Question

Challenging integral:

Integrate from 0--> infinity the argument: (e^(-z*x^2)*x^(n+1)*J_n(c*x) dx)

Answer 1

@Zarkon

Answer 2

@TuringTest

Answer 3

@joemath314159

Answer 4

\[\Large \int_0^\infty e^{-zx^2}x^{n+1}J_ncx dx\]What is \(\large J_n\)? A constant?

Answer 5

Oh wait, it's a function...?
\[\Large \int_0^\infty e^{-zx^2}x^{n+1}J_n(cx) dx\]

Answer 6

yea its the nth order bessel function

Answer 7

I use the recursion relation: J_n = SUM ( (-1)^k / (gamma(k+1) *gamma(k+n+1)) *(1/2x)^(2k+n) where gamma is the gamma function. Just the definition of the J_n function. Not sure how to integrate it...

Answer 8

Is it true that the Bessel functions are normalized so that
\[\int_0^\infty J_n(x) dx=1 ?\]

Answer 9

yup they are

Answer 10

and you are using
\[\large J_n(x) = \sum_{k=0}^\infty \frac{(-1)^k}{\Gamma(k+1)\Gamma(k+n+1)}\left( \frac{1}{2}x\right) ^{2k+n}\]

Answer 11

yep

Answer 12

how did you adjust for J_n(cx) ? Is is just
\[\large J_n(cx) = \sum_{k=0}^\infty \frac{(-1)^k}{\Gamma(k+1)\Gamma(k+n+1)}\left( \frac{c}{2}x\right) ^{2k+n}\]

Answer 13

I believe so yea

Answer 14

Is this from QM or probability ???

Answer 15

electromagnatism

Answer 16

I don't claim to know anything about EM. Usually there is some trick when working with distributions of this type, so that you don't have to go through all the integration mess. However, have you tried this approach (see pdf attached)

Answer 17

About the messy repeated integration by parts see
http://www.colby.edu/chemistry/PChem/notes/Integral.pdf

Answer 18

humm....

Answer 19

Thanks that helped. I did a bit of re-shuffling of some terms and got the right answer.

Answer 20

cool!