How to prove that $\int (1-x^2)^{k-1} \ dx= x \cdot {_2F_1}(\frac 12, 1-k, \frac 32, x^2)$?

Question

I am reading a paper and there is written that

$$\int (1-x^2)^{k-1} \ dx = x \cdot {_2F_1}\left(\frac 12, 1-k, \frac 32, x^2\right)$$

where $ {_2F_1}$ is the hypergeometric function

(this should hold for $x \in [-1, 1]$)

How is this result proven?

I tried computing the taylor expansion of $(1-x^2)^{k-1}$ is neighborhood of $0$, then integrate term by term and find $a,b,c,z$ such that our geometric function has the same form. This may be doable but I find it hard to compute the $\displaystyle \frac{d^n}{x^n}\left(1-x^2\right)^{k-1}$

Are there easier ways? Thank you! :-)

Answer 1

With $X=x^2$ : $$\int (1-x^2)^{k-1}dx=\frac{1}{2}\int (1-X)^{k-1}X^{-1/2}dX=\frac{1}{2}B_{X}(\frac{1}{2},k)$$ $B$ is the Incomplete Beta function.

The Incomplete Beta function is a particular hypergeometric function, with relationship : $$ _2F_1(a,b;b+1;X)=bX^{-b}B_X(b,1-a)$$ In this case $a=1-k$ and $b=\frac{1}{2}$: $$ _2F_1(1-k,\frac{1}{2};\frac{1}{2}+1;X)=\frac{1}{2}X^{-\frac{1}{2}}B_{X}(\frac{1}{2},k)$$

$$ _2F_1(1-k,\frac{1}{2};\frac{3}{2};x^2)=\frac{1}{2}x^{-1}B_{X}(\frac{1}{2},k)$$

$$\frac{1}{2}B_{X}(\frac{1}{2},k)=x \: _2F_1(1-k,\frac{1}{2};\frac{3}{2};x^2)=x \: _2F_1(\frac{1}{2},1-k;\frac{3}{2};x^2)=\int (1-x^2)^{k-1}dx$$