AdS/CFT scalar 3-point function integral from Feynman parametrization? In this paper the scalar 3-point function in AdS/CFT is obtained by performing the following integral:

The authors comment that they obtain the result by Feynman parameter integration. For practice I wanted to reproduce the same result, so I looked up the Wikipedia page on Feynman parameters:

The trouble here is the condition Re$(\alpha_j)>0$, while the Re$(a)>0$ in the above integral (22) would lead to a negative real part of an alpha.
Are there any other variations of Feynman parametrization known, which would cover the case at hand?
Thanks for any suggestion!
EDIT:
Pretending that the real part constraint is not an issue and just applying the rule I get the same result as in the paper times the following:
$$\frac{\Gamma \left(c-\frac{d}{2}\right) \Gamma \left(\frac{1}{2} (-2 c+d+2)\right) \Gamma (-a+b+c-d) \Gamma (a-b-c+d+1)}{\Gamma
   \left(\frac{1}{2}-\frac{a}{2}\right) \Gamma \left(\frac{a}{2}+\frac{1}{2}\right) \Gamma \left(-\frac{a}{2}+b-\frac{d}{2}+\frac{1}{2}\right) \Gamma
   \left(\frac{a}{2}-b+\frac{d}{2}+\frac{1}{2}\right)}$$
which is equivalent to
$$-\frac{\cos \left(\frac{\pi  a}{2}\right)  \cos \left(\frac{1}{2} \pi  (a-2 b+d)\right)}{\sin \left(\pi  c-\frac{\pi  d}{2}\right)\sin (\pi  (a-b-c+d))} $$
And I should mention that here again I had to ignore the constraint $\Re(C)>\Re(A+B)$ when evaluating the special case of the Gauss hypergeometric function.
The fact that two instances of "greater than" constraints had to be violated makes me hope that there might be a proper Feynman parameter approach in which all convergence criteria are met and no wacky extra factors appear.
Perhaps someone knows a proper reference?
 A: A few months ago, I also needed the result of this computation of the three-point contact Witten diagram by
Freedman, Mathur, Matusis and Rastelli. I was also flabbergasted by the lack of detail in their article since all they said was "it is easily done by conventional Feynman parameter methods". So I figured out my own derivation.
It is probably not what FMMR had in mind and not the most elegant approach but it does the job and is mathematically rigorous.
It goes as follows.
Let
$$
U:=\int_{0}^{\infty}dz_0\int_{\mathbb{R}^d} dz\ \frac{z_0^a}{(z_0^2+(z-x)^2)^b(z_0^2+(z-y)^2)^c}\ .
$$
Assuming $b,c>0$, we insert the identity $\frac{1}{A^\alpha}=\frac{1}{\Gamma(\alpha)}\int_{0}^{\infty}ds\ s^{\alpha-1} e^{-sA}$ twice to get
$$
U=\frac{1}{\Gamma(b)\Gamma(c)}\int_{0}^{\infty}dz_0\int_{\mathbb{R}^d} dz
\int_{0}^{\infty}ds
\int_{0}^{\infty}dt\ z_0^a s^{b-1}t^{c-1}\exp(-V_1)
$$
with
$$
V_1:=s(z_0^2+(z-x)^2)+t(z_0^2+(z-y)^2)
$$
$$
=s(z_0^2+x^2)+t(z_0^2+y^2)+(s+t)z^2-2(sx+ty)z\ .
$$
Thus
$$
\int_{\mathbb{R}^d} dz\ \exp(-V_1)=e^{-s(z_0^2+x^2)-t(z_0^2+y^2)}
\int_{\mathbb{R}^d} dz\ \exp \left(
-(s+t) 
\left( z-\frac{sx+ty}{s+t} \right)^2
+\frac{(sx+ty)^2}{s+t}
\right)
$$
$$
=\exp(V_2)
\times\left(\frac{\pi}{s+t}\right)^{\frac{d}{2}}
$$
with
$$
V_2:=\frac{(sx+ty)^2}{s+t}-s(z_0^2+x^2)-t(z_0^2+y^2)\ ,
$$
and after computing the Gaussian integral on $z$.
A bit of algebra shows
$$
(s+t)V_2=-(s+t)^2z_0^2-st(x-y)^2\ .
$$
Plugging back into $U$ we get
$$
U=\frac{\pi^{\frac{d}{2}}}{\Gamma(b)\Gamma(c)}
\int_{0}^{\infty}dz_0\int_{0}^{\infty}ds\int_{0}^{\infty}dt\ 
z_0^{a}s^{b-1}t^{c-1}(s+t)^{-\frac{d}{2}}\exp\left(
\frac{-(s+t)^2z_0^2-st(x-y)^2}{s+t}
\right)\ .
$$
Change variables from $z_0$ to $z_0^2$ and integrate over the latter. Provided $a>-1$, we now obtain
$$
U=\frac{\pi^{\frac{d}{2}}\Gamma\left(\frac{a+1}{2}\right)}{2\Gamma(b)\Gamma(c)}
\int_{0}^{\infty}ds\int_{0}^{\infty}dt\ 
s^{b-1}t^{c-1}(s+t)^{-\frac{a+1+d}{2}}\exp\left(
-\frac{st}{s+t}(x-y)^2
\right)\ .
$$
Change variables: ${\rm old}\ s=\frac{1}{(x-y)^2}\times\ {\rm new}\ s$ and ${\rm old}\ t=\frac{1}{(x-y)^2}\times\ {\rm new}\ t$. Hence,
$$
U=\frac{\pi^{\frac{d}{2}}\Gamma\left(\frac{a+1}{2}\right)}{2\Gamma(b)\Gamma(c)}\times
|x-y|^{a+1+d-2b-2c}
\times W
$$
with
$$
W:=\int_{0}^{\infty}ds\int_{0}^{\infty}dt\ 
s^{b-1}t^{c-1}(s+t)^{-\frac{a+1+d}{2}}\exp\left(
-\frac{st}{s+t}
\right)\ .
$$
We use the bivariate change of variable formula to integrate over $0<u<\infty$ and $0<v<1$ related to the old variables by
$$
\left(\begin{array}{c}s \\ t\end{array}\right)=
\left(\begin{array}{c} vu \\ (1-v)u\end{array}\right)\ .
$$
The Jacobian is
$$
\left|\begin{array}{cc} \frac{\partial s}{\partial u} & \frac{\partial s}{\partial v} \\
\frac{\partial t}{\partial u} & \frac{\partial t}{\partial v}\end{array}\right|=-u\ .
$$
As a result
$$
W=\int_{0}^{\infty}du\int_{0}^{1}dv\ 
v^{b-1}(1-v)^{c-1}u^{-\frac{a+d+1}{2}+b+c-1}\ \exp[-uv(1-v)]\ .
$$
Provided $-\frac{a+d+1}{2}+b+c>0$, we can integrate on $u$ so that
$$
W=\Gamma\left(-\frac{a+d+1}{2}+b+c\right)\times \int_{0}^{1}dv\ 
v^{\frac{a+d+1}{2}-c-1}(1-v)^{\frac{a+d+1}{2}-b-1}\ .
$$
Of course, the last beta integral gives a ratio of Gamma functions, so in all 
$$
U:=\frac{\pi^{\frac{d}{2}}}{2}|x-y|^{a+1+d-2b-2c}
\frac{\Gamma\left(\frac{a+1}{2}\right)\Gamma\left(-\frac{a+d+1}{2}+b+c\right)
\Gamma\left(\frac{a+d+1}{2}-c\right)\Gamma\left(\frac{a+d+1}{2}-b\right)}{\Gamma\left(b\right)\Gamma\left(c\right)\Gamma\left(a+d+1-b-c\right)}
$$
which is valid when all the Gamma function arguments are positive (or complex with positive real part).
To see that these constraints are not contradictory, it is better to switch to the more symmetric scaling dimension variables
$$
\left\{\begin{array}{ccl}
\Delta_1 & = & b \\
\Delta_2 & = & c \\
\Delta_3 & = & a+d+1-b-c
\end{array}
\right.
$$
The constraints amount to $\Delta_1,\Delta_2,\Delta_3$ being the lengths of a nondegenerate triangle which is sufficiently big, i.e., $\Delta_1+\Delta_2+\Delta_3>d$.
I should also mention that there are more sophisticated techniques for computing Witten diagrams. See for instance the TASI lectures by Penedones. 
