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Three-term integral

  1. Jul 23, 2009 #1
    Hello

    I've been having some difficulty solving a three-term integral of the form:


    [tex]\int^{\beta_{r}}_{\beta_{0}}\frac{y^{\tau-3+\frac{\eta}{3-\kappa}}}{\left(1-y^{2}\right)^{3-\frac{\tau}{2}}}\left(1+\left(\frac{1}{(3-\kappa)N}-\frac{1}{\beta_{0}}\right)y\right)^{-\frac{\eta}{3-\kappa}}dy[/tex]

    Where [tex]y < 1[/tex] and [tex]\beta_{0}, \tau, \eta, \kappa, N[/tex] are constants.

    I've tried expanding the term:
    [tex]\left(1-y^{2}\right)^{3-\frac{\tau}{2}}}[/tex]
    As a taylor series and then integrating term by term, although for this to be accurate this tends to require an expansion up to an order of 40 which cannot be done using Mathematica. It is also required that this integral be done non-numerically and expressed in terms of the limits of the integral.

    It is also quite difficult expressing the solution using hypergeometric functions (eg 2F1) using these limits.

    Any help would be greatly appreciated!
     
  2. jcsd
  3. Jul 23, 2009 #2
    dude you have way to many different constants in there

    is this correct:

    [tex] \int^{\beta_{r}}_{\beta_{0}}\frac{y^k}{\left(1-y^{2}\right)^n}\left(1+ly\right)^{m}dy
    [/tex]

    is [itex]\frac{\eta}{3-\kappa}[/itex] and integer? a positive integer? what about all the screwy looking exponents you have? can you expand using the binomial formula and then integrate those term by term?
     
  4. Jul 23, 2009 #3
    I would take this form to an integral table. It's most likely too complex to find there though.
     
  5. Jul 23, 2009 #4
    like i said - it depends on the exponents
     
  6. Jul 24, 2009 #5
    Thanks for the replies,

    The integral could be expressed in that simplified form and [tex]\frac{\eta}{3-\kappa}[/tex] would be be a negative, non-integer value. I could expand either the term [tex]\left(1-y^{2}\right)^{n}[/tex] or [tex]\left(1+ly\right)^{m}[/tex] and integrate term by term, although for this to be accurate it tends to require an expansion up to an order of 40. I was hoping there might be an analytical solution to this or a better approximation, although I can see this might be unlikely in this case.
    I've tried looking through an integral table although I've not had much luck so far.
     
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