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(http://functions.wolfram.com/HypergeometricFunctions/Hypergeometric2F1/07/01/01/):

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

which for my case turns out to be: $$\tag{3}{}_2F_{1}\left(\frac{1}{2},1,\frac{D-1}{2};A^2\right)=\frac{\Gamma\left(\frac{D-1}{2}\right)}{\Gamma\left(\frac{D-3}{2}\right)}\int_0^1 dt\; \left(1-t\right)^{\frac{D-4}{2}}(1-t A^2 )^{-\frac{1}{2}}$$

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

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

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

The above identity doesn't seem to be true as can be checked for $D=4$ and $A=0$ in which case the LHS is $2$ whereas the RHS is $1$. Where is the mistake?

For convenience, the code for LHS and RHS are:

LHS=Integrate[(1 - u^2)^((D - 4)/2)/(1 + A u), {u, -1, 1}, 
 Assumptions -> Abs[A] < 1 && D > 2]
RHS=(Sqrt[\[Pi]] Gamma[(D - 2)/2])/
 Gamma[(D - 3)/2] Integrate[(1 - u)^((D - 4)/2)/(1 - u A^2)^(
  1/2), {u, 0, 1}, Assumptions -> Abs[A] < 1 && D > 2]

Note that I posted a duplicate of this question in Math SE(https://math.stackexchange.com/questions/2827189/an-integral-identity-involving-gamma-function) as well.

  • 2
    $\begingroup$ A malicious comment: what is the differential operator in the exponent in the first expression supposed to do? A straightforward comment: D is a built-in command - do not use it as a name for symbols, either user-defined or undefined. $\endgroup$ – corey979 Jun 21 '18 at 18:39

$c-b-1=\frac{D-5}{2}$ not $\frac{D-4}{2}$ in this case.

  • $\begingroup$ What a blunder! Thanks for pointing it out. $\endgroup$ – Subho Jun 21 '18 at 13:34

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