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I started with the integral:

$$\tag{1} \int_{-1}^1 \frac{\left(1-u^2\right)^\frac{D-4}{2}\; du}{1+A u} $$

I evaluated the integral in Mathematica:

Integrate[(1 - u^2)^((D - 4)/2)/(1 + A u), {u, -1, 1}]

In the appropriate parameter space, the result came out as:

$$\tag{2} \frac{\sqrt{\pi}\; \Gamma\left(\frac{D-2}{2}\right)}{\Gamma\left(\frac{D-1}{2}\right)} {}_2F_1\left(\frac{1}{2},1,\frac{D-1}{2},A^2\right)$$

Thereafter, I re-wrote the hypergeometric function in its integral representation, using the formula:

$${}_2F_{1}\left(a,b,c;z\right)=\frac{\Gamma(c)}{\Gamma(b)\Gamma(c-b)}\int_0^1 t^{b-1}(1-t)^{c-b-1}(1-tz)^{-a} \;dt$$

which for my case turns out to be: $$\tag{3}{}_2F_{1}\left(\frac{1}{2},1,\frac{D-1}{2};A^2\right)=\frac{\Gamma\left(\frac{D-1}{2}\right)}{\Gamma\left(\frac{D-3}{2}\right)}\int_0^1 dt\; \left(1-t\right)^{\frac{D-5}{2}}(1-t A^2 )^{-\frac{1}{2}}$$

Using (1), (2) and (3), I arrive at the following identity:

$$\int\limits_{-1}^1 \frac{(1-u^2)^{\frac{D-4}{2}}du}{1+A u}=\frac{\sqrt{\pi}\Gamma\left(\frac{D-2}{2}\right)}{\Gamma\left(\frac{D-3}{2}\right)} \int\limits_0^1 dt\; (1-t)^{\frac{D-5}{2}}(1-t A^2)^{-\frac{1}{2}}$$

where $A,D\in \mathbb{R}\;$, $|A|<1$ and $D>2$

How to prove this identity?

Edit 1:

Note that I have re-posted this question in Mathematica SE as well(https://mathematica.stackexchange.com/questions/175757/an-integral-identity-involving-gamma-function).

Edit 2:

The identity that I had given earlier had a factor wrong. I corrected it.

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closed as off-topic by Stefan4024, Math1000, Jose Arnaldo Bebita-Dris, user99914, user223391 Jun 24 '18 at 0:28

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  • $\begingroup$ @ComplexYetTrivial Ah, I see there is a problem. This came up while I was playing around with some identities. Maybe I will elaborate the post to include the background details. $\endgroup$ – Subho Jun 21 '18 at 12:04
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Here is a quick way to see this: Let, $\beta=\frac{d-4}{2}$ and $$I=\int_{-1}^{1}\!\mathrm{d}u~\frac{(1-u^2)^\beta}{1+\alpha u}.$$ Letting $v=-u$, one obtains $$I=\int_{-1}^{1}\!\mathrm{d}v~\frac{(1-v^2)^\beta}{1-\alpha v}.$$ Renaming $v=u$ and adding the two representations of $I$, we have $$2I=\int_{-1}^{1}\!\mathrm{d}u~(1-u^2)^\beta\Big[\frac{1}{1+ \alpha u}+\frac{1}{1-\alpha u}\Big]=2\int_{-1}^{1}\!\mathrm{d}u~\frac{(1-u^2)^\beta}{1-(\alpha u)^2}=4\int_0^{1}\!\mathrm{d}u~\frac{(1-u^2)^\beta}{1-(\alpha u)^2}.$$ The last step results form the symmetry of the integrand. Next, letting $t=u^2$, we have $$I=\int_0^{1}\frac{\mathrm{d}t}{t^{1/2}}~\frac{(1-t)^\beta}{1-\alpha^2t}.$$ Now, comparing this result with the integral representation of the Hypergeometric function (above), we see that $$z=\alpha^2,\qquad b=\frac{1}{2},\qquad c=\beta+\frac{3}{2},\qquad a=1.$$

Cheers!

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