## mukushla Group Title i need help with an integral$\int x J_n^2(x) \ \text{d}x$ one year ago one year ago

1. mukushla Group Title

@mahmit2012

$J _{n}$ is just any arbitary function ?

3. mukushla Group Title

bessel function of first kind

i am out then

5. experimentX Group Title

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

so the bessel function is the solution to this curve

can we find the intergral of these

9. mukushla Group Title

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

10. vf321 Group Title

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

11. vf321 Group Title

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

12. vf321 Group Title

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

13. mukushla Group Title

can u show me the steps with wolfram?

14. vf321 Group Title

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

15. mahmit2012 Group Title

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16. mahmit2012 Group Title

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17. mahmit2012 Group Title

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18. mahmit2012 Group Title

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