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

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1. mukushla Group Title

@mahmit2012

2. Jonask Group Title

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

3. mukushla Group Title

bessel function of first kind

4. Jonask Group Title

i am out then

5. experimentX Group Title

*

6. Jonask Group Title

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

7. Jonask Group Title

so the bessel function is the solution to this curve

8. Jonask Group Title

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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