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?


  • $\begingroup$ Does the integral vanish (or is bounded) along the contour in the L/R half-plane as, say, the radius of the contour goes to infinity? Hard for me to tell one way or another, but I would guess no. $\endgroup$ – Ron Gordon Dec 23 '13 at 16:35
  • $\begingroup$ I think it does vanish, but I don't know how to proof it. I parameterized an arc in the half-plane and evaluated the integral with a finite radius numerically. The contribution from the arc decreases, if the radius is increased. So I guess it will vanish, if the radius goes to infinity. $\endgroup$ – mario Dec 23 '13 at 17:18
  • $\begingroup$ Perhaps you could share what you know specifically about that in the question. $\endgroup$ – Ron Gordon Dec 23 '13 at 17:19

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 $$ I = -\frac{2\pi^2}3 - \frac{2\gamma_E \pi^2}{3} - \zeta(3) \approx -11.5797\,. $$


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.


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