# A How to get this result for this integral?

1. Mar 27, 2017

### ShayanJ

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. Mar 27, 2017

### dextercioby

The series expansion of the 1/sqrt can be put in correspondence with the Gauss Hypergeometric function. Just look at the definition of 2F1.

3. Mar 29, 2017

### Fred Wright

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.