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.


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

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  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Stefan4024, Math1000, Jose Arnaldo Bebita-Dris, Community, Community
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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

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 $$z=\alpha^2,\qquad b=\frac{1}{2},\qquad c=\beta+\frac{3}{2},\qquad a=1.$$



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