How to get this result for this integral?

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The integral discussed is \(\displaystyle \int \frac {dy}{y^d} \left( \frac 1 {\sqrt{1-y^{2d}}}-1\right)\), which evaluates to \(\frac{y^{1-d}\left[ y^{2d} \ _2F_1\left( \frac 1 2,\frac{d+1}{2d};\frac {3d+1}{2d};y^{2d} \right)+(d+1)\left( \sqrt{1-y^{2d}}-1 \right) \right]}{1-d^2}\). The solution involves using the series expansion of \(\frac{1}{\sqrt{1-y^{2d}}}\) in relation to the Gauss Hypergeometric function, specifically \(_2F_1\). A substitution of \(t=y^{2d}\) and the application of Kummer's first formula are essential steps in deriving the solution.

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ShayanJ
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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?

Thanks
 
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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.
 
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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