# 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

## 1 Answer

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 $6I=0 \implies I=0$

P.S- I know it doesn't answer your question, but it was too big/clumsy for a comment.

• How about the convergence of the integrals? The above is in Euler's style. – user64494 Jun 30 '15 at 3:27
• I never worry about convergence until a singularity hits me on the head. – marty cohen Jun 30 '15 at 4:43
• @marty cohen: Then all that is built on sand. – user64494 Jun 30 '15 at 5:26
• Ah, but take that sand, add water, and it turns into mud. Form that mud into blocks, let it dry in the sun, and it turns into bricks which can then be used to construct a magnificent building. – marty cohen Jun 30 '15 at 14:02