You could teach Mathematica about the poles of $\Gamma$ (it can already compute the residues). This is done with a "divisor" object in mathematics, defined to be an integral linear combination of the zeros and poles (positive for the zeros, negative for the poles). The following implementation computes its coefficients for products and quotients of Gamma functions. It really only needs to know that $\Gamma$ has simple poles at all non-positive integers (which is on the first line of the definition); the rest tells it how to decompose the products and powers (which includes quotients, which are $-1$ powers):
divisor[Gamma[x_]] := -Boole[x <= 0 && x \[Element] Integers];
divisor[Times[x_, y__]] := divisor[Times[x]] + divisor[Times[y]];
divisor[Power[x_Gamma, n_Integer]] := n divisor[x];
divisor[x_] := 0 (* Everything else, for now *)
Let's encapsulate the right hand side of the integral equation in the question:
f[d_, n1_, n2_] := Gamma[d/2 - n1] Gamma[d/2 - n2] Gamma[n1 + n2 - d/2] /
(Gamma[n1] Gamma[n2] Gamma[d/2 - n1 - n2])
I will assume that $\nu_1$ and $\nu_2$ are positive integers in the following calculation, which characterizes the points where $f$ might have a pole:
FullSimplify[Reduce[divisor[f[d, n1, n2]] < 0, d],
Assumptions -> n1 \[Element] Integers && n2 \[Element] Integers && n1 > 0 && n2 > 0]
$\left(-\frac{d}{2}\in \text{Integers}\&\&d>2 (\text{n1}+\text{n2})\right)\|\left(\left(\frac{d}{2}\left|\frac{d}{2}\right|\frac{d}{2}\right)\in \text{Integers}\&\&(d\leq 2 \text{n1}\|\text{n1}>\text{n2})\&\&(d\leq 2 \text{n2}\|\text{n1}\leq \text{n2})\right)$
Although that's a little redundant, it's readable: there are indeed poles wherever $d/2$ is integral and does not exceed the larger of $\nu_1$ and $\nu_2$ or exceeds their sum. We may compute the residues at such points using Residue
. Here is a bunch of them computed at once for a particular choice of $\nu_1$ and $\nu_2$ to illustrate the previous result:
With[{n1 = 2, n2 = 1}, Residue[f[d, n1, n2], {d, #}] & /@ Range[-2, 8]]
$\{-24, 0, 12, 0, -4, 0, 0, 0, 0, 4\}$
If other assumptions need to be made about $d$, $\nu_1$, or $\nu_2$, modify the assumptions in Reduce
and FullSimplify
accordingly.
Series
? $\endgroup$d
simply means $d=\epsilon+3$ and letting $\epsilon\rightarrow 3$. $\endgroup$