TITLE: Derivative of an expectation (using integrals)
QUESTION [6 upvotes]: I am working through an economics paper and I need to take the derivative of the following function:
$h\left(\overline{\omega}\right) = \int^{\infty}_{\overline{\omega}} \omega \Phi \left(d\omega\right)$
Even though I don't understand it well, I can do the derivative for the case
$g\left(\overline{\omega}\right) = \int^{\overline{\omega}}_{0} \omega \Phi \left(d\omega\right)$
where the derivative is simply
$g'\left(\overline{\omega}\right) = \overline{\omega} \phi \left(\overline{\omega}\right)$
But for $h\left(\overline{\omega}\right)$ where the upper bound is $\infty$ I really have no idea of what to do. 
Can anyone help me? Any explanation or even a pointer to where I can learn those things would be greatly appreciated.

REPLY [5 votes]: If the integral is sufficiently nice by which I mean it doesn't blow up to infinity, we can write $h(\bar{\omega}) = \int_{l}^{\infty} \omega d(\Phi(w)) - \int_{l}^{\bar{\omega}} \omega d(\Phi(w))$.
where $l$ is some constant number. It could be $0$ or $\infty$ based on your problem.
Now the first integral is just a constant, and the second integral is similar to $g(\bar{\omega})$

REPLY [4 votes]: Note that $h(\bar \omega ) + g(\bar \omega )$ is constant, and use the result for $g(\bar \omega )$.

REPLY [3 votes]: Since $$h(\overline{\omega})=-\int_{\infty}^{\overline{\omega}}\omega\Phi(d\omega),$$
$h'(\overline{\omega})=-\overline{\omega}\phi(\overline{\omega})$.