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Recently I came across the following problem: I know the exact result of the integral

$$ \mathcal{V}=\int_0^1 (z(1-z))^{-\epsilon} \int_0^1 (1-y)^{1-2\epsilon}y^{-1-\epsilon}\left[-(1+z)-\epsilon(1-z)\right] dydz $$

which leads to

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

It is easy to obtain using the Euler beta function. Instead of performing the exact calculation and expanding in $\epsilon$ afterwards, I wanted to make use of a method, that is often employed in theoretical physics. The idea reads as follows: Consider an integral of the form

$$ \int_0^1\int_0^1 \frac{F(x,y)}{x^{1+\epsilon}y^{1+\epsilon}}dxdy $$

where $F(x,y)$ donotes a funtion that behaves non-singular in the integration domain (for $\epsilon=0$), such as in our explicit example. We can rewrite it as

\begin{align} \int_0^1\int_0^1 \frac{F(x,y)}{x^{1+\epsilon}y^{1+\epsilon}}dxdy &= \int_0^1\int_0^1 \frac{F(x,y)-F(x,0)-F(0,y)+F(0,0)}{x^{1+\epsilon}y^{1+\epsilon}}dxdy \\ &+\int_0^1\int_0^1 \frac{F(x,0)-F(0,0)}{x^{1+\epsilon}y^{1+\epsilon}}dxdy \\ &+\int_0^1\int_0^1 \frac{F(0,y)-F(0,0)}{x^{1+\epsilon}y^{1+\epsilon}}dxdy \\ &+\int_0^1\int_0^1 \frac{F(0,0)}{x^{1+\epsilon}y^{1+\epsilon}}dxdy. \end{align}

For the first summand we expand the integrand in $\epsilon$ up to order $\epsilon^0$ and integrate. For the second summand we expand everything but $\frac{1}{y^{1+\epsilon}}$ up to order $\epsilon^0$ and integrate. For the third summand we expand everything but $\frac{1}{x^{1+\epsilon}}$ up to order $\epsilon^0$ and integrate. For the fourth summand we expand just $F(0,0)$ up to order $\epsilon^0$ and integrate.

Unfortunately what I get is

$$ \mathcal{V}=\frac{3}{2\epsilon}+\frac{3}{2}+\mathcal{O}(\epsilon) $$

and not the desired result. Any suggestions how to improve the method? By the way, does anyone know the name of the method described above?

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