MHB How Can Bessel Functions Be Integrated Using Recurrence Relations?

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.
 
I have the equation ##F^x=m\frac {d}{dt}(\gamma v^x)##, where ##\gamma## is the Lorentz factor, and ##x## is a superscript, not an exponent. In my textbook the solution is given as ##\frac {F^x}{m}t=\frac {v^x}{\sqrt {1-v^{x^2}/c^2}}##. What bothers me is, when I separate the variables I get ##\frac {F^x}{m}dt=d(\gamma v^x)##. Can I simply consider ##d(\gamma v^x)## the variable of integration without any further considerations? Can I simply make the substitution ##\gamma v^x = u## and then...

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