Derivative of a definite integral. I'm studying about Fourier Transform and the author wrote this:
$f(b) = \displaystyle \int_{0}^{\infty} e^{-ax^2}cos(bx)dx$ 
(b is the variable here)
Therefore:
$f'(b) = \displaystyle - \int_{0}^{\infty} xe^{-ax^2}sin(bx)dx$
Well, I cannot see this straightforward, and I am having troubles trying to see this. Can anyone help me?

Use Leibniz's formula for derivation under integral sign:
$$\frac{\partial f(a,b)}{\partial b}=\frac{\partial }{\partial b}\int_{0}^{\infty}\exp{(-ax^2)}\cos{(bx)}\,dx=\int_{0}^{\infty}\exp{(-ax^2)}\frac{\partial \cos{(bx)}}{\partial b}\,dx$$
The rest is obvious.
Hope it helps
