Can You Solve These Tricky 2D Integrals on a Unit Circle?

[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.

 

[Mark44]

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.

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

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.

 

[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...

 

[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

 

[Mark44]

\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

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.

 

[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.

 

[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.

 

[GFauxPas]

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

 

[presto]

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

 

[Saitama]

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

 

[presto]

I told that already:

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.

Therefore the disk mass density is just: rho(r) = 2arccosr/r, so a force at a point (a,0):

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
...