Mar 13, 2012 #1 alexmahone Messages 303 Reaction score 0 Find $\displaystyle\int x^2J_0(x)$ in terms of higher Bessel functions and $\displaystyle\int J_0(x)$. Last edited: Mar 13, 2012
Find $\displaystyle\int x^2J_0(x)$ in terms of higher Bessel functions and $\displaystyle\int J_0(x)$.
Mar 13, 2012 #2 Mathwebster Messages 185 Reaction score 0 $J_0(x)$ 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 $J_0' = - J_1$
$J_0(x)$ 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 $J_0' = - J_1$
Mar 13, 2012 #3 chisigma Gold Member MHB Messages 1,627 Reaction score 0 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 $\chi$ $\sigma$
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 $\chi$ $\sigma$