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At equal argument and large integer order, the Bessel function of the first kind has phase $n(\sin t-t)=-n{t}^3/6+\cdots$ near the endpoint. The cubic stationary endpoint yields the displayed positive constant. It is also the zero-argument value of the Airy function in the uniform Bessel turning-point approximation.

Let $a\in C_c^{\infty }(I)$ and let $\varphi$ be a smooth real phase on an interval $I$. Assume $x_0$ is the only stationary point of $\varphi$ on the amplitude support and ${\varphi }^{\prime \prime }(x_0)\ne 0$. As $t\to +\infty$,The compact support avoids endpoint terms. A local quadratic coordinate change reduces the central contribution to a Fresnel integral; repeated integration by parts controls the nonstationary remainder.

The additive remainder is deliberate. If $a(x_0)=0$, an asymptotic equivalence to the zero expression would be meaningless, whereas the displayed estimate still says that the integral is $O(t^{-3/2})$. For several separated nondegenerate stationary points, localize and add their contributions. A cubic stationary endpoint or a nonzero boundary amplitude is outside these hypotheses.

The Bessel turning-point asymptotic comes from a cubic stationary endpoint, not an ordinary quadratic stationary point. Near $\vartheta =0$,Thus the contributing width is $\vartheta =O(n^{-1/3})$. On writing $\vartheta =(6/n{)}^{1/3}s$, the leading integral isThe oscillatory integral is understood with a vanishing damping factor. Substitution $u=s^3$ and the Gamma function Fourier integral giveThereforeThe equality uses the Gamma reflection formula. Contributions away from the degenerate endpoint are smaller. The $n^{-1/3}$ scale explains why the preceding $n^{-1/2}$ formula cannot be extended directly to zero angle.