all.
So, I'm working on the following problem for a statistical mechanics course.
A physical measurement of $x$ gives results for $-\infty < x < \infty$ with probability distribution (or density) given approximately by the distribution \begin{equation}\label{prob2} p(x) = \frac{a}{\pi (a^2 + x^2)} \end{equation} with $a<0$.
One part of the problem asks me to verify that the average value and the variance of $x$ are not well-defined. I'm a bit stuck at the end of the problem, and I think it is because of how I am defining what a well-defined function is. Here is my work, I will mention the part where I am stuck.
The average value of a probability distribution is given by the expectation value. Namely, \begin{align} \label{expect1} E[x] = \int_{-\infty}^{\infty} x p(x) dx \end{align} Using the Cauchy probability function and integrating according to the equation above, we see that \begin{align} E[x] = \frac{a}{\pi} \int_{-\infty}^{\infty} \frac{x}{x^2 + a^2} dx \end{align} For the expectation value to hold any meaning, it must converge. This means that $E[x]$ is 'well-behaved' and converges to a non-infinite value. So, according to the definition of convergence, \begin{align} \lim_{t \xrightarrow{} \infty} \int_{0}^{t} x p(x) dx \end{align} must hold for our function to be well-behaved. Here I will show that the Cauchy distribution is not well-defined at its first moment, which subsequently means that its variance ($\text{Var}(x) = E[x^2] - E[x]^2$) also is not not well-defined.
Taking the integral, \begin{align} \int_0^t \frac{x}{x^2 + a^2}dx &= \frac{1}{a^2} \int_0^t \frac{x}{\frac{x^2}{a^2} + 1} dx \end{align} Making a change of variables such that $u = \frac{x^2}{a^2} \Rightarrow{} \frac{2}{a^2}du = xdx$ gives \begin{align} \frac{1}{2} \int_0^{t^2 / a^2} \frac{1}{u + 1} du &= \frac{1}{2} \ln{(u+1)}_{\big |_0^{t^2 /a^2}} \\ &= \frac{1}{2}\ln{\bigg (\frac{t^2}{a^2} + 1 \bigg )} \end{align} Taking the limit as $t \rightarrow \infty$, \begin{align} \lim_{t \rightarrow \infty} \frac{1}{2}\ln{\bigg (\frac{t^2}{a^2} + 1 \bigg )} = \infty \end{align} In a similar fashion, we can see that for $-\infty < t < 0$, \begin{align} \int_t^0 \frac{x}{x^2 + a^2}dx = -\frac{1}{2}\ln{\bigg (\frac{t^2}{a^2} + 1 \bigg )} \end{align} Which, for the limit as $t \rightarrow -\infty$, \begin{align} \lim_{t \rightarrow -\infty} -\frac{1}{2}\ln{\bigg (\frac{t^2}{a^2} + 1 \bigg )} = -\infty \end{align}
This is where I am stuck. I do not see how the limits of this function as $t \rightarrow 0$ from either side is not well-defined, as the limits seem to show that the value at $x=0$ is zero. I read that a function that is not well-defined is not one-to-one, but I do not see how this is the case here.
Any help would be appreciated! Thank you.