# How to find $I=\int_{-4}^4\int_{-3}^3 \int_{-2}^2 \int_{-1}^1 \frac{x_1-x_2+x_3{-}x_4}{x_1+x_2+x_3+x_4} \, dx_1 \, dx_2 \, dx_3 \, dx_4$

How can I find this integral $$I=\int_{-4}^4\int_{-3}^3 \int_{-2}^2 \int_{-1}^1 \frac{x_1-x_2+x_3{-}x_4}{x_1+x_2+x_3+x_4} \, dx_1 \, dx_2 \, dx_3 \, dx_4$$

• Maple says undefined, calculating it numerically. Have you tried a change of variables $$u=x_1+x_2+x_3+x_4, v=x_1-x_2-x_3+x_4, w=x_3,z=x_4?$$ – user64494 Jun 30 '15 at 2:59
• Hint: $I=a\ln2+b\ln3+c\ln5$. – Lucian Jun 30 '15 at 3:04
• Mathematica outputs $$28 i \pi + 1536 \log2 - 144 \log3 -2000/5 \log5$$ – user64494 Jun 30 '15 at 3:07
• Have you tried $y_1=x_1+x_2$, $y2_=x_3+x_4$, $y_3=x_1-x_2$, $y_4=x_3-x_4$? – Michael Galuza Jun 30 '15 at 3:31
• Is there any reason to believe that it converges? You have non-isolated singularities in the domain (the entirety of the plane $x_1=-x_2, x_3=-x_4$) so at best you can look at PVs. – Steven Stadnicki Jun 30 '15 at 3:56

I couldn't resist as if your limits were same for each variable say varying $a$ to $b$, for any $a$ and $b$ , the following trick I saw somewhere, is applicable, but in your case, there is no short cut, I guess.
$$I=\int_{a}^b\int_{a}^b \int_{a}^b \int_{a}^b \frac{x_1-x_2+x_3{-}x_4}{x_1+x_2+x_3+x_4}dx_1dx_2dx_3dx_4 = \int_{a}^b\int_{a}^b \int_{a}^b \int_{a}^b \frac{-x_1-x_2+x_3+x_4}{x_1+x_2+x_3+x_4}dx_1dx_2dx_3dx_4=\int_{a}^b\int_{a}^b \int_{a}^b \int_{a}^b \frac{x_1-x_2-x_3+x_4}{x_1+x_2+x_3+x_4}dx_1dx_2dx_3dx_4 = \int_{a}^b\int_{a}^b \int_{a}^b \int_{a}^b \frac{-x_1+x_2{-}x_3+x_4}{x_1+x_2+x_3+x_4}dx_1dx_2dx_3dx_4=\int_{a}^b\int_{a}^b \int_{a}^b \int_{a}^b \int_{-a}^a \frac{x_1+x_2-x_3-x_4}{x_1+x_2+x_3+x_4}dx_1dx_2dx_3dx_4 = \int_{a}^b\int_{a}^b \int_{a}^b \int_{a}^b \frac{-x_1+x_2+x_3-x_4}{x_1+x_2+x_3+x_4}dx_1dx_2dx_3dx_4$$
Adding them all gives $6I=0 \implies I=0$