I asked the question on math.stackexchange but didn't get an answer so I came here.

I tried to compute with Wolfram Mathematica the following integral
$$I=\int_0^\pi\int_{-\infty}^\infty x e^{-\mathrm jx\cos(\theta-\varphi)}f(\alpha\cos(\theta-\psi))\mathrm \, \mathrm dx\mathrm \, \mathrm d\theta$$
where $-\pi\leq\varphi,\psi\leq\pi$.

Assuming that the inner intergral is by definition the derivative of the Dirac delta function I get:
$$I=2\pi \mathrm j\int_0^\pi\delta'(\cos(\theta-\varphi))f(\alpha\cos(\theta-\psi)) \, \mathrm \, \mathrm d\theta$$
Wolfram Mathematica tells me that
$$I_\mathrm{wolfram}=0$$
Then I tried to do it by hand. If I use the definition of a delta function:
$$
\begin{eqnarray}
\delta^{'} (\cos(\theta-\varphi))&=&\left[ \sum_i \frac{\delta(\theta-\varphi-\frac{\pi}{2}-i\pi)}{|\sin(\frac{\pi}{2}+i\pi)|}\right]^{'}=\sum_i \delta^{'}\left(\theta-\left(\varphi+\frac{\pi}{2}+i\pi\right)\right)
\end{eqnarray}
$$
$$
\begin{eqnarray}
I_\mathrm{me}&=&2\pi \mathrm j\sum_i\int_0^\pi\delta'\left(\theta-\left(\varphi+\frac{\pi}{2}+i\pi\right)\right)f(\alpha\cos(\theta-\psi)) \, \mathrm d\theta=\\
&=&-2\pi \mathrm j \sum_i f'(\alpha\cos(\varphi+\frac{\pi}{2}+i\pi-\psi))u(\pi -2 \varphi ) u(2 \varphi +\pi )
\end{eqnarray}
$$
where $u(\cdot)$ is a Heaviside function. Here I used the fact that $$\int f(x)\delta'(x-a)\mathrm dx=-\int f'(x)\delta(x-a)\mathrm dx=-f'(a)$$

And $I_\mathrm{me}$ is actually not $0$.

So where's the catch?