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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

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

with

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

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

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

Hence

\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?

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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 $$I = \frac{1}{\epsilon^2} -\frac{\pi^2}{2} + O(\epsilon).$$

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  • $\begingroup$ 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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