Using the Leibniz Rule

Question

Consider

$$F(z) = \int_0^\infty\frac{1-\cos(yz)}{y}\ {e^{-y}} dy$$

How do I use the Leibniz Rule to solve this? I've tried to follow the general rule but can't seem to get the answer which I am meant to prove, which is:

$$\frac{dF}{dz} = \int_0^\infty \sin(yz){e^{-y}} dy$$

Thanks!

Answer 1

$F(z) = \int_0^\infty\frac{1-cos(yz)}{y}\ {e^{-y}}dy$

$\implies F(z) = \lim_{n \rightarrow +\infty}\int_0^n\frac{1-cos(yz)}{y}\ {e^{-y}}dy $

Then, Leibniz Formula gives us:

$\frac{dF}{dz} = \lim_{n \rightarrow +\infty} \int_0^n \frac{d}{dz}\frac{1-cos(yz)}{y}\ {e^{-y}}dy = \lim_{n \rightarrow +\infty} \int_0^n \frac{ysen(yz)}{y}\ {e^{-y}}dy = \lim_{n \rightarrow +\infty}\int_0^n sen(yz){e^{-y}}dy = \int_0^\infty sen(yz){e^{-y}}dy$