Integral representations and a range bound for integer Bessel functions

For fixed $z$ and $\mathrm{\mid }t\mathrm{\mid }=1$, both exponential series in $\exp \text{\hspace{0.17em}⁣}(\frac{z}{2}(t-t^{-1}))=\left({\sum }_{p\ge 0}\frac{(z\mathrm{/}2{)}^p{t}^p}{p!}\right)\left({\sum }_{q\ge 0}\frac{(-z\mathrm{/}2{)}^q{t}^{-q}}{q!}\right)$ have absolute tails bounded independently of $t$, by [F3]. Their product therefore has an absolutely and uniformly convergent double series on the unit circle; grouping its terms by the Laurent exponent $p-q$ is legitimate by [F4] and the absolute bound. For $n\ge 0$, the coefficient of $t^n$ is ${\sum }_{q\ge 0}\frac{(-1{)}^q(z\mathrm{/}2{)}^{2q+n}}{(q+n)!q!}=J_n(z)$ by [F1]. Replacing the minus sign in the first exponential by a plus sign gives the coefficient ${\sum }_{q\ge 0}\frac{(z\mathrm{/}2{)}^{2q+n}}{(q+n)!q!}=I_n(z)\mathrm{.}$

For every integer $m$, direct integration of the real and imaginary parts gives ${\int }_0^{2\pi }{e}^{im\theta }d\theta =\{\begin{array}{ll}2\pi ,& m=0,\\ 0,& m\mathrm{\ne }0.\end{array}$ For $m\mathrm{\ne }0$, the sine and cosine antiderivatives and their $2\pi$-periodicity from [F5] give zero. Integrating each finite Laurent partial sum against $e^{-in\theta }$ therefore extracts exactly its $t^n$ coefficient. The uniform convergence in step 1.1 permits passage to the full integral by [F6], applied componentwise. Consequently each coefficient is $\frac{1}{2\pi }{\int }_0^{2\pi }G(e^{i\theta })e^{-in\theta }d\theta ,$ where $G$ is the corresponding generating function.

For real $x$, Euler's formula in [F5] gives $G_J(e^{i\theta })=\exp (ix\sin \theta )$ and $G_I(e^{i\theta })=\exp (x\cos \theta )$. In the first coefficient integral, the real part is $\cos (x\sin \theta -n\theta )$; the substitution $\theta \mapsto 2\pi -\theta$ on $[\pi ,2\pi ]$, justified by [F7], shows that the imaginary parts cancel and the real parts on the two half intervals agree. Thus the coefficient is ${\pi }^{-1}{\int }_0^{\pi }\cos (n\theta -x\sin \theta )d\theta$. For the second integral, $\cos (2\pi -\theta )=\cos \theta$ and the sine Fourier terms cancel under the same substitution, giving ${\pi }^{-1}{\int }_0^{\pi }{e}^{x\cos \theta }\cos (n\theta )d\theta$. This proves both representations.

For $x\ge 0$, every term in the series for $I_n(x)$ is nonnegative, so $I_n(x)\ge 0$. The integral gives $e^{-x}{I}_n(x)\le {\pi }^{-1}{\int }_0^{\pi }{e}^{-x(1-\cos \theta )}d\theta \le 1$, since $1-\cos \theta \ge 0$.