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?
On
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.
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