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mukushla

  • 2 years ago

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

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  1. mukushla
    • 2 years ago
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    @mahmit2012

  2. Jonask
    • 2 years ago
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    \[J _{n}\] is just any arbitary function ?

  3. mukushla
    • 2 years ago
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    bessel function of first kind

  4. Jonask
    • 2 years ago
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    i am out then

  5. experimentX
    • 2 years ago
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    *

  6. Jonask
    • 2 years ago
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    \[x^2\frac{ d^2y }{ dx^2 }+x \frac{ dy }{ dx }+(x^2-n^2)y=0\]

  7. Jonask
    • 2 years ago
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    so the bessel function is the solution to this curve

  8. Jonask
    • 2 years ago
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    can we find the intergral of these

  9. mukushla
    • 2 years ago
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    emm...im just trying !! i dont know :(

  10. vf321
    • 2 years ago
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    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
    • 2 years ago
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    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
    • 2 years ago
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    \[\int x\left(\sum_{m=0}^\infty \frac{(-1)^m}{m!(m+n)!}(\frac{1}{2}x)^{2m+n}\right)^2 dx\] - forgot the dx

  13. mukushla
    • 2 years ago
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    can u show me the steps with wolfram?

  14. vf321
    • 2 years ago
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    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
    • 2 years ago
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    |dw:1350273811145:dw|

  16. mahmit2012
    • 2 years ago
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    |dw:1350274006370:dw|

  17. mahmit2012
    • 2 years ago
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    |dw:1350274137879:dw|

  18. mahmit2012
    • 2 years ago
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    |dw:1350274335219:dw|

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