# Derivative of the Fourier integrals calculated in a point

I have considered the following definitions for the Fourier integral pairs:

$\Phi(\omega) = \frac{1}{(2\pi)^3} \int_{\mathbb{R}^3} e^{ix\omega \cos\theta}R(x)x^2 dx \sin\theta d\theta d\phi$

$R(x) = \int_{\mathbb{R}^3} e^{-ix\omega \cos\theta}\Phi(\omega)\omega^2 d\omega \sin\theta d\theta d\phi$

If one takes the derivative from both sides of the first equation with respect to $\omega$, we will have:

$\Phi^{\prime}(\omega) = \frac{1}{(2\pi)^3} \int_{\mathbb{R}^3} (ix \cos\theta) e^{ix\omega \cos\theta}R(x)x^2 dx \sin\theta d\theta d\phi$

If now we evaluate the value of the derivative at $\omega = 0$ we have simply $\Phi^{\prime}(\omega=0) = 0$.

The paris of the functions $R(x) = \frac{1}{(1+\pi^2 x^2/4)^2}$ and $\Phi(\omega) = \frac{1}{\pi^4} e^{\frac{-2\omega}{\pi}}$ satisfy the pair of Fourier integrals, but $\Phi^{\prime}(0)$ is not equal to $0$. Does someone have an idea about why this problem happens?

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Here is the one-dimensional version of what you encountered. Let $\Phi(x)=e^{-|x|}$. Since $\Phi$ is even, its Fourier integral representation involves only cosine function $\cos t x$. When differentiated, the cosine turns to $-t \sin t x$ which is $0$ at $x=0$. So, the differentiation under integral sign results in the integral being zero when $x=0$. But $\Phi'(0)$ does not exist.
If you dig deeper, you'll find that in this and similar examples the formal differentiation results in the average of one-sided derivatives, in the same way how the Fourier series converges to $\frac12(f(x^+)+f(x^-))$ at a jump discontinuity. In higher dimensions, the property of $\Phi$ being even corresponds to rotational symmetry, and the same averaging effect occurs for any function that depends on the distance to the origin only.