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A How to get this result for this integral?

  1. Mar 27, 2017 #1


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    I used wolframalpha to calculate an integral and here's what I got:

    ## \displaystyle \int \frac {dy}{y^d} \left( \frac 1 {\sqrt{1-y^{2d}}}-1\right)=\frac{y^{1-d}\left[ y^{2d} \ _2F_1\left( \frac 1 2,\frac{d+1}{2d};\frac {3d+1}{2d};y^{2d} \right)+(d+1)\left( \sqrt{1-y^{2d}}-1 \right) \right]}{1-d^2} ##

    Does anyone have any idea how to arrive at this solution?

  2. jcsd
  3. Mar 27, 2017 #2


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    The series expansion of the 1/sqrt can be put in correspondence with the Gauss Hypergeometric function. Just look at the definition of 2F1.
  4. Mar 29, 2017 #3
    I suggest that you make the substitution, ##t=y^{2d}## and use the integral representation of of the hypergeometric function. Then apply Kummer's first formula to your result.
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