Why is $\int_0^{2\pi} e^{i\,k\rho[\sin\alpha\cos\alpha-\sin\theta\cos(\phi-\beta)]}\mathrm{d}\beta = 2\pi J_0(k\rho\xi)$? The following is an integral in Jackson Classical Electrodynamics (3rd ed.). In equation (10.112) the integral
$$
\int_0^{2\pi} e^{i\,k\rho[\sin\alpha\cos\alpha-\sin\theta\cos(\phi-\beta)]}\mathrm{d}\beta\,,
$$
is solved by defining the function
$$
\xi \equiv (\sin^2\theta+\sin^2\alpha-2\sin\theta\sin\alpha\cos\phi)^{1/2}\,.
$$
The integral is then transformed into
\begin{equation}
\int_0^{2\pi}e^{-i\,k\rho \xi \cos\beta'}\mathrm{d}\beta' = 2\pi J_0(k\rho\xi)\,.
\end{equation}
I understand why the last integral is given by a Bessel function, but I don't understand the substitution that leads to it. 
Numerically, I checked that the first integral is indeed equal to the Bessel function. However, I checked numerically that the exponents in the integrals are not equal. But when plotting them as functions of $\beta$ I see that both are $\cos$ functions, with the same amplitude but different phases. That explains at least why the two integrals yield the same result.
 A: After a good night's sleep and with the help of Leucippus' comment, I was able to figure it out:
The integration is over a disk in the $(x,y)$-plane with the area element $\rho\mathrm{d}\rho\mathrm{d}\beta$. The integral in question is the angular integration. The phase is the scalar product $\vec x\cdot\vec k_0-\vec x\cdot\vec k=(\vec k_0-\vec k)\cdot\vec x$ with the vectors
$$
\vec x = \rho\left( \begin{array}{c} \sin\beta\\ \sin\beta\\ 0 \end{array}\right),\,
\vec k_0 = k\left( \begin{array}{c} \sin\alpha\\ 0\\ \cos\alpha \end{array}\right),\,
\vec k = k\left( \begin{array}{c} \sin\theta\cos\phi\\ \sin\theta\sin\phi\\ \cos\theta \end{array}\right).
$$
For the purpose here, the scalar product is most conveniently evaluated as
$$
(\vec k_0-\vec k)\cdot\vec x = |\vec k_0-\vec k|\;|\vec x|\,\cos\measuredangle(\vec k_0-\vec k,\vec x)\,.
$$
Since integration is only in the $(x,y)$-plane, only those two components of $\vec k$ and $\vec k_0$ contribute. The length of the (projected) $\vec k_0-\vec k$ is then given by $k\,\xi$. This can be seen by either calculating the norm or from the figure and the law of cosines. 

Thus, the phase reads indeed
$$
(\vec k_0-\vec k)\cdot\vec x = k\,\rho\,\xi \cos\measuredangle(\vec k_0-\vec k,\vec x)\,.
$$
Next, notice that we integrate over the whole azimuthal angle $\beta=\measuredangle((1,0)^T,\vec x)$. It is, therefore, irrelevant whether we start at 0 angle with the $x$-axis and end at $2\pi$ after one revolution, or we start at 0 angle with the axis defined by the constant vector $\vec k-\vec k_0$ (dashed arrow in the figure) and at $2\pi$ after one revolution. The new integration angle is thus defined by $\beta'\equiv\measuredangle(\vec k_0-\vec k,\vec x)$ and runs from 0 to $2\pi$. Putting all together gives indeed
$$
\int_0^{2\pi} e^{i\,k\rho[\sin\alpha\cos\alpha-\sin\theta\cos(\phi-\beta)]}\mathrm{d}\beta = \int_0^{2\pi}e^{i\,k\rho \xi \cos\beta'}\mathrm{d}\beta'\,,
$$
which then evaluates to the Bessel function.
