Question

i need help with an integral\[\int x J_n^2(x) \ \text{d}x\]

Answer 1

@mahmit2012

Answer 2

\[J _{n}\]
is just any arbitary function ?

Answer 3

bessel function of first kind

Answer 4

i am out then

Answer 5

*

Answer 6

\[x^2\frac{ d^2y }{ dx^2 }+x \frac{ dy }{ dx }+(x^2-n^2)y=0\]

Answer 7

so the bessel function is the solution to this curve

Answer 8

can we find the intergral of these

Answer 9

emm...im just trying !! i dont know :(

Answer 10

Well my good friend wikipedia tells us that we can expand the Bessel Function as follows with taylor series:
http://upload.wikimedia.org/math/1/b/2/1b23400208b273377e8cdec7d82f0242.png

Answer 11

So then the integral becomes:
\[\int x\sum_{m=0}^\infty \frac{(-1)^m}{m!(m+n)!}(\frac{1}{2}x)^{2m+n}\]for integer orders

Answer 12

\[\int x\left(\sum_{m=0}^\infty \frac{(-1)^m}{m!(m+n)!}(\frac{1}{2}x)^{2m+n}\right)^2 dx\] - forgot the dx

Answer 13

can u show me the steps with wolfram?

Answer 14

No, I can't. But you do see the answer, right? My guess is that to get the answer you have to write \((\sum...)^2\) as a single sum using binomial expansion or something, then you can integrate it easily since it's just a polynomial. Then you reconstruct the sum as three separate sums and get the nice answer that alpha got:
http://www.wolframalpha.com/input/?i=integrate+x+BesselJ [n%2C+x]^2+dx