I would like to determine a closed-form expression for the following symbolic integral
$$ \int_{-1/2}^{1/2} \!\!\!\! \mathrm{d} x \int_{-1/2}^{1/2} \!\!\!\! \mathrm{d} y \, \frac{1 + b x + c y}{1 + e x + f y + i \eta} \, ,$$
where the coefficients appearing in this expression are such that
$$ \begin{cases} \displaystyle b,c,e,f \in \mathbb{R} \, , \\ \displaystyle \eta \in \mathbb{R}\, , \, \eta \neq 0 \, . \end{cases} $$
As a consequence, this integral is always well-defined since its denominator is never equal to 0. To compute this integral in Mathematica, I used the instruction :
int = Integrate[(1 + b x + c y)/(1 + e x + f y + I \[Eta]), {x, -1/2,1/2}, {y, -1/2, 1/2}, Assumptions -> {b \[Element] Reals, c \[Element] Reals, e \[Element] Reals, f \[Element] Reals, \[Eta] \[Element] Reals, \[Eta] != 0}]
After a quite long calculation (260s) on my computer with Mathematica 10.0.2, I obtained a result of the form
int = ConditionalExpression[......,(f<-2 && (2+f<e<0 || (e>0 && 2+e+f<0))) || (-2<f<0 && ((2+e+f>0 && e<0) || 0<e<2+f)) || (0<f<2 && (-2+f<e<0 || (e>0 && e+f<2))) || (f>2 && ((e+f>2 && e<0) || 0<e<-2+f))]]
Why do I obtain a result with a ConditionalExpression, while my integral should always be correctly defined ? How should I proceed to convince Integrate
that this integral is always well-defined ?
I noticed that the result inside the ConditionalExpression
involves terms of the form ArcTan[(2 \[Eta])/(-2+e+f)]
, which partially explain the constraints obtained.
Another issue is that the result inside the ConditionalExpression also involves complex logarithmic terms of the form Log[2-e-f+2 I \[Eta]]
. I know that the complex logarithm has a branch-cut so that the result may not be robust when being evaluated.
Would it also be possible to obtain a result that would not involve expression with branch-cuts, so that the results would always be straightforwardly well defined ?