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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.

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    @Sivaram: $\Phi (d\omega )$ is basically the same as $d\Phi (\omega )$.2010-11-26

3 Answers 3

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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 $g(\bar{\omega})$

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    @Sivaram: sorry, now I finally get it. It is $1-\int_0^{\overline{\omega}} \omega \Phi(\omega)$, but when I take the derivative the one drops out, so it is just $- \overline{\omega} \phi (\overline{\omega})$. Duh! Thanks so much again :)2010-11-26
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Note that $h(\bar \omega ) + g(\bar \omega )$ is constant, and use the result for $g(\bar \omega )$.

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Since $h(\overline{\omega})=-\int_{\infty}^{\overline{\omega}}\omega\Phi(d\omega),$ h'(\overline{\omega})=-\overline{\omega}\phi(\overline{\omega}).

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    I can't accept your answer yet (need to wait 4 minutes) but I will do it soon. Thanks heaps!2010-11-26