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  How can we reproduce the exact solution of the integral of the splitting function, which play a role in the dipole factorisation formulae?

+ 4 like - 0 dislike

I have a problem performing the following integration provided in the paper by Catani and Seymour (https://arxiv.org/abs/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



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


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)}.


\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)

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\\

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

& \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.

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

asked Oct 27, 2018 in Mathematics by Schnarco (20 points) [ revision history ]
edited Oct 28, 2018 by Schnarco

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