# Thread: Stuck on evaluating an integral for expect, in order to show that it's non-convergent

1. ## Stuck on evaluating an integral for expect, in order to show that it's non-convergent

Hi,

My question is more related to calculus here but it has a probability component so I decided to stick to the main thread

I have a pdf defined as $\frac{k}{1+y^2}, y \in R$;

I found $k = \frac{1}{\arctan(y)}$

Then I need to show that the random variable Y has a Cauchy distribution; for this I tried to find E[X], and here I get stuck:

$E[X] = \int_{-\infty}^{+\infty} yf(y)dy = \int_{-\infty}^{+\infty} \frac{y}{arctan(y)(1+y^2)}dy = ???$

- without that $y$ in the numerator I can do some nice substitutions but in this form...

I would really appreciate your help guys, in evaluating this integral,
Thank you

2. ## Re: Stuck on evaluating an integral for expect, in order to show that it's non-conver

You've messed up.

In the first part I assume $k$ is just a constant to ensure that the pdf integrates to 1.

It's not a function of $y$, it's a constant.

$k = \dfrac{1}{\displaystyle{\int_{-\infty}^{\infty}}~\dfrac{dy}{1+y^2}}$

3. ## Re: Stuck on evaluating an integral for expect, in order to show that it's non-conver

yes, that's right, k is a constant

4. ## Re: Stuck on evaluating an integral for expect, in order to show that it's non-conver

Originally Posted by dokrbb
yes, that's right, k is a constant
and it's value is?

5. ## Re: Stuck on evaluating an integral for expect, in order to show that it's non-conver

oh, it's $\frac{1}{\pi}$

then my integral becomes $\int_{-infty}^{infty} \frac{y}{\pi(1+y^2)}dy$
let $u = 1+x^2, du = 2xdx$
$= \frac{1}{2\pi}\int_{-infty}^{infty} \frac{1}{u}du = \frac{1}{2\pi} lim_{a \rightarrow \infty}log(u) \bigg]_{-a}^{a} = \frac{1}{2\pi} lim_{a \rightarrow \infty}log(1+x^2) \bigg]_{-a}^{0} + \frac{1}{2\pi} lim_{a \rightarrow \infty}log(1+x^2) \bigg]_{0}^{a}$

- both parts tend to $\infty$, thus, and expected value does not exist

well done