How to prove that ${_2F_1}(1-k, 1, k+1, -1) = k\cdot {_2F_1}(\frac 12, 1-k, \frac 32, 1)$

Question

This is related to this question

Since we know that $$\int (1-x^2)^{k-1} \ dx = x \cdot {_2F_1}\left(\frac 12, 1-k, \frac 32, x^2\right)$$ (see linked question) then it's clear that

$$\int_0^1 (1-x^2)^{k-1} \ dx = {_2F_1}\left(\frac 12, 1-k, \frac 32, 1\right)$$

If we use the relationship $$\int_0^1 x^{b-1}(1-x)^{c-b-1}(1-zx)^{-a} \ dx = B(b, c-b) \cdot {_2F_1}(a,b,c,z)$$

where $B$ is the beta function, we get that

$$\int_0^1 (1-x^2)^{k-1} \ dx = \frac 1k \cdot {_2F_1}(1-k, 1, k+1, -1)$$

Which implies

$$ {_2F_1}(1-k, 1, k+1, -1) = k\cdot {_2F_1}\left(\frac 12, 1-k, \frac 32, 1\right)$$

Is there an easier way to prove this relationship? How could one see that this is correct without doing what is written before?

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In general dealing with the hypergeometric function seems a mess. I tried using various (linear) transformation, but with no use. I can't use Kummer's theorem to simplify the LHS, and in general I have no idea how to manipulate the hypergeometric.

Answer 1

Would you like to do it by showing $$ {_2F_1}(1-k, 1, k+1, -1) = \frac{\sqrt{\pi}\;\Gamma(k+1)}{\Gamma(k+\frac{1}{2})} \\ {_2F_1}\left(\frac 12, 1-k, \frac 32, 1\right)=\frac{\sqrt{\pi}\;\Gamma(k)}{\Gamma(k+\frac{1}{2})} $$ or would that be the same proof?

See $z=-1$ and $z=1$