How Can Bessel Functions Be Integrated Using Recurrence Relations?

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Discussion Overview

The discussion revolves around the integration of Bessel functions, specifically focusing on the integral of the product of \(x^2\) and the Bessel function \(J_0(x)\). Participants explore methods to express this integral in terms of higher Bessel functions and the integral of \(J_0(x)\).

Discussion Character

  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant requests the integral \(\int x^2 J_0(x)\) in terms of higher Bessel functions and \(\int J_0(x)\).
  • Another participant cites the differential equation satisfied by \(J_0(x)\) and presents an integration approach leading to a relationship involving \(\int J_0 dx\) and \(\int x^2 J_0 dx\).
  • A different participant introduces a general formula for integrating products of powers of \(x\) and Bessel functions, applying integration by parts to derive an expression for \(\int x^2 J_0(x) dx\) that includes terms involving \(J_1(x)\) and \(\int J_0(x) dx\).

Areas of Agreement / Disagreement

The discussion presents multiple approaches to the problem without a consensus on a single method or solution. Participants offer different perspectives and techniques for integrating the Bessel function.

Contextual Notes

Some assumptions regarding the properties of Bessel functions and the constants of integration are not explicitly stated, which may affect the interpretation of the results.

alexmahone
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Find $\displaystyle\int x^2J_0(x)$ in terms of higher Bessel functions and $\displaystyle\int J_0(x)$.
 
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$J_0(x)$ satisfies

$x^2 J_0'' + x J_0' + x^2J_0 = 0$

Integrating gives

$\displaystyle \int \left(x^2 J_0'' + xJ_0'\right) dx + \int x^2 J_0dx = c$

or

$\displaystyle x^2 J_0' - x J_0 + \int J_0 dx + \int x^2 J_0 dx = c$

then use the fact the $J_0' = - J_1$
 
Using the general formula...

$\displaystyle \int x^{n}\ J_{n-1}(x)\ dx = x^{n}\ J_{n}(x)$ (1)

... and taking into account that $\displaystyle J^{'}_{0}(x)= - J_{1}(x)$ , integration by parts gives You ...

$\displaystyle \int x^{2}\ J_{0}(x)\ dx = \int x\ x\ J_{0}(x)\ dx= x^{2}\ J_{1}(x) - \int x\ J_{1}(x)\ dx = x^{2}\ J_{1}(x) + x\ J_{0}(x) - \int J_{0}(x)\ dx$ (2)

Kind regards

$\chi$ $\sigma$
 
Thanks.
 

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