I need help solving the following integral:
$$I = \frac{1}{2 \pi i} \int_{c-i\infty}^{c+i\infty} \mathrm{d}p \hspace{2pt} m^{d-2p} \Gamma(-p)\Gamma(p-\frac{5}{2})A(p)$$
where $A(p)$ is an analytic function of $p$ everywhere. In an old thread there was a reference to the book "Assymptotics and Mellin-Barnes Integrals." On page 96 of that reference (in section 3.3.4 - Gamma Function Integrals) there is a potentially helpful formula:
\begin{equation}\frac{1}{2 \pi i} \int_{c-i\infty}^{c+i\infty} \mathrm{d}s \hspace{2pt} \Gamma(s+a)\Gamma(b-s)z^{-s} = z^a(1+z)^{-a-b}\Gamma(a+b) \hspace{1in} (1)\end{equation}
which is an application of Parseval's formula
$$\int_0^\infty f(x)g(x)\mathrm{d}x = \frac{1}{2 \pi i} \int_{c-i\infty}^{c+i\infty} F(1-s)G(s)ds$$
where F and G are the Mellin transforms. My $m^2$ translates to $z$ nicely. However, $(1)$ is only valid for $Re(a+b)>0$, so $b=0$, $a=-5/2$ wont work. Also, it doesn't say how do to deal with an extra function ($A(p)$), although I think I could solve that with some extra use of Parseval's formula (considering the product of Gamma's as one function, I'd just need a function that gives two Gamma's as its Mellin transform). Any ideas?