Symbolic integration of potential over a disc : branch cut problem?
$begingroup$
Context
I am trying to explore the geometry of a crystal made of irregular bubbles.
See animation here.
very vaguely in the spirit of this post (it is in fact motivated by cosmology and galaxy formation).
So I give myself an interaction potential (which is both attractive and repulsive at large and small distances resp.)
pot[r_] = 1/r^2 + r^2
looking like this
Plot[pot[r], {r, 0.1, 5}]
and I integrate it over a Disk
int= Integrate[ pot[Sqrt[(x - x0)^2 + (y - y0)^2]], {x0, y0} ∈
Disk[{0, 0}, 1]]
(* π (x^2+y^2+1/2) *)
which incidentally looks suspicious, because it is lacking a repulsion near the disc.
But if I take a specific value for {x,y}
rxy = Thread[{x, y} -> {2, 3}]
and carry out the integration numerically
NIntegrate[
pot[Sqrt[(x - x0)^2 + (y - y0)^2]] /. rxy, {x0, y0} ∈
Disk[{0, 0}, 1], PrecisionGoal -> 6]
(* 42.663 *)
I get a different answer from
int /. rxy
(* 42.4115 *)
Indeed if I do the replacement First
Integrate[pot[Sqrt[(x - x0)^2 + (y - y0)^2]] /. rxy, {x0, y0} ∈ Disk[{0, 0}, 1]]
(* π (27/2+log(13/12)) *)
N[%]
(* 42.663 *)
So mathematica seems to be doing the general integration wrong.
Questions
Is this a bug? Any workaround?
Check
Indeed I can check by integrating numerically radially away from the edge of the disk that the potential generated by the disc is repulsive at close distance:
dat = ParallelTable[
NIntegrate[
pot[Sqrt[(x - x0)^2 + (y - y0)^2]] /. {x -> r Cos[t],
y -> r Sin[t]} /. t -> Pi/4, {x0, y0}∈
Disk[{0, 0}, 1], PrecisionGoal -> 8],
{r, 1.01, 2, 0.025}];
dat // ListLinePlot
calculus-and-analysis numerical-integration bugs symbolic
asked Nov 28 '18 at 7:06
chrischris
12.4k442113
$endgroup$
add a comment |
$begingroup$
Context
I am trying to explore the geometry of a crystal made of irregular bubbles.
See animation here.
very vaguely in the spirit of this post (it is in fact motivated by cosmology and galaxy formation).
So I give myself an interaction potential (which is both attractive and repulsive at large and small distances resp.)
pot[r_] = 1/r^2 + r^2
looking like this
Plot[pot[r], {r, 0.1, 5}]
and I integrate it over a Disk
int= Integrate[ pot[Sqrt[(x - x0)^2 + (y - y0)^2]], {x0, y0} ∈
Disk[{0, 0}, 1]]
(* π (x^2+y^2+1/2) *)
which incidentally looks suspicious, because it is lacking a repulsion near the disc.
But if I take a specific value for {x,y}
rxy = Thread[{x, y} -> {2, 3}]
and carry out the integration numerically
NIntegrate[
pot[Sqrt[(x - x0)^2 + (y - y0)^2]] /. rxy, {x0, y0} ∈
Disk[{0, 0}, 1], PrecisionGoal -> 6]
(* 42.663 *)
I get a different answer from
int /. rxy
(* 42.4115 *)
Indeed if I do the replacement First
Integrate[pot[Sqrt[(x - x0)^2 + (y - y0)^2]] /. rxy, {x0, y0} ∈ Disk[{0, 0}, 1]]
(* π (27/2+log(13/12)) *)
N[%]
(* 42.663 *)
So mathematica seems to be doing the general integration wrong.
Questions
Is this a bug? Any workaround?
Check
Indeed I can check by integrating numerically radially away from the edge of the disk that the potential generated by the disc is repulsive at close distance:
dat = ParallelTable[
NIntegrate[
pot[Sqrt[(x - x0)^2 + (y - y0)^2]] /. {x -> r Cos[t],
y -> r Sin[t]} /. t -> Pi/4, {x0, y0}∈
Disk[{0, 0}, 1], PrecisionGoal -> 8],
{r, 1.01, 2, 0.025}];
dat // ListLinePlot
calculus-and-analysis numerical-integration bugs symbolic
asked Nov 28 '18 at 7:06
chrischris
12.4k442113
$endgroup$
2
$begingroup$
Indeed,Integrateappears to have a problem with the repulsive part;Integrate[ 1/((x - x0)^2 + (y - y0)^2), {x0, y0} [Element] Disk[{0, 0}, 1]]returns0which is obviously wrong. I'd say, this is a bug. Please inform Wolfram Support.
$endgroup$
– Henrik Schumacher
Nov 28 '18 at 8:16
add a comment |
$begingroup$
Context
I am trying to explore the geometry of a crystal made of irregular bubbles.
See animation here.
very vaguely in the spirit of this post (it is in fact motivated by cosmology and galaxy formation).
So I give myself an interaction potential (which is both attractive and repulsive at large and small distances resp.)
pot[r_] = 1/r^2 + r^2
looking like this
Plot[pot[r], {r, 0.1, 5}]
and I integrate it over a Disk
int= Integrate[ pot[Sqrt[(x - x0)^2 + (y - y0)^2]], {x0, y0} ∈
Disk[{0, 0}, 1]]
(* π (x^2+y^2+1/2) *)
which incidentally looks suspicious, because it is lacking a repulsion near the disc.
But if I take a specific value for {x,y}
rxy = Thread[{x, y} -> {2, 3}]
and carry out the integration numerically
NIntegrate[
pot[Sqrt[(x - x0)^2 + (y - y0)^2]] /. rxy, {x0, y0} ∈
Disk[{0, 0}, 1], PrecisionGoal -> 6]
(* 42.663 *)
I get a different answer from
int /. rxy
(* 42.4115 *)
Indeed if I do the replacement First
Integrate[pot[Sqrt[(x - x0)^2 + (y - y0)^2]] /. rxy, {x0, y0} ∈ Disk[{0, 0}, 1]]
(* π (27/2+log(13/12)) *)
N[%]
(* 42.663 *)
So mathematica seems to be doing the general integration wrong.
Questions
Is this a bug? Any workaround?
Check
Indeed I can check by integrating numerically radially away from the edge of the disk that the potential generated by the disc is repulsive at close distance:
dat = ParallelTable[
NIntegrate[
pot[Sqrt[(x - x0)^2 + (y - y0)^2]] /. {x -> r Cos[t],
y -> r Sin[t]} /. t -> Pi/4, {x0, y0}∈
Disk[{0, 0}, 1], PrecisionGoal -> 8],
{r, 1.01, 2, 0.025}];
dat // ListLinePlot
calculus-and-analysis numerical-integration bugs symbolic
asked Nov 28 '18 at 7:06
chrischris
12.4k442113
$endgroup$
Context
I am trying to explore the geometry of a crystal made of irregular bubbles.
See animation here.
very vaguely in the spirit of this post (it is in fact motivated by cosmology and galaxy formation).
So I give myself an interaction potential (which is both attractive and repulsive at large and small distances resp.)
pot[r_] = 1/r^2 + r^2
looking like this
Plot[pot[r], {r, 0.1, 5}]
and I integrate it over a Disk
int= Integrate[ pot[Sqrt[(x - x0)^2 + (y - y0)^2]], {x0, y0} ∈
Disk[{0, 0}, 1]]
(* π (x^2+y^2+1/2) *)
which incidentally looks suspicious, because it is lacking a repulsion near the disc.
But if I take a specific value for {x,y}
rxy = Thread[{x, y} -> {2, 3}]
and carry out the integration numerically
NIntegrate[
pot[Sqrt[(x - x0)^2 + (y - y0)^2]] /. rxy, {x0, y0} ∈
Disk[{0, 0}, 1], PrecisionGoal -> 6]
(* 42.663 *)
I get a different answer from
int /. rxy
(* 42.4115 *)
Indeed if I do the replacement First
Integrate[pot[Sqrt[(x - x0)^2 + (y - y0)^2]] /. rxy, {x0, y0} ∈ Disk[{0, 0}, 1]]
(* π (27/2+log(13/12)) *)
N[%]
(* 42.663 *)
So mathematica seems to be doing the general integration wrong.
Questions
Is this a bug? Any workaround?
Check
Indeed I can check by integrating numerically radially away from the edge of the disk that the potential generated by the disc is repulsive at close distance:
dat = ParallelTable[
NIntegrate[
pot[Sqrt[(x - x0)^2 + (y - y0)^2]] /. {x -> r Cos[t],
y -> r Sin[t]} /. t -> Pi/4, {x0, y0}∈
Disk[{0, 0}, 1], PrecisionGoal -> 8],
{r, 1.01, 2, 0.025}];
dat // ListLinePlot
calculus-and-analysis numerical-integration bugs symbolic
calculus-and-analysis numerical-integration bugs symbolic
asked Nov 28 '18 at 7:06
chrischris
12.4k442113
asked Nov 28 '18 at 7:06
chrischris
12.4k442113
edited Nov 28 '18 at 8:19
chris
asked Nov 28 '18 at 7:06
chrischris
12.4k442113
asked Nov 28 '18 at 7:06
chrischris
12.4k442113
asked Nov 28 '18 at 7:06
chrischris
12.4k442113
12.4k442113
2
$begingroup$
Indeed,Integrateappears to have a problem with the repulsive part;Integrate[ 1/((x - x0)^2 + (y - y0)^2), {x0, y0} [Element] Disk[{0, 0}, 1]]returns0which is obviously wrong. I'd say, this is a bug. Please inform Wolfram Support.
$endgroup$
– Henrik Schumacher
Nov 28 '18 at 8:16
add a comment |
2
$begingroup$
Indeed,Integrateappears to have a problem with the repulsive part;Integrate[ 1/((x - x0)^2 + (y - y0)^2), {x0, y0} [Element] Disk[{0, 0}, 1]]returns0which is obviously wrong. I'd say, this is a bug. Please inform Wolfram Support.
$endgroup$
– Henrik Schumacher
Nov 28 '18 at 8:16
2
2
$begingroup$
Indeed,
Integrate appears to have a problem with the repulsive part; Integrate[ 1/((x - x0)^2 + (y - y0)^2), {x0, y0} [Element] Disk[{0, 0}, 1]] returns 0 which is obviously wrong. I'd say, this is a bug. Please inform Wolfram Support.$endgroup$
– Henrik Schumacher
Nov 28 '18 at 8:16
$begingroup$
Indeed,
Integrate appears to have a problem with the repulsive part; Integrate[ 1/((x - x0)^2 + (y - y0)^2), {x0, y0} [Element] Disk[{0, 0}, 1]] returns 0 which is obviously wrong. I'd say, this is a bug. Please inform Wolfram Support.$endgroup$
– Henrik Schumacher
Nov 28 '18 at 8:16
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
It's worth noting that the integrals will evaluate separately!
totalPot[x_]=Integrate[(x-x0)^2+(y0)^2,{x0,y0}∈ Disk[{0,0},1]]+
Integrate[1/((x-x0)^2+(y0)^2),{x0,y0}∈ Disk[{0,0},1],Assumptions->{x>1}];
N[totalPot[Sqrt[2^2 + 3^2]]]
(* 42.663 *)
The exact form of the potential being
$$frac{1}{2} pi left(2 r^2-2 log left(r^2-1right)+4 log (r)+1right)$$
Where I made sure to use the manifest rotational symmetry to put y=0, and also added an assumption that x is greater than 1 to avoid any issues with divergences in the 1/r^2 case.
Since it's of physical interest, to put units back in, if I take the potential to be an energy density $k_1 r^2+k_2/r^2$ and the disk is of radius $R$, I find:
$$E(r)=k_1 frac{pi}{2}(R^4+2 R^2 r^2)-k_2 pi log(1-frac{R^2}{r^2})$$
As noted by Henrik in the comments, this looks like a bug & should be reported to wolfram support.
edited Nov 28 '18 at 8:53
chris
12.4k442113
answered Nov 28 '18 at 8:12
DavidDavid
1135
$endgroup$
$begingroup$
thanks! Would you know how to do the integral over an elliptic disc?int= Integrate[ pot[Sqrt[(x - x0)^2 + (y - y0)^2]], {x0, y0} ∈ Disk[{0, 0}, {1,2}]]
$endgroup$
– chris
Nov 28 '18 at 8:31
$begingroup$
@chris hmm I don't. In 3D for the 1/r potential and an ellipsoid I know that the result is pretty complicated with no nice answer. But maybe it's easier here.
$endgroup$
– David
Nov 28 '18 at 10:33
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It's worth noting that the integrals will evaluate separately!
totalPot[x_]=Integrate[(x-x0)^2+(y0)^2,{x0,y0}∈ Disk[{0,0},1]]+
Integrate[1/((x-x0)^2+(y0)^2),{x0,y0}∈ Disk[{0,0},1],Assumptions->{x>1}];
N[totalPot[Sqrt[2^2 + 3^2]]]
(* 42.663 *)
The exact form of the potential being
$$frac{1}{2} pi left(2 r^2-2 log left(r^2-1right)+4 log (r)+1right)$$
Where I made sure to use the manifest rotational symmetry to put y=0, and also added an assumption that x is greater than 1 to avoid any issues with divergences in the 1/r^2 case.
Since it's of physical interest, to put units back in, if I take the potential to be an energy density $k_1 r^2+k_2/r^2$ and the disk is of radius $R$, I find:
$$E(r)=k_1 frac{pi}{2}(R^4+2 R^2 r^2)-k_2 pi log(1-frac{R^2}{r^2})$$
As noted by Henrik in the comments, this looks like a bug & should be reported to wolfram support.
edited Nov 28 '18 at 8:53
chris
12.4k442113
answered Nov 28 '18 at 8:12
DavidDavid
1135
$endgroup$
$begingroup$
thanks! Would you know how to do the integral over an elliptic disc?int= Integrate[ pot[Sqrt[(x - x0)^2 + (y - y0)^2]], {x0, y0} ∈ Disk[{0, 0}, {1,2}]]
$endgroup$
– chris
Nov 28 '18 at 8:31
$begingroup$
@chris hmm I don't. In 3D for the 1/r potential and an ellipsoid I know that the result is pretty complicated with no nice answer. But maybe it's easier here.
$endgroup$
– David
Nov 28 '18 at 10:33
add a comment |
$begingroup$
It's worth noting that the integrals will evaluate separately!
totalPot[x_]=Integrate[(x-x0)^2+(y0)^2,{x0,y0}∈ Disk[{0,0},1]]+
Integrate[1/((x-x0)^2+(y0)^2),{x0,y0}∈ Disk[{0,0},1],Assumptions->{x>1}];
N[totalPot[Sqrt[2^2 + 3^2]]]
(* 42.663 *)
The exact form of the potential being
$$frac{1}{2} pi left(2 r^2-2 log left(r^2-1right)+4 log (r)+1right)$$
Where I made sure to use the manifest rotational symmetry to put y=0, and also added an assumption that x is greater than 1 to avoid any issues with divergences in the 1/r^2 case.
Since it's of physical interest, to put units back in, if I take the potential to be an energy density $k_1 r^2+k_2/r^2$ and the disk is of radius $R$, I find:
$$E(r)=k_1 frac{pi}{2}(R^4+2 R^2 r^2)-k_2 pi log(1-frac{R^2}{r^2})$$
As noted by Henrik in the comments, this looks like a bug & should be reported to wolfram support.
edited Nov 28 '18 at 8:53
chris
12.4k442113
answered Nov 28 '18 at 8:12
DavidDavid
1135
$endgroup$
$begingroup$
thanks! Would you know how to do the integral over an elliptic disc?int= Integrate[ pot[Sqrt[(x - x0)^2 + (y - y0)^2]], {x0, y0} ∈ Disk[{0, 0}, {1,2}]]
$endgroup$
– chris
Nov 28 '18 at 8:31
$begingroup$
@chris hmm I don't. In 3D for the 1/r potential and an ellipsoid I know that the result is pretty complicated with no nice answer. But maybe it's easier here.
$endgroup$
– David
Nov 28 '18 at 10:33
add a comment |
$begingroup$
It's worth noting that the integrals will evaluate separately!
totalPot[x_]=Integrate[(x-x0)^2+(y0)^2,{x0,y0}∈ Disk[{0,0},1]]+
Integrate[1/((x-x0)^2+(y0)^2),{x0,y0}∈ Disk[{0,0},1],Assumptions->{x>1}];
N[totalPot[Sqrt[2^2 + 3^2]]]
(* 42.663 *)
The exact form of the potential being
$$frac{1}{2} pi left(2 r^2-2 log left(r^2-1right)+4 log (r)+1right)$$
Where I made sure to use the manifest rotational symmetry to put y=0, and also added an assumption that x is greater than 1 to avoid any issues with divergences in the 1/r^2 case.
Since it's of physical interest, to put units back in, if I take the potential to be an energy density $k_1 r^2+k_2/r^2$ and the disk is of radius $R$, I find:
$$E(r)=k_1 frac{pi}{2}(R^4+2 R^2 r^2)-k_2 pi log(1-frac{R^2}{r^2})$$
As noted by Henrik in the comments, this looks like a bug & should be reported to wolfram support.
edited Nov 28 '18 at 8:53
chris
12.4k442113
answered Nov 28 '18 at 8:12
DavidDavid
1135
$endgroup$
It's worth noting that the integrals will evaluate separately!
totalPot[x_]=Integrate[(x-x0)^2+(y0)^2,{x0,y0}∈ Disk[{0,0},1]]+
Integrate[1/((x-x0)^2+(y0)^2),{x0,y0}∈ Disk[{0,0},1],Assumptions->{x>1}];
N[totalPot[Sqrt[2^2 + 3^2]]]
(* 42.663 *)
The exact form of the potential being
$$frac{1}{2} pi left(2 r^2-2 log left(r^2-1right)+4 log (r)+1right)$$
Where I made sure to use the manifest rotational symmetry to put y=0, and also added an assumption that x is greater than 1 to avoid any issues with divergences in the 1/r^2 case.
Since it's of physical interest, to put units back in, if I take the potential to be an energy density $k_1 r^2+k_2/r^2$ and the disk is of radius $R$, I find:
$$E(r)=k_1 frac{pi}{2}(R^4+2 R^2 r^2)-k_2 pi log(1-frac{R^2}{r^2})$$
As noted by Henrik in the comments, this looks like a bug & should be reported to wolfram support.
edited Nov 28 '18 at 8:53
chris
12.4k442113
answered Nov 28 '18 at 8:12
DavidDavid
1135
edited Nov 28 '18 at 8:53
chris
12.4k442113
edited Nov 28 '18 at 8:53
chris
12.4k442113
edited Nov 28 '18 at 8:53
chris
12.4k442113
12.4k442113
answered Nov 28 '18 at 8:12
DavidDavid
1135
answered Nov 28 '18 at 8:12
DavidDavid
1135
answered Nov 28 '18 at 8:12
DavidDavid
1135
1135
$begingroup$
thanks! Would you know how to do the integral over an elliptic disc?int= Integrate[ pot[Sqrt[(x - x0)^2 + (y - y0)^2]], {x0, y0} ∈ Disk[{0, 0}, {1,2}]]
$endgroup$
– chris
Nov 28 '18 at 8:31
$begingroup$
@chris hmm I don't. In 3D for the 1/r potential and an ellipsoid I know that the result is pretty complicated with no nice answer. But maybe it's easier here.
$endgroup$
– David
Nov 28 '18 at 10:33
add a comment |
$begingroup$
thanks! Would you know how to do the integral over an elliptic disc?int= Integrate[ pot[Sqrt[(x - x0)^2 + (y - y0)^2]], {x0, y0} ∈ Disk[{0, 0}, {1,2}]]
$endgroup$
– chris
Nov 28 '18 at 8:31
$begingroup$
@chris hmm I don't. In 3D for the 1/r potential and an ellipsoid I know that the result is pretty complicated with no nice answer. But maybe it's easier here.
$endgroup$
– David
Nov 28 '18 at 10:33
$begingroup$
thanks! Would you know how to do the integral over an elliptic disc?
int= Integrate[ pot[Sqrt[(x - x0)^2 + (y - y0)^2]], {x0, y0} ∈ Disk[{0, 0}, {1,2}]]$endgroup$
– chris
Nov 28 '18 at 8:31
$begingroup$
thanks! Would you know how to do the integral over an elliptic disc?
int= Integrate[ pot[Sqrt[(x - x0)^2 + (y - y0)^2]], {x0, y0} ∈ Disk[{0, 0}, {1,2}]]$endgroup$
– chris
Nov 28 '18 at 8:31
$begingroup$
@chris hmm I don't. In 3D for the 1/r potential and an ellipsoid I know that the result is pretty complicated with no nice answer. But maybe it's easier here.
$endgroup$
– David
Nov 28 '18 at 10:33
$begingroup$
@chris hmm I don't. In 3D for the 1/r potential and an ellipsoid I know that the result is pretty complicated with no nice answer. But maybe it's easier here.
$endgroup$
– David
Nov 28 '18 at 10:33
add a comment |
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$begingroup$
Indeed,
Integrateappears to have a problem with the repulsive part;Integrate[ 1/((x - x0)^2 + (y - y0)^2), {x0, y0} [Element] Disk[{0, 0}, 1]]returns0which is obviously wrong. I'd say, this is a bug. Please inform Wolfram Support.$endgroup$
– Henrik Schumacher
Nov 28 '18 at 8:16