Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

A How to get this result for this integral?

  1. Mar 27, 2017 #1

    ShayanJ

    User Avatar
    Gold Member

    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?

    Thanks
     
  2. jcsd
  3. Mar 27, 2017 #2

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: How to get this result for this integral?
  1. How to get the result? (Replies: 0)

Loading...