I am looking at the bessel function $$J_n(a)=\frac{1}{2\pi}\int_{[0,2\pi]}\exp(ia\cos(x)-inx)dx$$ I wish to show that $$|J_n|\leq Ke^{-Cn}$$ The hint is to rewrite $J_n$ as a contour integral, which I have done as $$J_n(a)=\frac{1}{2\pi}\int_{C}\frac{e^{ia(z+1/z)/2}}{z^{n+1}}dz$$ Where $C=S^1$ is the unit circle, so $z(x)=e^{ix}$ for $x\in[0,2\pi]$.
But now I'm stuck. I don't see how we can get an exponential bound out of this. Any help is appreciated. I would like to use the residue theorem somehow, but I don't see a tenable way forward.
Just integrate along $|z|=r$ (instead of $|z|=1$), where $r>1$ is arbitrary. You get $$|J_n(a)|\leqslant Mr^{-n},\qquad M=\sup_{|z|=r}|e^{ia(z+1/z)/2}|.$$
Letting $r$ vary (with $n$) yields a stronger bound: for real $a>0$, and $n>a$, the optimum is $$r=\frac{n+\sqrt{n^2-a^2}}{a}\implies|J_n(a)|\leqslant\left(\frac{a}{n+\sqrt{n^2-a^2}}\right)^n e^{\sqrt{n^2-a^2}}.$$