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I would like to to express the result of my integration just in terms of Gamma functions. The following integral is at hand:

$$ \int_0^1dz\int_0^1dy(z(1-z))^{-\epsilon}(1-y)^{1-2\epsilon}y^{-1-\epsilon}(z(1-z)-2) $$

yielding the result with mathematica

$$ \frac{\sqrt{\pi}2^{2\epsilon-3}(7\epsilon-11)\Gamma(2-2\epsilon)\Gamma(1-\epsilon)\Gamma(-\epsilon)}{\Gamma(2-3\epsilon)\Gamma(\frac{5}{2}-\epsilon)}, $$

where one can rewrite the $\sqrt{\pi}2^{2\epsilon-3}$ in terms of Gamma functions. By employing

Series[%, {\[Epsilon], 0, 0}]

we obtain

$$ \frac{11}{6\epsilon}+\frac{1}{6}\left(4+11\gamma+22\log(2)+11\psi^{(0)}\left(\frac{5}{2}\right)\right)+\mathcal{O}(\epsilon^1). $$

Is it possible to express the result of the integration in terms of Gamma functions or to express the last line as $\frac{11}{6\epsilon}+\frac{50}{9}+\mathcal{O}(\epsilon^1)$?

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Use FullSimplify with Series

(int = Assuming[Re[ϵ] < 1, Integrate[
    (z (1 - z))^-ϵ (1 - y)^(1 - 2 ϵ) y^(-1 - ϵ) (z (1 - z) - 2), 
    {z, 0, 1}, {y, 0, 1}] // Simplify]) // TraditionalForm

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Series[int, {ϵ, 0, 0}] // FullSimplify

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Series[int, {ϵ, 0, 1}] // FullSimplify

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EDIT: Note that since

Gamma[1/2]

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then

rule = Gamma[1/2] -> 8/15*Inactive[Gamma][7/2]

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(int2 = int /. rule) // TraditionalForm

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int == int2 // Activate

(* True *)
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