= Solution
For $t=u+iv$, the exponent is
$$
ix\cosh t=-x\sinh u\sin v+ix\cosh u\cos v.
$$
Its imaginary part must remain equal to its value $x$ at the saddle $t=0$. The descending branch connecting the required ends is therefore
$$
\boxed{\cosh u\cos v=1,\qquad v=\operatorname{sgn}(u)\arccos(\operatorname{sech}u).}
$$
Along it $\sin v=\tanh u$, so the real part is $-x(\cosh u-\operatorname{sech}u)$, strictly negative away from zero for $x>0$. The branch approaches $v=\pm\pi/2$ at the two infinities and has tangent $t=e^{i\pi/4}s$ at the saddle, oriented towards increasing $s$.
Locally $\cosh t=1+t^2/2+O(t^4)$, hence the <steepest descent method> gives
$$
\int_Ce^{ix\cosh t}dt=e^{ix+i\pi/4}\int_{-\infty}^\infty e^{-xs^2/2}ds\,[1+O(x^{-1})]
=e^{ix+i\pi/4}\sqrt{\frac{2\pi}{x}}\,[1+O(x^{-1})].
$$
Outside a small saddle neighborhood the descending contour has strictly smaller real exponent; inside it the Gaussian rescaling $s=x^{-1/2}r$ controls the higher terms. Multiplication by $1/(i\pi)$ and taking the real part yield
$$
\boxed{J_0(x)=\sqrt{\frac2{\pi x}}\cos\left(x-\frac\pi4\right)+O(x^{-3/2})\quad(x\to+\infty).}
$$
This additive asymptotic form specifies the leading term even near the zeros of the oscillating cosine, where a literal pointwise ratio interpretation of the printed $\sim$ is unsuitable.
Back to article page
