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

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  • $\begingroup$ Maple says undefined, calculating it numerically. Have you tried a change of variables $$u=x_1+x_2+x_3+x_4, v=x_1-x_2-x_3+x_4, w=x_3,z=x_4?$$ $\endgroup$
    – user64494
    Jun 30, 2015 at 2:59
  • $\begingroup$ Hint: $I=a\ln2+b\ln3+c\ln5$. $\endgroup$
    – Lucian
    Jun 30, 2015 at 3:04
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    $\begingroup$ Mathematica outputs $$28 i \pi + 1536 \log2 - 144 \log3 -2000/5 \log5$$ $\endgroup$
    – user64494
    Jun 30, 2015 at 3:07
  • $\begingroup$ Have you tried $y_1=x_1+x_2$, $y2_=x_3+x_4$, $y_3=x_1-x_2$, $y_4=x_3-x_4$? $\endgroup$
    – Michael Galuza
    Jun 30, 2015 at 3:31
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    $\begingroup$ 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. $\endgroup$
    – Steven Stadnicki
    Jun 30, 2015 at 3:56

1 Answer 1

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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.

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  • $\begingroup$ How about the convergence of the integrals? The above is in Euler's style. $\endgroup$
    – user64494
    Jun 30, 2015 at 3:27
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    $\begingroup$ I never worry about convergence until a singularity hits me on the head. $\endgroup$
    – marty cohen
    Jun 30, 2015 at 4:43
  • $\begingroup$ @marty cohen: Then all that is built on sand. $\endgroup$
    – user64494
    Jun 30, 2015 at 5:26
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    $\begingroup$ 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. $\endgroup$
    – marty cohen
    Jun 30, 2015 at 14:02

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