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?
The condition $Re(a+b)$ for equation (1) ensures the convergence of the integral if it is performed along a line with $-a<c<b$. To get the value of the integral with $a=-5/2$, it is necessary to follow another integration path that goes around some poles of the gamma functions. The path you have to select depends on $A$. For instance with $A(p)=1$, you will find in the table of inverse Mellin transforms given in Erdélyi, Magnus, Oberhettinger, Tricomi, Table of integrals transforms, the equation that you need, number 7.3(16) (chapter 7). For $0<c<1/2$, the formula should be used with $\alpha=-5/2$, $\beta=0$, $h=3$ and $k=1$ and the inverse Mellin transform provides the identity: $$ \frac1{2\pi\mathrm{i}} \int_{c-\mathrm{i}\infty}^{c+\mathrm{i}\infty} \Gamma(-s)\Gamma\left(s-\frac 52\right)z^{-s}\,\mathrm{d}s=\\ \Gamma\left(-\frac52\right)z^{-5/2}(1+z)^{5/2}-\Gamma\left(-\frac52\right)z^{-5/2}+\Gamma\left(-\frac32\right)z^{-3/2}-\frac12\Gamma\left(-\frac12\right)z^{-1/2}-\Gamma\left(-\frac52\right).$$ The integers $h$ and $k$ have to be chosen such that $0<a+h+c<1$, $0<b+k-c<1$. This because the poles of the gamma functions are the negative integers: $h$ and $k$ count how many poles you have to go around for $\Gamma(s-5/2)$ and $\Gamma(-s)$ respectively.
So to compute $I$ in the case $A(p)\neq 1$, you need to know which domain you want to use, it will give you the poles you have to take into account.