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I'm working on a problem in complex analysis that I don't know how to approach. The problem is as follows:
Let the Bessel functions $J_n$, for integer $n$, be defined by
$$e^\left(\frac{w}{2}(z-\frac{1}{z}\right)=\sum_{n=-\infty}^{+\infty}J_{n}(w)z^n. $$
Show that
$$J_{n} = \frac{1}{2\pi}\int_{0}^{2\pi}\exp(iw\sin\theta-in\theta)d\theta. $$
Any pointers welcome - thanks!
Hint: The right hand side of the defining equation is a Laurent series. How do you compute the coefficients of a Laurent series for a given function (e.g., the left hand side of the defining equation) via contour integrals?
The functions we are looking for are called Bessel functions. With $$ \exp \left(\frac{z}{2}\left(\zeta-\frac{1}{\zeta}\right)\right)=\sum_{n=-\infty}^{\infty} J_{n}(z) \zeta^{n} $$ follows from Cauchy's integral formulas $$ J_n(z)=\frac{1}{2 \pi i} \int_{|\zeta|=1} \frac{1}{\zeta^{n+1}} \exp\left(\frac{z}{2}\left(\zeta-\frac{1}{\zeta}\right)\right) d\zeta $$ The integrand is an integer function in $z$, so $J_n(z)$ is also integer. With $\zeta(t)=\exp(it)$, it follows that $$ \begin{aligned} J_n(z) & =\frac{1}{2 \pi} \int_0^{2 \pi} \exp(-i(n+1)) \exp\left(\frac{z}{2}\left(\exp(it)-\exp(-it)\right)\right) \exp(it) d t\\\ &=\frac{1}{2 \pi} \int_0^{2 \pi} \exp(-i n t) \exp(i z \sin t) d t =\frac{1}{2 \pi} \underbrace{\int_{-\pi}^\pi \cos (z \sin t-n t) d t}_{=: I_1}+\frac{i}{2 \pi} \underbrace{\int{-\pi}^\pi \sin (z \sin t-n t) d t}_{=: I_2} \\ & =\frac{1}{\pi} \int_0^\pi \cos (n t-z \sin t) d t \end{aligned} $$ since the integrand of $I_1$ is even and that of $I_2$ is odd. The exponential series yields $$ \begin{aligned} \exp \left(\frac{z}{2}\left(\zeta-\frac{1}{\zeta}\right)\right) & =\sum_{k=0}^{\infty} \frac{1}{k !}\left(\frac{z}{2}\right)^k\left(\sum_{j=0}^k\binom{k}{j}(-1)^j \zeta^{k-2 j}\right) \\ & =\sum_{k=0}^{\infty} \frac{1}{k !}\left(\frac{z}{2}\right)^k\left(\sum_{j=0}^k\binom{k}{j}(-1)^{k-j} \zeta^{2 j-k}\right) \end{aligned} $$ We are looking for the coefficient $J_n(z)$ of $\zeta^n$ in the Laurent series expansion. Rearrangement is allowed because of absolute convergence. The first series representation yields for $n \geq 0$ (set $k=2 j+n$ ): $$ J_n(z)=\sum_{j=0}^{\infty} \frac{1}{(2 j+n) !}\binom{2j+n}{j}(-1)^j\left(\frac{z}{2}\right)^{2 j+n}=\sum_{k=0}^{\infty} \frac{(-1)^k}{k !(n+k) !}\left(\frac{z}{2}\right)^{n+2 k} . $$ The second row representation yields for $n<0$ (with $k=2 j-n)$ : $$ J_n(z) =\sum_{j=0}^{\infty} \frac{1}{(2 j-n) !}\binom{2j-n}{j}(-1)^{j-n}\left(\frac{z}{2}\right)^{2 j-n}=(-1)^n \sum_{j=0}^{\infty} \frac{(-1)^j}{(j-n) ! j !}\left(\frac{z}{2}\right)^{2 j-n}=(-1)^n J_{-n}(z) . $$
