NIntegrate: How can I solve this integral numerically? NIntegrate fails while Integrate works
4
$begingroup$
I know the exact solution of the principal value of this integral is equal to zero:
$int_{-1}^{1}int_{-1}^{1}frac{x^2}{sqrt{1-x^2}}frac{sqrt{1-y^2}}{y-x}dydx=0$
doing:
Integrate[x^2/Sqrt[1 - x^2] Sqrt[1 - y^2]/(y - x), {x, -1, 1}, {y, -1, 1},
PrincipalValue -> True]
but I want to do it numerically and it doesn't work:
NIntegrate[x^2/Sqrt[1 - x^2] Sqrt[1 - y^2]/(y - x), {x, -1, 1}, {y, -1, 1}]
This is the error message returned:
How can I get Mathematica to solve this problem numerically?
numerical-integration
edited 2 days ago
mjw
1,04110
asked 2 days ago
Javier AlaminosJavier Alaminos
213
New contributor
Javier Alaminos is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
4
$begingroup$
I know the exact solution of the principal value of this integral is equal to zero:
$int_{-1}^{1}int_{-1}^{1}frac{x^2}{sqrt{1-x^2}}frac{sqrt{1-y^2}}{y-x}dydx=0$
doing:
Integrate[x^2/Sqrt[1 - x^2] Sqrt[1 - y^2]/(y - x), {x, -1, 1}, {y, -1, 1},
PrincipalValue -> True]
but I want to do it numerically and it doesn't work:
NIntegrate[x^2/Sqrt[1 - x^2] Sqrt[1 - y^2]/(y - x), {x, -1, 1}, {y, -1, 1}]
This is the error message returned:
How can I get Mathematica to solve this problem numerically?
numerical-integration
edited 2 days ago
mjw
1,04110
asked 2 days ago
Javier AlaminosJavier Alaminos
213
New contributor
Javier Alaminos is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
1
$begingroup$
The issue is that the integrand approaches infinity asx->±1,x->y, andy->x. That kind of behavior is toxic to numerical methods: you need to reason out a way to deal with it, not merely probe it numerically.PrincipalValue -> Truegives you access to automated reasoning in this case, and you've solved the problem that way. Do you have a different problem you're trying to solve?
$endgroup$
– John Doty
2 days ago
$begingroup$
Yes, I'm trying to solve a similar integral, when x^2 is multplied by exp^(-i*b*(x + y)). So, firstly I wanted to try to solve this known integral.
$endgroup$
– Javier Alaminos
2 days ago
2
$begingroup$
Use option Exclusions -> {-1, 1, y + x == 0}]
$endgroup$
– user18792
2 days ago
$begingroup$
All that is built on the sand because the PrincipalValue option for multivariate integrals is undocumented.
$endgroup$
– user64494
2 days ago
add a comment |
4
4
4
$begingroup$
I know the exact solution of the principal value of this integral is equal to zero:
$int_{-1}^{1}int_{-1}^{1}frac{x^2}{sqrt{1-x^2}}frac{sqrt{1-y^2}}{y-x}dydx=0$
doing:
Integrate[x^2/Sqrt[1 - x^2] Sqrt[1 - y^2]/(y - x), {x, -1, 1}, {y, -1, 1},
PrincipalValue -> True]
but I want to do it numerically and it doesn't work:
NIntegrate[x^2/Sqrt[1 - x^2] Sqrt[1 - y^2]/(y - x), {x, -1, 1}, {y, -1, 1}]
This is the error message returned:
How can I get Mathematica to solve this problem numerically?
numerical-integration
edited 2 days ago
mjw
1,04110
asked 2 days ago
Javier AlaminosJavier Alaminos
213
New contributor
Javier Alaminos is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I know the exact solution of the principal value of this integral is equal to zero:
$int_{-1}^{1}int_{-1}^{1}frac{x^2}{sqrt{1-x^2}}frac{sqrt{1-y^2}}{y-x}dydx=0$
doing:
Integrate[x^2/Sqrt[1 - x^2] Sqrt[1 - y^2]/(y - x), {x, -1, 1}, {y, -1, 1},
PrincipalValue -> True]
but I want to do it numerically and it doesn't work:
NIntegrate[x^2/Sqrt[1 - x^2] Sqrt[1 - y^2]/(y - x), {x, -1, 1}, {y, -1, 1}]
This is the error message returned:
How can I get Mathematica to solve this problem numerically?
numerical-integration
numerical-integration
edited 2 days ago
mjw
1,04110
asked 2 days ago
Javier AlaminosJavier Alaminos
213
New contributor
Javier Alaminos is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 days ago
mjw
1,04110
asked 2 days ago
Javier AlaminosJavier Alaminos
213
New contributor
Javier Alaminos is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 days ago
mjw
1,04110
edited 2 days ago
mjw
1,04110
edited 2 days ago
mjw
1,04110
1,04110
asked 2 days ago
Javier AlaminosJavier Alaminos
213
New contributor
Javier Alaminos is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
Javier AlaminosJavier Alaminos
213
asked 2 days ago
Javier AlaminosJavier Alaminos
213
213
New contributor
Javier Alaminos is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Javier Alaminos is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Javier Alaminos is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
$begingroup$
The issue is that the integrand approaches infinity asx->±1,x->y, andy->x. That kind of behavior is toxic to numerical methods: you need to reason out a way to deal with it, not merely probe it numerically.PrincipalValue -> Truegives you access to automated reasoning in this case, and you've solved the problem that way. Do you have a different problem you're trying to solve?
$endgroup$
– John Doty
2 days ago
$begingroup$
Yes, I'm trying to solve a similar integral, when x^2 is multplied by exp^(-i*b*(x + y)). So, firstly I wanted to try to solve this known integral.
$endgroup$
– Javier Alaminos
2 days ago
2
$begingroup$
Use option Exclusions -> {-1, 1, y + x == 0}]
$endgroup$
– user18792
2 days ago
$begingroup$
All that is built on the sand because the PrincipalValue option for multivariate integrals is undocumented.
$endgroup$
– user64494
2 days ago
add a comment |
1
$begingroup$
The issue is that the integrand approaches infinity asx->±1,x->y, andy->x. That kind of behavior is toxic to numerical methods: you need to reason out a way to deal with it, not merely probe it numerically.PrincipalValue -> Truegives you access to automated reasoning in this case, and you've solved the problem that way. Do you have a different problem you're trying to solve?
$endgroup$
– John Doty
2 days ago
$begingroup$
Yes, I'm trying to solve a similar integral, when x^2 is multplied by exp^(-i*b*(x + y)). So, firstly I wanted to try to solve this known integral.
$endgroup$
– Javier Alaminos
2 days ago
2
$begingroup$
Use option Exclusions -> {-1, 1, y + x == 0}]
$endgroup$
– user18792
2 days ago
$begingroup$
All that is built on the sand because the PrincipalValue option for multivariate integrals is undocumented.
$endgroup$
– user64494
2 days ago
1
1
$begingroup$
The issue is that the integrand approaches infinity as
x->±1, x->y, and y->x. That kind of behavior is toxic to numerical methods: you need to reason out a way to deal with it, not merely probe it numerically. PrincipalValue -> True gives you access to automated reasoning in this case, and you've solved the problem that way. Do you have a different problem you're trying to solve?$endgroup$
– John Doty
2 days ago
$begingroup$
The issue is that the integrand approaches infinity as
x->±1, x->y, and y->x. That kind of behavior is toxic to numerical methods: you need to reason out a way to deal with it, not merely probe it numerically. PrincipalValue -> True gives you access to automated reasoning in this case, and you've solved the problem that way. Do you have a different problem you're trying to solve?$endgroup$
– John Doty
2 days ago
$begingroup$
Yes, I'm trying to solve a similar integral, when x^2 is multplied by exp^(-i*b*(x + y)). So, firstly I wanted to try to solve this known integral.
$endgroup$
– Javier Alaminos
2 days ago
$begingroup$
Yes, I'm trying to solve a similar integral, when x^2 is multplied by exp^(-i*b*(x + y)). So, firstly I wanted to try to solve this known integral.
$endgroup$
– Javier Alaminos
2 days ago
2
2
$begingroup$
Use option Exclusions -> {-1, 1, y + x == 0}]
$endgroup$
– user18792
2 days ago
$begingroup$
Use option Exclusions -> {-1, 1, y + x == 0}]
$endgroup$
– user18792
2 days ago
$begingroup$
All that is built on the sand because the PrincipalValue option for multivariate integrals is undocumented.
$endgroup$
– user64494
2 days ago
$begingroup$
All that is built on the sand because the PrincipalValue option for multivariate integrals is undocumented.
$endgroup$
– user64494
2 days ago
add a comment |
2 Answers
2
active
oldest
votes
2
$begingroup$
The main problem is the point x=y. In principle, it seems that there the integral is singular. If you agree to get the principal value of it, you may exclude this point by a regularization as follows
NIntegrate[
x^2/Sqrt[1 - x^2] Sqrt[1 - y^2]/Sqrt[(y - x)^2 + i^-2], {x, -1 + 1/i,
1 - 1/i}, {y, -1 + 1/i, 1 - 1/i}, Method -> "AdaptiveMonteCarlo"]
where i is a large number. Then you may increase i and check the convergence of the integral:
lst = Table[{1/i,
NIntegrate[
x^2/Sqrt[1 - x^2] Sqrt[1 - y^2]/
Sqrt[(y - x)^2 + i^-2], {x, -1 + 1/i, 1 - 1/i}, {y, -1 + 1/i,
1 - 1/i}, Method -> "AdaptiveMonteCarlo"]}, {i, 1000, 100000,
1000}] // N;
ListLogPlot[lst /. {x_, y_} -> {1/x, y}, Frame -> True,
FrameLabel -> {Style["Number i", 16], Style["Integral", 16]}]
yielding this
One can further a few other methods which may eventually enable a more accurate estimate of the integral.
Have fun!
answered 2 days ago
Alexei BoulbitchAlexei Boulbitch
21.9k2570
$endgroup$
3
$begingroup$
I don't understand what you're plotting. The value of this integral is 0 and your result is around 12.
$endgroup$
– Javier Alaminos
2 days ago
$begingroup$
You are right, it is not the same integral, since I tookSqrt[(x-y)^2+eps^2]instead ofx-y.
$endgroup$
– Alexei Boulbitch
2 days ago
$begingroup$
So, to solve the original integral what do I have to do?
$endgroup$
– Javier Alaminos
2 days ago
$begingroup$
Note that $sqrt{(y-x)^2 + epsilon^2}$ approaches $|y - x|$ as $epsilon to 0$, not the $y - x$ that's in the original integrand.
$endgroup$
– Michael Seifert
2 days ago
add a comment |
2
$begingroup$
As the others say,simply integrate by avoiding singular points?
Fixed.
Try other integral.
target = Compile[{{x, _Real}, {y, _Real}},
x/[Sqrt](1 - x) [Sqrt](1 - y^2)/(y - x)];
integrand[x_, y_] := If[Or[(1 - x) == 0, y == x], 0, target[x, y]];
Quiet@NIntegrate[integrand[x, y], {x, -1., 1.}, {y, -1., 1.}]
=>
-4.06259
Integration by manual
.
Plus @@ Flatten@
Table[integrand[x, y]*0.001*0.001, {x, -1., 1., 0.001}, {y, -1., 1.,
0.001}]
=>
-3.99866
Integration by NIntegrate
N@Integrate[
x/Sqrt[1 - x] Sqrt[1 - y^2]/(y - x), {x, -1, 1}, {y, -1, 1},
PrincipalValue -> True]
=>
-4.14669
the question's integral.
target = Compile[{{x, _Real}, {y, _Real}},
x^2/[Sqrt](1 - x^2) [Sqrt](1 - y^2)/(y - x)];
integrand[x_, y_] := If[Or[(1 - x^2) == 0, y == x], 0, target[x, y]];
Quiet@NIntegrate[integrand[x, y], {x, -1., 1.}, {y, -1., 1.}]
=>
-0.4542
By manual.
Plus @@ Flatten@
Table[integrand[x, y]*0.1*0.1, {x, -1., 1., 0.1}, {y, -1., 1., 0.1}]
=>
-8.88178*10^-16
By other method.
Quiet@NIntegrate[integrand[x, y], {x, -1., 1.}, {y, -1., 1.},
Method -> "LocalAdaptive"]
=>
7.73766*10^-17
For now, we can see that the integral value is close to zero.
answered 2 days ago
XminerXminer
33818
$endgroup$
$begingroup$
But I can't avoid the singularity $x==y$, because for example if I have $x$ instead of $x^2$ the result of integral is $-pi^2/2$, and with your code the result is always 0.
$endgroup$
– Javier Alaminos
2 days ago
$begingroup$
@JavierAlaminos as your point,my code was always 0. it's mainly because my code returnsNothingwhen the condition met,I think. so just modified.
$endgroup$
– Xminer
2 days ago
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
2
$begingroup$
The main problem is the point x=y. In principle, it seems that there the integral is singular. If you agree to get the principal value of it, you may exclude this point by a regularization as follows
NIntegrate[
x^2/Sqrt[1 - x^2] Sqrt[1 - y^2]/Sqrt[(y - x)^2 + i^-2], {x, -1 + 1/i,
1 - 1/i}, {y, -1 + 1/i, 1 - 1/i}, Method -> "AdaptiveMonteCarlo"]
where i is a large number. Then you may increase i and check the convergence of the integral:
lst = Table[{1/i,
NIntegrate[
x^2/Sqrt[1 - x^2] Sqrt[1 - y^2]/
Sqrt[(y - x)^2 + i^-2], {x, -1 + 1/i, 1 - 1/i}, {y, -1 + 1/i,
1 - 1/i}, Method -> "AdaptiveMonteCarlo"]}, {i, 1000, 100000,
1000}] // N;
ListLogPlot[lst /. {x_, y_} -> {1/x, y}, Frame -> True,
FrameLabel -> {Style["Number i", 16], Style["Integral", 16]}]
yielding this
One can further a few other methods which may eventually enable a more accurate estimate of the integral.
Have fun!
answered 2 days ago
Alexei BoulbitchAlexei Boulbitch
21.9k2570
$endgroup$
3
$begingroup$
I don't understand what you're plotting. The value of this integral is 0 and your result is around 12.
$endgroup$
– Javier Alaminos
2 days ago
$begingroup$
You are right, it is not the same integral, since I tookSqrt[(x-y)^2+eps^2]instead ofx-y.
$endgroup$
– Alexei Boulbitch
2 days ago
$begingroup$
So, to solve the original integral what do I have to do?
$endgroup$
– Javier Alaminos
2 days ago
$begingroup$
Note that $sqrt{(y-x)^2 + epsilon^2}$ approaches $|y - x|$ as $epsilon to 0$, not the $y - x$ that's in the original integrand.
$endgroup$
– Michael Seifert
2 days ago
add a comment |
2
$begingroup$
The main problem is the point x=y. In principle, it seems that there the integral is singular. If you agree to get the principal value of it, you may exclude this point by a regularization as follows
NIntegrate[
x^2/Sqrt[1 - x^2] Sqrt[1 - y^2]/Sqrt[(y - x)^2 + i^-2], {x, -1 + 1/i,
1 - 1/i}, {y, -1 + 1/i, 1 - 1/i}, Method -> "AdaptiveMonteCarlo"]
where i is a large number. Then you may increase i and check the convergence of the integral:
lst = Table[{1/i,
NIntegrate[
x^2/Sqrt[1 - x^2] Sqrt[1 - y^2]/
Sqrt[(y - x)^2 + i^-2], {x, -1 + 1/i, 1 - 1/i}, {y, -1 + 1/i,
1 - 1/i}, Method -> "AdaptiveMonteCarlo"]}, {i, 1000, 100000,
1000}] // N;
ListLogPlot[lst /. {x_, y_} -> {1/x, y}, Frame -> True,
FrameLabel -> {Style["Number i", 16], Style["Integral", 16]}]
yielding this
One can further a few other methods which may eventually enable a more accurate estimate of the integral.
Have fun!
answered 2 days ago
Alexei BoulbitchAlexei Boulbitch
21.9k2570
$endgroup$
3
$begingroup$
I don't understand what you're plotting. The value of this integral is 0 and your result is around 12.
$endgroup$
– Javier Alaminos
2 days ago
$begingroup$
You are right, it is not the same integral, since I tookSqrt[(x-y)^2+eps^2]instead ofx-y.
$endgroup$
– Alexei Boulbitch
2 days ago
$begingroup$
So, to solve the original integral what do I have to do?
$endgroup$
– Javier Alaminos
2 days ago
$begingroup$
Note that $sqrt{(y-x)^2 + epsilon^2}$ approaches $|y - x|$ as $epsilon to 0$, not the $y - x$ that's in the original integrand.
$endgroup$
– Michael Seifert
2 days ago
add a comment |
2
2
2
$begingroup$
The main problem is the point x=y. In principle, it seems that there the integral is singular. If you agree to get the principal value of it, you may exclude this point by a regularization as follows
NIntegrate[
x^2/Sqrt[1 - x^2] Sqrt[1 - y^2]/Sqrt[(y - x)^2 + i^-2], {x, -1 + 1/i,
1 - 1/i}, {y, -1 + 1/i, 1 - 1/i}, Method -> "AdaptiveMonteCarlo"]
where i is a large number. Then you may increase i and check the convergence of the integral:
lst = Table[{1/i,
NIntegrate[
x^2/Sqrt[1 - x^2] Sqrt[1 - y^2]/
Sqrt[(y - x)^2 + i^-2], {x, -1 + 1/i, 1 - 1/i}, {y, -1 + 1/i,
1 - 1/i}, Method -> "AdaptiveMonteCarlo"]}, {i, 1000, 100000,
1000}] // N;
ListLogPlot[lst /. {x_, y_} -> {1/x, y}, Frame -> True,
FrameLabel -> {Style["Number i", 16], Style["Integral", 16]}]
yielding this
One can further a few other methods which may eventually enable a more accurate estimate of the integral.
Have fun!
answered 2 days ago
Alexei BoulbitchAlexei Boulbitch
21.9k2570
$endgroup$
The main problem is the point x=y. In principle, it seems that there the integral is singular. If you agree to get the principal value of it, you may exclude this point by a regularization as follows
NIntegrate[
x^2/Sqrt[1 - x^2] Sqrt[1 - y^2]/Sqrt[(y - x)^2 + i^-2], {x, -1 + 1/i,
1 - 1/i}, {y, -1 + 1/i, 1 - 1/i}, Method -> "AdaptiveMonteCarlo"]
where i is a large number. Then you may increase i and check the convergence of the integral:
lst = Table[{1/i,
NIntegrate[
x^2/Sqrt[1 - x^2] Sqrt[1 - y^2]/
Sqrt[(y - x)^2 + i^-2], {x, -1 + 1/i, 1 - 1/i}, {y, -1 + 1/i,
1 - 1/i}, Method -> "AdaptiveMonteCarlo"]}, {i, 1000, 100000,
1000}] // N;
ListLogPlot[lst /. {x_, y_} -> {1/x, y}, Frame -> True,
FrameLabel -> {Style["Number i", 16], Style["Integral", 16]}]
yielding this
One can further a few other methods which may eventually enable a more accurate estimate of the integral.
Have fun!
answered 2 days ago
Alexei BoulbitchAlexei Boulbitch
21.9k2570
answered 2 days ago
Alexei BoulbitchAlexei Boulbitch
21.9k2570
answered 2 days ago
Alexei BoulbitchAlexei Boulbitch
21.9k2570
answered 2 days ago
Alexei BoulbitchAlexei Boulbitch
21.9k2570
21.9k2570
3
$begingroup$
I don't understand what you're plotting. The value of this integral is 0 and your result is around 12.
$endgroup$
– Javier Alaminos
2 days ago
$begingroup$
You are right, it is not the same integral, since I tookSqrt[(x-y)^2+eps^2]instead ofx-y.
$endgroup$
– Alexei Boulbitch
2 days ago
$begingroup$
So, to solve the original integral what do I have to do?
$endgroup$
– Javier Alaminos
2 days ago
$begingroup$
Note that $sqrt{(y-x)^2 + epsilon^2}$ approaches $|y - x|$ as $epsilon to 0$, not the $y - x$ that's in the original integrand.
$endgroup$
– Michael Seifert
2 days ago
add a comment |
3
$begingroup$
I don't understand what you're plotting. The value of this integral is 0 and your result is around 12.
$endgroup$
– Javier Alaminos
2 days ago
$begingroup$
You are right, it is not the same integral, since I tookSqrt[(x-y)^2+eps^2]instead ofx-y.
$endgroup$
– Alexei Boulbitch
2 days ago
$begingroup$
So, to solve the original integral what do I have to do?
$endgroup$
– Javier Alaminos
2 days ago
$begingroup$
Note that $sqrt{(y-x)^2 + epsilon^2}$ approaches $|y - x|$ as $epsilon to 0$, not the $y - x$ that's in the original integrand.
$endgroup$
– Michael Seifert
2 days ago
3
3
$begingroup$
I don't understand what you're plotting. The value of this integral is 0 and your result is around 12.
$endgroup$
– Javier Alaminos
2 days ago
$begingroup$
I don't understand what you're plotting. The value of this integral is 0 and your result is around 12.
$endgroup$
– Javier Alaminos
2 days ago
$begingroup$
You are right, it is not the same integral, since I took
Sqrt[(x-y)^2+eps^2] instead of x-y.$endgroup$
– Alexei Boulbitch
2 days ago
$begingroup$
You are right, it is not the same integral, since I took
Sqrt[(x-y)^2+eps^2] instead of x-y.$endgroup$
– Alexei Boulbitch
2 days ago
$begingroup$
So, to solve the original integral what do I have to do?
$endgroup$
– Javier Alaminos
2 days ago
$begingroup$
So, to solve the original integral what do I have to do?
$endgroup$
– Javier Alaminos
2 days ago
$begingroup$
Note that $sqrt{(y-x)^2 + epsilon^2}$ approaches $|y - x|$ as $epsilon to 0$, not the $y - x$ that's in the original integrand.
$endgroup$
– Michael Seifert
2 days ago
$begingroup$
Note that $sqrt{(y-x)^2 + epsilon^2}$ approaches $|y - x|$ as $epsilon to 0$, not the $y - x$ that's in the original integrand.
$endgroup$
– Michael Seifert
2 days ago
add a comment |
2
$begingroup$
As the others say,simply integrate by avoiding singular points?
Fixed.
Try other integral.
target = Compile[{{x, _Real}, {y, _Real}},
x/[Sqrt](1 - x) [Sqrt](1 - y^2)/(y - x)];
integrand[x_, y_] := If[Or[(1 - x) == 0, y == x], 0, target[x, y]];
Quiet@NIntegrate[integrand[x, y], {x, -1., 1.}, {y, -1., 1.}]
=>
-4.06259
Integration by manual
.
Plus @@ Flatten@
Table[integrand[x, y]*0.001*0.001, {x, -1., 1., 0.001}, {y, -1., 1.,
0.001}]
=>
-3.99866
Integration by NIntegrate
N@Integrate[
x/Sqrt[1 - x] Sqrt[1 - y^2]/(y - x), {x, -1, 1}, {y, -1, 1},
PrincipalValue -> True]
=>
-4.14669
the question's integral.
target = Compile[{{x, _Real}, {y, _Real}},
x^2/[Sqrt](1 - x^2) [Sqrt](1 - y^2)/(y - x)];
integrand[x_, y_] := If[Or[(1 - x^2) == 0, y == x], 0, target[x, y]];
Quiet@NIntegrate[integrand[x, y], {x, -1., 1.}, {y, -1., 1.}]
=>
-0.4542
By manual.
Plus @@ Flatten@
Table[integrand[x, y]*0.1*0.1, {x, -1., 1., 0.1}, {y, -1., 1., 0.1}]
=>
-8.88178*10^-16
By other method.
Quiet@NIntegrate[integrand[x, y], {x, -1., 1.}, {y, -1., 1.},
Method -> "LocalAdaptive"]
=>
7.73766*10^-17
For now, we can see that the integral value is close to zero.
answered 2 days ago
XminerXminer
33818
$endgroup$
$begingroup$
But I can't avoid the singularity $x==y$, because for example if I have $x$ instead of $x^2$ the result of integral is $-pi^2/2$, and with your code the result is always 0.
$endgroup$
– Javier Alaminos
2 days ago
$begingroup$
@JavierAlaminos as your point,my code was always 0. it's mainly because my code returnsNothingwhen the condition met,I think. so just modified.
$endgroup$
– Xminer
2 days ago
add a comment |
2
$begingroup$
As the others say,simply integrate by avoiding singular points?
Fixed.
Try other integral.
target = Compile[{{x, _Real}, {y, _Real}},
x/[Sqrt](1 - x) [Sqrt](1 - y^2)/(y - x)];
integrand[x_, y_] := If[Or[(1 - x) == 0, y == x], 0, target[x, y]];
Quiet@NIntegrate[integrand[x, y], {x, -1., 1.}, {y, -1., 1.}]
=>
-4.06259
Integration by manual
.
Plus @@ Flatten@
Table[integrand[x, y]*0.001*0.001, {x, -1., 1., 0.001}, {y, -1., 1.,
0.001}]
=>
-3.99866
Integration by NIntegrate
N@Integrate[
x/Sqrt[1 - x] Sqrt[1 - y^2]/(y - x), {x, -1, 1}, {y, -1, 1},
PrincipalValue -> True]
=>
-4.14669
the question's integral.
target = Compile[{{x, _Real}, {y, _Real}},
x^2/[Sqrt](1 - x^2) [Sqrt](1 - y^2)/(y - x)];
integrand[x_, y_] := If[Or[(1 - x^2) == 0, y == x], 0, target[x, y]];
Quiet@NIntegrate[integrand[x, y], {x, -1., 1.}, {y, -1., 1.}]
=>
-0.4542
By manual.
Plus @@ Flatten@
Table[integrand[x, y]*0.1*0.1, {x, -1., 1., 0.1}, {y, -1., 1., 0.1}]
=>
-8.88178*10^-16
By other method.
Quiet@NIntegrate[integrand[x, y], {x, -1., 1.}, {y, -1., 1.},
Method -> "LocalAdaptive"]
=>
7.73766*10^-17
For now, we can see that the integral value is close to zero.
answered 2 days ago
XminerXminer
33818
$endgroup$
$begingroup$
But I can't avoid the singularity $x==y$, because for example if I have $x$ instead of $x^2$ the result of integral is $-pi^2/2$, and with your code the result is always 0.
$endgroup$
– Javier Alaminos
2 days ago
$begingroup$
@JavierAlaminos as your point,my code was always 0. it's mainly because my code returnsNothingwhen the condition met,I think. so just modified.
$endgroup$
– Xminer
2 days ago
add a comment |
2
2
2
$begingroup$
As the others say,simply integrate by avoiding singular points?
Fixed.
Try other integral.
target = Compile[{{x, _Real}, {y, _Real}},
x/[Sqrt](1 - x) [Sqrt](1 - y^2)/(y - x)];
integrand[x_, y_] := If[Or[(1 - x) == 0, y == x], 0, target[x, y]];
Quiet@NIntegrate[integrand[x, y], {x, -1., 1.}, {y, -1., 1.}]
=>
-4.06259
Integration by manual
.
Plus @@ Flatten@
Table[integrand[x, y]*0.001*0.001, {x, -1., 1., 0.001}, {y, -1., 1.,
0.001}]
=>
-3.99866
Integration by NIntegrate
N@Integrate[
x/Sqrt[1 - x] Sqrt[1 - y^2]/(y - x), {x, -1, 1}, {y, -1, 1},
PrincipalValue -> True]
=>
-4.14669
the question's integral.
target = Compile[{{x, _Real}, {y, _Real}},
x^2/[Sqrt](1 - x^2) [Sqrt](1 - y^2)/(y - x)];
integrand[x_, y_] := If[Or[(1 - x^2) == 0, y == x], 0, target[x, y]];
Quiet@NIntegrate[integrand[x, y], {x, -1., 1.}, {y, -1., 1.}]
=>
-0.4542
By manual.
Plus @@ Flatten@
Table[integrand[x, y]*0.1*0.1, {x, -1., 1., 0.1}, {y, -1., 1., 0.1}]
=>
-8.88178*10^-16
By other method.
Quiet@NIntegrate[integrand[x, y], {x, -1., 1.}, {y, -1., 1.},
Method -> "LocalAdaptive"]
=>
7.73766*10^-17
For now, we can see that the integral value is close to zero.
answered 2 days ago
XminerXminer
33818
$endgroup$
As the others say,simply integrate by avoiding singular points?
Fixed.
Try other integral.
target = Compile[{{x, _Real}, {y, _Real}},
x/[Sqrt](1 - x) [Sqrt](1 - y^2)/(y - x)];
integrand[x_, y_] := If[Or[(1 - x) == 0, y == x], 0, target[x, y]];
Quiet@NIntegrate[integrand[x, y], {x, -1., 1.}, {y, -1., 1.}]
=>
-4.06259
Integration by manual
.
Plus @@ Flatten@
Table[integrand[x, y]*0.001*0.001, {x, -1., 1., 0.001}, {y, -1., 1.,
0.001}]
=>
-3.99866
Integration by NIntegrate
N@Integrate[
x/Sqrt[1 - x] Sqrt[1 - y^2]/(y - x), {x, -1, 1}, {y, -1, 1},
PrincipalValue -> True]
=>
-4.14669
the question's integral.
target = Compile[{{x, _Real}, {y, _Real}},
x^2/[Sqrt](1 - x^2) [Sqrt](1 - y^2)/(y - x)];
integrand[x_, y_] := If[Or[(1 - x^2) == 0, y == x], 0, target[x, y]];
Quiet@NIntegrate[integrand[x, y], {x, -1., 1.}, {y, -1., 1.}]
=>
-0.4542
By manual.
Plus @@ Flatten@
Table[integrand[x, y]*0.1*0.1, {x, -1., 1., 0.1}, {y, -1., 1., 0.1}]
=>
-8.88178*10^-16
By other method.
Quiet@NIntegrate[integrand[x, y], {x, -1., 1.}, {y, -1., 1.},
Method -> "LocalAdaptive"]
=>
7.73766*10^-17
For now, we can see that the integral value is close to zero.
answered 2 days ago
XminerXminer
33818
edited 2 days ago
answered 2 days ago
XminerXminer
33818
answered 2 days ago
XminerXminer
33818
answered 2 days ago
XminerXminer
33818
33818
$begingroup$
But I can't avoid the singularity $x==y$, because for example if I have $x$ instead of $x^2$ the result of integral is $-pi^2/2$, and with your code the result is always 0.
$endgroup$
– Javier Alaminos
2 days ago
$begingroup$
@JavierAlaminos as your point,my code was always 0. it's mainly because my code returnsNothingwhen the condition met,I think. so just modified.
$endgroup$
– Xminer
2 days ago
add a comment |
$begingroup$
But I can't avoid the singularity $x==y$, because for example if I have $x$ instead of $x^2$ the result of integral is $-pi^2/2$, and with your code the result is always 0.
$endgroup$
– Javier Alaminos
2 days ago
$begingroup$
@JavierAlaminos as your point,my code was always 0. it's mainly because my code returnsNothingwhen the condition met,I think. so just modified.
$endgroup$
– Xminer
2 days ago
$begingroup$
But I can't avoid the singularity $x==y$, because for example if I have $x$ instead of $x^2$ the result of integral is $-pi^2/2$, and with your code the result is always 0.
$endgroup$
– Javier Alaminos
2 days ago
$begingroup$
But I can't avoid the singularity $x==y$, because for example if I have $x$ instead of $x^2$ the result of integral is $-pi^2/2$, and with your code the result is always 0.
$endgroup$
– Javier Alaminos
2 days ago
$begingroup$
@JavierAlaminos as your point,my code was always 0. it's mainly because my code returns
Nothing when the condition met,I think. so just modified.$endgroup$
– Xminer
2 days ago
$begingroup$
@JavierAlaminos as your point,my code was always 0. it's mainly because my code returns
Nothing when the condition met,I think. so just modified.$endgroup$
– Xminer
2 days ago
add a comment |
Javier Alaminos is a new contributor. Be nice, and check out our Code of Conduct.
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Javier Alaminos is a new contributor. Be nice, and check out our Code of Conduct.
Javier Alaminos is a new contributor. Be nice, and check out our Code of Conduct.
Javier Alaminos is a new contributor. Be nice, and check out our Code of Conduct.
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1
$begingroup$
The issue is that the integrand approaches infinity as
x->±1,x->y, andy->x. That kind of behavior is toxic to numerical methods: you need to reason out a way to deal with it, not merely probe it numerically.PrincipalValue -> Truegives you access to automated reasoning in this case, and you've solved the problem that way. Do you have a different problem you're trying to solve?$endgroup$
– John Doty
2 days ago
$begingroup$
Yes, I'm trying to solve a similar integral, when x^2 is multplied by exp^(-i*b*(x + y)). So, firstly I wanted to try to solve this known integral.
$endgroup$
– Javier Alaminos
2 days ago
2
$begingroup$
Use option Exclusions -> {-1, 1, y + x == 0}]
$endgroup$
– user18792
2 days ago
$begingroup$
All that is built on the sand because the PrincipalValue option for multivariate integrals is undocumented.
$endgroup$
– user64494
2 days ago