Expectation using CDF I have a problem understanding the solution of an exercise:
$F_x(x) = 1-(\frac{\sqrt{3}}{2}x + \frac{3}{2})^{-3}$ for $x \geq \frac{-1}{3} \sqrt{3}$, $0$ elsewhere
and i am asked to compute the expectation.
I know:
$\int{}^x_{\frac{-1}{3} \sqrt{3}}$ $f(x)dx$ $:= F(x)$
$\int{}^\infty_{\frac{-1}{3} \sqrt{3}}$ $f(x)dx$ $= 1$
$\int{}^\infty_{\frac{-1}{3} \sqrt{3}}$$xf(x)dx$ $:= E(X)$
Now he uses integration by parts:
$E(X)$ $=$ $\int{}^\infty_{\frac{-1}{3} \sqrt{3}}$$xf(x)dx= x(F(x)-1)-\int{}^\infty_{\frac{-1}{3} \sqrt{3}}$$1(F(x)-1)dx$
(first term evaluated at bounds)
why is it $(F(x)-1)$?
 A: Note that for any $c$ we have $\dfrac{d}{dx}\, (F(x)-c) = f(x)$. 
So we would then write 
$$E(X) = \int^\infty_{{-\sqrt{3}}/{3} }\, xf(x)dx=\int^\infty_{{-\sqrt{3}}/{3} }\,x\dfrac{d}{dx}(F(x)-c)\,dx$$ for some convenient choice of $c$.
You might think we can choose any $c$ we like for integration by parts (e.g. $c=0$ is simpler so why not choose that?) ... but we can't just choose any $c$ we like. Why?
Because of the very part you skipped over with a casual "first term evaluated at bounds". For the left and right sides of integration by parts to actually maintain an equality (i.e. for the step to be valid), the components must converge in the limit. Note that otherwise you could be writing a finite quantity is equal to the difference of two things that each go off to infinity, and next thing you've proved that 1=0.
Now let's look at it carefully. The term should be written like this:
$$\int^\infty_{{-\sqrt{3}}/{3} } x\dfrac{d}{dx}(F(x)-c)\,dx= x(F(x)-c)\Big{\vert}^\infty_{{-\sqrt{3}}/{3}}-\int^\infty_{{-\sqrt{3}}/{3} }(\dfrac{d}{dx} x)(F(x)-c)dx$$
We need both parts there to converge for this to work.
Now what does that upper limit of infinity in the integral really mean? $\int_a^\infty$ is just a shorthand for $\lim_{b\to\infty}\int_a^b$.
So let's be explicit with that first term:
$x(F(x)-c)\Big{\vert}^\infty_{{-\sqrt{3}}/{3}}\,$ is really $\:\lim_{b\to\infty} x(F(x)-c)\Big{\vert}^b_{{-\sqrt{3}}/{3}}$
At the lower end everything is fine. Now at the upper end you have a product of two terms there. $b$ and $F(b)-c$. For a given $c$, the second term is bounded as we let $b\to\infty$ (it won't exceed $1-c$), but the first term is $b$ and we've got $b$ increasing without limit here. The product can't be finite in the limit if the second term is non-zero in the limit (it may not be finite then either but that's easy to check$^\dagger$).
So the only possible value for $c$ is $1$ -- anything else will plainly leave us with an invalid step in our integration by parts. 
[If we examine the second term on the right hand side of the integration by parts step we strike a similar problem -- you need $c=1$ for it to be finite$^\ddagger$]
[If you'd tried to use any other value for $c$ before posting -- like $c=0$ say -- you would hopefully have noticed what the problem was. I am curious why you didn't try to see what would happen.]

$\dagger$ I will leave you to verify that the upper bound does then converge to a finite value for yourself.
$\ddagger$ and again, you should check it works for $c=1$.
