{u}\;du = +\infty \\ \frac{1}{2\pi} \int_{-\infty}^{-1} \frac{1}{u}\;du = -\infty \\ \frac{1}{2\pi} \int_{-1}^0 \frac{1}{u}\;du = -\infty $$ When we say $$ \frac{1}{2\pi} \int_{-\infty}^{\infty} \frac{1}{u} \; \mathrm{d}u $$ converges, we mean those four integrals all converge.

GEdgar
  • 111,679