I have a problem performing the following integration provided in the paper by Catani and Seymour (arXiv: hep-ph/9605323) page 27. Given is the integral

$$ \mathcal{V}=\int_0^1 (z(1-z))^{-\epsilon} \int_0^1 (1-y)^{1-2\epsilon}y^{-1-\epsilon}V(z;y) dydz $$


$$ V(z;y)=\frac{2}{1-z(1-y)}-(1+z)-\epsilon(1-z). $$

In the paper the exact result and an approximation is stated as follows:

$$ \mathcal{V}=\frac{\Gamma(1-\epsilon)^3}{\Gamma(1-3\epsilon)}\left[\frac{1}{\epsilon^2}+\frac{1}{\epsilon}\frac{3+\epsilon}{2(1-3\epsilon)}\right]=\frac{1}{\epsilon^2}+\frac{3}{2\epsilon}+5-\frac{\pi^2}{2}+\mathcal{O}(\epsilon). $$

To me it is obvious how to deal with the second and the third summand of $V$ under the integral. I simply expand and employ the definition of the Euler Beta function

$$ B(a,b)=\int_0^1 t^{a-1}(1-t)^{b-1} dt=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}. $$


\begin{align} \int_0^1 (z(1-z))^{-\epsilon} \int_0^1 (1-y)^{1-2\epsilon}y^{-1-\epsilon}\left[-(1+z)-\epsilon(1-z)\right] dydz & = \frac{\Gamma(1-\epsilon)^3}{\Gamma(1-3\epsilon)}\left[\frac{1}{\epsilon}\frac{3+\epsilon}{2(1-3\epsilon)}\right] \\ & = \frac{3}{2\epsilon}+5+\mathcal{O}(\epsilon) \end{align}

for the simple part. Do you have any suggestions how to treat the singular term of $V$ in order to get the exact result? In particular how to solve

$$ \int_0^1 (z(1-z))^{-\epsilon} \int_0^1 (1-y)^{1-2\epsilon}y^{-1-\epsilon}\left[\frac{2}{1-z(1-y)}\right] dydz=\frac{\Gamma(1-\epsilon)^3}{\Gamma(1-3\epsilon)}\left[\frac{1}{\epsilon^2}\right]=\frac{1}{\epsilon^2}-\frac{\pi^2}{2}+\mathcal{O}(\epsilon). $$

Thank you in advance.

Edit: After some effort in reverse engineering (basically rewriting the product of Gamma-functions in the exact result in terms of two Euler Beta functions) the initial problem boils down to showing the equality of

\begin{align} & \int_0^1 (z(1-z))^{-\epsilon} \int_0^1 y^{1-2\epsilon}(1-y)^{-1-\epsilon}\left[\frac{2}{1-zy}\right] dydz \\ &=\int_0^1 (t(1-t))^{-\epsilon} \int_0^1 s^{1-2\epsilon}(1-s)^{-1-\epsilon}\left[\frac{(1-s)}{t(1-t)s^2}\right] dsdt. \end{align}

I guess the solution is an integral transformation. Does anyone know how to transform in order to show the equality of the upper expressions?


I'm not sure the extent to which you would like a complete analytical solution to the integral but since the integrations over $y$ and $z$ decouple you can proceed by identifying, say the integration over $z$, as a Hypergeometric2F1 which has the following integral representation $$_2F_1(a,b,c;t) = \frac{\Gamma(c)}{\Gamma(b) \Gamma(c-b)} \int_0^1 \mathrm{d}z \,z^{b-1} (1-z)^{c-b-1} (1-zt)^{-a}\,\,\,\text{for}\,\,\,\text{Re}(c)> \text{Re}(b) > 0.$$ Let $I$ be the integral of interest. In your case, with $t = 1-y$, you have $$I = \frac{2\Gamma(1-\epsilon)^2}{\Gamma(2-2\epsilon)} \int_0^1 \mathrm{d}y\, (1-y)^{1-2\epsilon}y^{-1-\epsilon}\, _2F_1(1,1-\epsilon,2-2\epsilon;1-y).$$ Now, as $|t|<1$ for all $y \in (0,1)$, one is within the radius of convergence to permit use of the power series representation for the Hypergeometric2F1 in terms of so called Pochhammer symbols $$_2F_1(a,b,c;t) = \sum_{n=0}^{\infty} \frac{(a)_n (b)_n}{(c)_n} \frac{t^n}{n!},$$ where $(a)_n = \Gamma(a+n)/\Gamma(a)$ with $(1)_n := n!$. It follows $$I = 2\Gamma(1-\epsilon) \sum_{n=0}^{\infty} \frac{\Gamma(1-\epsilon+n)}{\Gamma(2-2\epsilon+n)} \int_0^1 \mathrm{d}y\, (1-y)^{1-2\epsilon+n}\,y^{-1-\epsilon}.$$ The integral over $y$ is an Euler Beta type so you can readily write $$I = 2\Gamma(1-\epsilon)\Gamma(-\epsilon) \sum_{n=0}^{\infty}\frac{\Gamma(1-\epsilon+n)}{\Gamma(2-2\epsilon+n)}\frac{\Gamma(2-2\epsilon+n)}{\Gamma(2+n-3\epsilon)} = 2\Gamma(1-\epsilon)\Gamma(-\epsilon) \sum_{n=0}^{\infty}\frac{\Gamma(1-\epsilon+n)}{\Gamma(2+n-3\epsilon)}$$

At this point the explicit summation over $n$ can be input into say mma, returning a combination of Gamma functions containing another $\Gamma(-\epsilon)$ which overlaps with the $\Gamma(-\epsilon)$ already present to provide the double pole in $\epsilon$. Indeed you find $$I = \frac{1}{\epsilon^2} -\frac{\pi^2}{2} + O(\epsilon).$$

  • $\begingroup$ Thanks a lot for your helpful answer! Nevertheless I was wondering whether or not there exists a way to evaluate the last sum of Gamma functions by hand. $\endgroup$ – Schnarco Oct 29 '18 at 13:13
  • $\begingroup$ Yes I spent a bit of time wondering too, trying to play with the various relations amongst gamma functions you can find in e.g Wikipedia but could not find a way. Regardless, I doubt the authors would have started to do the integral by hand if it can be solved by computer in the first instance. $\endgroup$ – CAF Oct 29 '18 at 17:38
  • $\begingroup$ This one is good. Definition and properties of the Beta function: $$ \frac{\Gamma(a+n)}{\Gamma(b+n)}=\frac{1}{\Gamma(b-a)}B(a+n,b-a) = \frac{1}{\Gamma(b-a)}\int_{0}^{1}(1-x)^{b-a-1}x^{a+n-1}\,dx. $$ If you sum both sides on $n\geq 0$, you end up with: $$ \sum_{n\geq 0}\frac{\Gamma(a+n)}{\Gamma(b+n)}=\frac{1}{\Gamma(b-a)}\int_{0}^{1}(1-x)^{b-a-2}x^{a-1}\,dx=\frac{B(a,b-a-1)}{\Gamma(b-a)}=\frac{\Gamma(a)\Gamma(b-a-1)}{\Gamma(b-1)\Gamma(b-a)}. $$ $\endgroup$ – Schnarco Oct 30 '18 at 18:49
  • $\begingroup$ Nice! So you have a complete analytical solution to your question now. $\endgroup$ – CAF Oct 30 '18 at 20:45

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