I want to try and potentially use a Dirichlet - Hypergeometric Function in order to compute the following integral. I would appreciate some help as I'm stuck on how to go about this is a multi-dimensional sense:
$$ \int_{0}^{\infty} \frac{(1+g)^{\delta - a - 1}}{\left(\prod_{i=1}^{q} 1 - g \lambda_{i} \right)^{\delta + n}} \; dg $$
Assume that $\delta \in \mathbb{R}^{+}, a \in [2,\infty) \subset \mathbb{R}, n \in \mathbb{N}, \lambda_i \in \mathbb{R} \; \forall i $ .
Furthermore, I'm not really interested in the $\lambda_i$s since they are deterministic constants. So if it is easier to write the denominator of the fraction of the integrand as:
$$\left(1 - \sum_{i=1}^{q} g^{i} \phi_{i} \right)^{\delta + n}$$ For some new set of constants $ \{\ \phi_{i} \}\ _{i=1}^{q}$, then this form can be used too.
Thanks in advance!
Hint:
$\int_0^\infty\dfrac{(1+g)^{\delta-a-1}}{\left(\prod\limits_{i=1}^q(1-g\lambda_i)\right)^{\delta+n}}dg$
$=\int_1^\infty\dfrac{g^{\delta-a-1}}{\prod\limits_{i=1}^q(1-(g-1)\lambda_i)^{\delta+n}}d(g-1)$
$=\int_1^\infty\dfrac{g^{\delta-a-1}}{\prod\limits_{i=1}^q(\lambda_i+1-\lambda_ig)^{\delta+n}}dg$
$=\int_1^0\dfrac{1}{g^{\delta-a-1}\prod\limits_{i=1}^q\left(\lambda_i+1-\dfrac{\lambda_i}{g}\right)^{\delta+n}}d\left(\dfrac{1}{g}\right)$
$=\int_0^1\dfrac{1}{g^{(1-q)\delta-a-nq+1}\prod\limits_{i=1}^q((\lambda_i+1)g-\lambda_i)^{\delta+n}}dg$
$=\dfrac{1}{\prod\limits_{i=1}^q\lambda_i^{\delta+n}}\int_0^1g^{(q-1)\delta+a+nq-1}\prod\limits_{i=1}^q\left(\left(\dfrac{1}{\lambda_i}+1\right)g-1\right)^{-\delta-n}dg$
which can express in terms of the Lauricella Functions