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I would greatly appreciate assistance with the following problem. show:

$$\int _0 ^\infty J_n(x)dx = 1; \forall n \in \mathbb{N}^+$$

for $J_o,$ use $$\mathscr{L}{J_o(at)} = \int _0 ^\infty e^{-pt}J_o(at)dt = (p^2 + a^2)^{- \frac{1}{2}}$$

By setting a = 1; p = 0; I obtain that which I was trying to prove. How would I generalize this to other orders of the Bessel function? I tried writing a generalized bessel function in closed form but this proved fruitless.

If I take the Laplace transform of a bessel function, then the only way to do that would be to do it term by term?

EDIT : I'm still not sure how one would actually take the Laplace Transform of such a function?

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I'm not entirely sure of what you mean by "term by term." Do you mean as a power series? – Ron Gordon Jan 8 '13 at 22:05
BTW you can find the LT of Bessels here: – Ron Gordon Jan 8 '13 at 22:06
Yes I did mean as a series. Thank you for that link, it is perfect – Cactus BAMF Jan 8 '13 at 22:37
I realize that that question mark was unnecessary – Cactus BAMF Jan 8 '13 at 23:05
up vote 2 down vote accepted

Hint: Using the integral representation of the Bessel function $$J_n(x) = \frac{1}{2\pi} \int_{-\pi}^\pi dt\, e^{-i(n t - x \sin t)}$$ we find \begin{eqnarray*} \int_0^\infty dx\, e^{-p x} J_n(x) &=& \frac{1}{2\pi} \int_{-\pi}^\pi dt\, \frac{e^{-i n t}}{p-i\sin t}. \end{eqnarray*} This integral can be handled using the methods of residue calculus. Let $z=e^{-i t}$ so the contour is the unit circle. Only one of the poles lies inside the contour.

Addendum: The integral in terms of $z$ is $$\frac{1}{\pi i} \int_\Gamma dz\, \frac{z^n}{z^2+2p z-1},$$ where $\Gamma$ is the unit circle traversed in the counterclockwise sense. The poles are $$z_\pm = -p \pm \sqrt{1+p^2}.$$ Only $z_+$ lies inside the unit circle, so we pick up the residue at $z=z_+$.

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So the pole is at $t = arcsin( \frac {p}{i})$ ? which of course must be less than one in magnitude – Cactus BAMF Jan 9 '13 at 0:59
Okay thanks a lot, I got it now – Cactus BAMF Jan 9 '13 at 1:04
@CactusBAMF: Glad to help. It may be too late, but I've added something above about your previous question. – user26872 Jan 9 '13 at 1:07

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