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mukushla
i need help with an integral\[\int x J_n^2(x) \ \text{d}x\]
\[J _{n}\] is just any arbitary function ?
bessel function of first kind
\[x^2\frac{ d^2y }{ dx^2 }+x \frac{ dy }{ dx }+(x^2-n^2)y=0\]
so the bessel function is the solution to this curve
can we find the intergral of these
emm...im just trying !! i dont know :(
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
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
\[\int x\left(\sum_{m=0}^\infty \frac{(-1)^m}{m!(m+n)!}(\frac{1}{2}x)^{2m+n}\right)^2 dx\] - forgot the dx
can u show me the steps with wolfram?
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
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