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?
 A: 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
A: Fix $a > 0$ and define
$$
            f(b)=\int_{0}^{\infty}e^{-ax^2}\cos(bx)dx,\;\;\; g(b)=-\int_{0}^{\infty}xe^{-ax^2}\sin(bx)dx.
$$
You can argue that $f$ and $g$ are continuous on $\mathbb{R}$ by looking at the convergence of the improper integrals. Then argue that you can interchange orders of integration to obtain
\begin{align}
     \int_{0}^{b}g(b')db' &= -\int_{0}^{b}\left(\int_{0}^{\infty}e^{-ax^2}x\sin(b'x)dx\right)db' \\
  &=-\int_{0}^{\infty}e^{-ax^2}\int_{0}^{b}x\sin(b'x)db'dx \\
  &=\int_{0}^{\infty}e^{-ax^2}\cos(bx)dx-\int_{0}^{\infty}e^{-ax^2}dx \\
  &=f(b)-f(0).
\end{align}
Because $g$ is continuous it follows that $f$ is continuously differentiable with
$$
       f'(b) = \frac{d}{db}\int_{0}^{b}g(b')db' = g(b),
$$
which is what you wanted to prove.
