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The Bessel Function $J_n(x)$ is defined, for a natural number $n$ and real number x, as

$J_n(x) = \frac{1}{2\pi}\int_0^{2\pi}\cos(n\theta-x\sin\theta)d\theta.$

By using contour integration with integrand $z^{n-1}\exp(\frac{-xz}{2})\exp(\frac{x}{2z})$, or otherwise, show that

$J_n(x)=\sum_{\substack{k=0}}^\infty\frac{(-1)^k}{k!(n+k)!}(\frac{x}{2})^{n+2k}$

Hi, can someone give me some hints or give a simple example to convert the integral to a sum form using integrand. I am not very familiar with how to use integrand.

Many many thanks!

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From the integral definition we have: $$J_n(x) = \frac{1}{2\pi}\Re\int_{0}^{2\pi}e^{-ix\sin\theta}e^{ni\theta}\,d\theta $$ and expanding the exponential function as a Taylor series we get: $$ [x^k]\,J_n(x) = \frac{1}{2\pi}\Re\int_{0}^{2\pi}\frac{(-i\sin\theta)^k}{k!}e^{ni\theta}\,d\theta, $$ so, by expressing $\sin\theta$ as $\frac{e^{i\theta}-e^{-i\theta}}{2i}$, using the binomial theorem and the identity: $$ \frac{1}{2\pi}\int_{0}^{2\pi}e^{ni\theta}e^{-ki\theta}\,d\theta = \delta_{n,k}$$ we easily prove our claim.

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  • $\begingroup$ I am kind of confused of x^k, can you write it more explicitly, and also how to use the integrand. thanks a lot!!! $\endgroup$ – kengkeng Dec 5 '14 at 2:27
  • $\begingroup$ $[x^k]\,f(x)$ is just a shorthand notation for "the coefficient of $x^k$ in the Taylor series of $f(x)$ around $x=0$". $\endgroup$ – Jack D'Aurizio Dec 5 '14 at 9:49

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