I want to calculate the integral $$ \int_0^1\int_0^1\int_0^1 {\rm d}r_1{\rm d}r_2{\rm d}r_3 \, r_1 r_2 r_3\int_0^\pi \int_0^\pi \int_0^\pi {\rm d}\phi_1{\rm d}\phi_2{\rm d}\phi_3 \\ \left| r_1r_2\sin(\phi_2-\phi_1) + r_2r_3\sin(\phi_3-\phi_2) + r_3r_1\sin(\phi_1-\phi_3)\right| $$ The problem obviously is the absolute value. I have tried for ages but don't get anywhere :-(

I tried to split the integral, but that leads to non-integrable terms, but maybe there is a trick using the symmetry?

I'm thinking to somehow transform the $\phi$ integrals and do them first.

The absolute value actually is two times the area of a triangle with vertices $P_1,P_2,P_3$ inside a semi-circle.

The transform mentioned by Yuriy seems a bit complicated: r1r2r3 to xyz transformation

  • 1
    $\begingroup$ If I'm not mistaken (it's hard to check numerically), we can make a substitution $x=r_1 r_2$, $y=r_2 r_3$, $z=r_3 r_1$, which gives us the same ranges as before, and simplifies the integrand: $$|a r_1 r_2+br_2 r_3 +c r_3 r_1| r_1 r_2 r_3 dr_1 dr_2 dr_3 = \frac{1}{2} |a x+b y +c z| dx dy dz$$ If I found the Jacobian correctly, of course $\endgroup$ – Yuriy S Oct 14 '18 at 1:13
  • 1
    $\begingroup$ @YuriyS Interesting observation. I would like to point out that the range of $x, y, z$ is a bit complicated since we must have $xy \leq z$, $yz \leq x$ and $zx \leq y$. $\endgroup$ – Sangchul Lee Oct 14 '18 at 1:40
  • $\begingroup$ @SangchulLee, oh, that explains why my numerical check failed $\endgroup$ – Yuriy S Oct 14 '18 at 8:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.