Hard integrals in 2D

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1. Oct 17, 2014

presto

I can't compute the integral:
$$\int \frac{\arccos(\sqrt{x^2+y^2})}{\sqrt{x^2+y^2}}\frac{x-a}/{(\sqrt{(x-1)^2+y^2})^3 dxdy$$
on an unit circle: r < 1.

for const: a = 0.01, 0.02, ect. up to 1 or 2.

I used a polar coordinates, but the values jump dramatically in some places (around the 'a' values), despite the function is quite smooth.

My intention is mainly this solution of a problem:
in this integral sits a singularity at a point (a, 0), and I'm looking for the effective method to eliminate it.

Greetings for everybody.

Last edited by a moderator: Oct 17, 2014
2. Oct 18, 2014

Staff: Mentor

I can't even tell what you're trying to integrate. Your LaTeX is pretty mangled.

3. Oct 18, 2014

presto

$$\int \frac{\arccos(\sqrt{x^2+y^2})}{\sqrt{x^2+y^2}}\frac{x-a}/{(\sqrt{(x-1)^2+y^2})^3 dxdy$$
on an unit circle: r < 1.

for const: a = 0.01, 0.02, ect. up to 1 or 2.

I don't konow what you latex use in the forum...

4. Oct 18, 2014

presto

$\int \frac{\arccos(\sqrt{x^2+y^2})}{\sqrt{x^2+y^2}}\frac{x-a}{(\sqrt{(x-1)^2+y^2})}^3 dxdy=\int_0^1 \int_0^{2\pi} \arccos(r)\frac{r\cos(f)-a}{(\sqrt{r^2-2ar\cos(f)+a^2)}^3} df dr$

5. Oct 18, 2014

Staff: Mentor

How did this integral come up? The reason I'm asking is that if this is an integral you wrote as part of another problem, it could be that your derivation is wrong.

6. Oct 18, 2014

presto

This is just a gravitational force of a disk with a mass density: rho(t) = 2arccos(r)/r.

I want to compare a gravity force of an isothermal sphere (density 1/r^2) to a disk, which we get by compressing the sphere to the plane.

7. Oct 21, 2014

presto

Maybe even could to calculate a simplified variant of the integral.

$\int_0^1 \int_0^{2\pi} \arccos(r)\frac{r\cos(f)-1}{(\sqrt{r^2-2r\cos(f)+1})^3} df dr$

I think that's equal to pi^2, but not sure.

8. Oct 21, 2014

GFauxPas

Maybe try a Weierstrass substitution?
$u = \tan (\frac f 2), \cos f = \frac {1-u^2}{1+u^2},\frac {\mathrm df}{\mathrm du} = \frac {2}{1 + u^2}$

9. Oct 21, 2014

presto

No, any such substitutions don't work here... I ask about a numerical solution of this integral.

10. Oct 22, 2014

Saitama

@presto: Can you post the exact problem which led you to the above integral?

11. Oct 22, 2014

presto

$dg = dm/d^2 cos(d,r) = \rho dS/d^2 \cos(d, r);\ where:\ cos(d,r) = (x-a)/d;\ and\ d^2 = (x-a)^2 + y^2; r^2 = x^2+y^2$