Timeline for Indefinite integration of multiplication of two Bessel function
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 28, 2014 at 10:07 | comment | added | bordart | It is the Hankel transform when $a = 0$, as I know. | |
| Jan 28, 2014 at 9:45 | comment | added | Nicola Ciccoli | Isn't it the Hankel transform of the Bessel function itself? Then this should give rise to so called Zernike polynomials. Have a look at formula (7) in arxiv.org/pdf/1007.0667.pdf | |
| Jan 27, 2014 at 23:06 | answer | added | Suvrit | timeline score: 0 | |
| Jan 27, 2014 at 22:23 | answer | added | Gerald Edgar | timeline score: 1 | |
| Jan 27, 2014 at 21:03 | history | edited | Carlo Beenakker | CC BY-SA 3.0 |
it's an indefinite integral, since "a" is arbitrary
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| Jan 27, 2014 at 19:56 | comment | added | bordart | let us continue this discussion in chat | |
| Jan 27, 2014 at 19:51 | comment | added | bordart | So we are going to argue about the terminology I use? Yes, you probably know much more than me. So please, if you can help, just help to integrate this. | |
| Jan 27, 2014 at 19:47 | comment | added | GH from MO | An integral cannot be "solved". Also, the expression you wrote is completely explicit. | |
| Jan 27, 2014 at 19:35 | comment | added | bordart | OK, let me explain it in this way. Can you solve the integral? Find any F(x) in an explicit way, that you may put it on the right side. | |
| Jan 27, 2014 at 19:11 | comment | added | GH from MO | It is not clear what you mean by "analytic solution". We are talking about an integral (not an equation), which is analytic (complex differentiable) in $a$. The word "solution" makes no sense in this context. | |
| Jan 27, 2014 at 19:03 | comment | added | bordart | I mean analytic solution. Sorry for confusion. | |
| Jan 27, 2014 at 18:52 | comment | added | GH from MO | It is not clear what you mean by "analytic expression". The integral that you wrote is an analytic expression (in my vocabulary). | |
| Jan 27, 2014 at 18:40 | history | edited | Gerald Edgar | CC BY-SA 3.0 |
deleted 8 characters in body
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| Jan 27, 2014 at 16:57 | comment | added | Suvrit | The following may be helpful: For $a=0$ we have a known formula; writing $\int_0^\infty-\int_0^a$ we get a formula for your case; might be possible because $\int_0^a J_\mu(x)J_{\mu+1}(x)dx = \sum_{k \ge 0} J_{\mu+k+1}(a)^2$, though haven't given it more thought. | |
| Jan 27, 2014 at 16:49 | review | First posts | |||
| Jan 27, 2014 at 17:03 | |||||
| Jan 27, 2014 at 16:32 | history | asked | bordart | CC BY-SA 3.0 |