Problem with Mellin Barnes type integral Using the Mellin Barnes technique for a certain Feynman integral, I arrive at
$$
  I=
  \frac1{2\pi i}
  \int\limits_{-i\infty}^{i\infty} dz\;
  \Gamma^4\left(\frac12 + z\right)
  \Gamma^4\left(\frac12 - z\right)
  \psi\left(\frac12 - z\right)\,,
$$
where $\psi(x)$ is the digamma-function. This integral evaluates to
$-11.57972$ numerically (Mathematica).
The standard way to solve an integral of this type would be to close the
integration contour in the left or right complex halfplane and sum up the
residues. This leads to
$$
  I =
  -\sum\limits_{n=0}^\infty
  \left(
    \frac{2\pi^2}3 \psi^{(1)}(1+n) +
    \frac16 \psi^{(3)}(1+n)
  \right)\,,
$$
where $\psi^{(m)}(x)$ is the polygamma function of order $m$. Unfortunately this series doesn't converge.
So my questions are: Why does it fail? Is there a way to solve this integral?
Thanks!
 A: The integral can be solved using an integral representation of the
digamma function
$$
  \psi(x) = \int\limits_0^\infty dt
    \left(
      \frac{e^{-t}}t -
      \frac{e^{-xt}}{1-e^{-t}}
    \right)\,.
$$
Euler's reflection formula and the substitution $z\rightarrow ix$ lead to
$$
  I = \frac{\pi^3}{2} \int\limits_0^\infty dt 
      \int\limits_{-\infty}^{\infty} dx\;
    \left(
      \frac{e^{-t}}{t\cosh^4(\pi x)}  -
      \frac{e^{-t/2}}{1-e^{-t}} \frac{e^{ixt}}{\cosh^4(\pi x)}
    \right)\,.
$$
As the imaginary part is odd in $x$, only the real part of $e^{ixt}$ remains:
$$
  I = \frac{\pi^3}{2} \int\limits_0^\infty dt 
      \int\limits_{-\infty}^{\infty} dx\;
    \left(
      \frac{e^{-t}}{t\cosh^4(\pi x)}  -
      \frac{e^{-t/2}}{1-e^{-t}} \frac{\cos(xt)}{\cosh^4(\pi x)}
    \right)\,.
$$
The $x$-integration yields
$$
  I = \int\limits_0^\infty dt 
    \left(
      \frac{2\pi^2}{3} \frac{e^{-t}}{t}  -
      \frac{e^{-t/2}}{1-e^{-t}} \frac{t(4\pi^2+t^2)}{12\sinh\left(\frac t2\right)}
    \right)\,.
$$
And finally the $t$-integration yields
$$
  I = -\frac{2\pi^2}3 - \frac{2\gamma_E \pi^2}{3} - \zeta(3) \approx -11.5797\,.
$$
A: You can use Euler's reflection formula to transform your integrand to 
$$\frac{\pi^4}{\sin(\pi(\frac12 - z)} \psi(\frac12 - z) = \frac{\pi^4 \psi(\frac12-z)}{\cos^4 \pi z}.$$ This is not quite an answer, but should make it easier to compute the residues.
