Bessel's integral

Remarks

The generating function $\exp\left( \frac{z}{2} (t - t^{-1}) \right) = \sum_{k \in \Z} J_k(z)\, t^k$ with $t = e^{i\tau}$ gives $e^{iz \sin \tau} = \sum_{k} J_k(z)\, e^{ik\tau}$, the Jacobi–Anger expansion. So $J_n(z)$ is the $n$-th Fourier coefficient $\frac{1}{2\pi} \int_{-\pi}^{\pi} e^{i(z \sin \tau - n\tau)}\, d\tau$. Writing $e^{i(z \sin \tau - n\tau)} = \cos(n\tau - z\sin\tau) - i \sin(n\tau - z\sin\tau)$, the sine term is odd in $\tau$ and the cosine term is even, which gives the formula on $[0, \pi]$. Example: for $n = 0$ and $z = 0$ the integral is $\frac{1}{\pi} \int_0^\pi 1\, d\tau = 1 = J_0(0)$. The hypothesis that $n$ is an integer is needed: for non-integer $\nu$ the same integral defines the Anger function, which differs from $J_\nu$. For $\Re z > 0$ and any $\nu$, Schläfli's extension reads $J_\nu(z) = \frac{1}{\pi} \int_0^\pi \cos(\nu\tau - z \sin\tau)\, d\tau - \frac{\sin(\nu\pi)}{\pi} \int_0^\infty e^{-z \sinh t - \nu t}\, dt$.

Going deeper

For a student

The integral turns many questions about $J_n$ into questions about an oscillatory integral. Applying stationary phase to it gives the large-$x$ asymptotics of $J_n(x)$. Differentiating under the integral sign checks Bessel's equation directly. It is also the bridge between Bessel functions and Fourier series of $e^{iz\sin\tau}$, which is how they enter Kepler's equation and frequency modulation.