I'm trying to determine the general PDF and Mean for the Pareto distribution description of the size of TCP packets, given that distribution's CDF:
$$ F(x) = \begin{cases} 1-\left(\frac{k}{x}\right)^a, & x > k\\ 0, & \text{else.} \end{cases}$$
I found the PDF easily by taking the derivative.
$$\text{PDF} = F'(x) = \begin{cases} \frac{ak^a}{x^{a+1}}, & x > k\\ 0, & \text{else.} \end{cases}$$
The last part however, finding the mean, has stumped me, as the only way I can think of the find it given this information is via the following definition of the mean:
$$ \int_{-\infty}^{\infty}xf(x)\,dx,$$
$f(x)$ being the PDF. This results in the following integral:
$$\int_{-\infty}^{\infty}x\frac{ak^a}{x^{a+1}}dx = \left.\frac{ak^ax^{1-a}}{1-a}\right|_{-\infty}^{\infty}$$
But the solutions manual has the mean as equal to $\frac{ak}{a-1}$, which I have to assume is a logical simplification of the antiderivative I calculated, but I can't figure out how to bridge that gap.
In other words, how does:
$$\left.\frac{ak^ax^{1-a}}{1-a}\right|_{-\infty}^{\infty}$$ simplify to $$\frac{ak}{a-1}?$$
Or am I not looking at this correctly?
Your bounds are wrong. You neglected the condition $x>k.$ Very important.
Now, rework it and you will see that $$E[X] = \int_{k}^\infty x\cdot\frac{ak^a}{x^{a+1}}\,dx = \frac{ak}{a-1}.$$