How Can Bessel Functions Be Integrated Using Recurrence Relations?

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SUMMARY

The integration of the function $\displaystyle\int x^2 J_0(x)$ using recurrence relations is established through the differential equation $x^2 J_0'' + x J_0' + x^2 J_0 = 0$. By applying integration by parts and the relationship $J_0' = -J_1$, the integral can be expressed as $\displaystyle x^{2} J_{1}(x) + x J_{0}(x) - \int J_{0}(x) dx$. This method effectively utilizes the general formula for Bessel functions, specifically $\displaystyle \int x^{n} J_{n-1}(x) dx = x^{n} J_{n}(x)$.

PREREQUISITES
  • Understanding of Bessel functions, specifically $J_0(x)$ and $J_1(x)$
  • Familiarity with integration techniques, particularly integration by parts
  • Knowledge of differential equations and their solutions
  • Basic grasp of recurrence relations in mathematical functions
NEXT STEPS
  • Study the properties and applications of Bessel functions in mathematical physics
  • Learn advanced integration techniques involving special functions
  • Explore the derivation and implications of recurrence relations for Bessel functions
  • Investigate the use of Bessel functions in solving partial differential equations
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Mathematicians, physicists, and engineers who require a deeper understanding of Bessel functions and their applications in solving integrals and differential equations.

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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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