Derive the expected value for a Pareto distribution?

The density is $$ f(x) = \frac{d}{dx} F(x) = \frac{d}{dx} \Pr(X\le x) = \frac{d}{dx} (1-\Pr(X> x)). $$ The expected value is $$ \int_1^\infty xf(x)\,dx. $$

Later addendum in response to comments:

In the posted question, we are told that for $x\ge 1$ we have $\Pr(X>x) = x^{-a}$. It follows that for $x=1$, $\Pr(X>x)=1^{-a}=1$, so this random variable is always $\ge 1$.

Above I wrote $\dfrac{d}{dx}(1-\Pr(X>x))$. Now we can see that that is equal to $$ \frac{d}{dx}(1-x^{-a}) = -(-ax^{-a-1}) = ax^{-a-1}. $$ Therefore this is the density on the interval $(1,\infty)$, and the density is $0$ everywhere else. Thus the expected value is $$ \int_1^\infty xf(x)\,dx = \int_1^\infty x\,ax^{-a-1}\,dx = a\int_1^\infty x^{-a}\,dx $$ $$ =a\left[\frac{x^{-a+1}}{-a+1}\right]_1^\infty = 0 - a\left(\frac{1}{-a+1}\right) = \frac{a}{a-1}. $$

We evaluate the integral $$\int_0^\infty \Pr(X\gt x)\,dx$$ of the post.

Note that if $0\le x\lt 1$, then $\Pr(X\gt x)=1$. And if $x\ge 1$, then $\Pr(X\gt x)=x^{-a}$. Since $\Pr(X\gt x)$ is given by two different formulas, it is natural to break up the integral at $x=1$.

The integral of $\Pr(X\gt x)$, from $0$ to $1$, is $1$.

By a standard integral calculation, $$\int_1^\infty x^{-a}\,dx=\frac{1}{a-1}.$$ So $E(X)=1+\dfrac{1}{a-1}=\dfrac{a}{a-1}$.