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:



enter image description here



How can I get Mathematica to solve this problem numerically?






numerical-integration







share|improve this question









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Javier Alaminos is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$








  • 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










  • $begingroup$
    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
















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:



enter image description here



How can I get Mathematica to solve this problem numerically?






numerical-integration







share|improve this question









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 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






  • 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














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:



enter image description here



How can I get Mathematica to solve this problem numerically?






numerical-integration







share|improve this question









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:



enter image description here



How can I get Mathematica to solve this problem numerically?






numerical-integration




numerical-integration






share|improve this question









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.











share|improve this question









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.









share|improve this question




share|improve this question








edited 2 days ago









mjw

1,04110




1,04110






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




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 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






  • 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




    $begingroup$
    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






  • 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










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



enter image description here



One can further a few other methods which may eventually enable a more accurate estimate of the integral.



Have fun!






share|improve this answer









$endgroup$









  • 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 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










  • $begingroup$
    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





















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.






share|improve this answer











$endgroup$













  • $begingroup$
    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 returns Nothing when the condition met,I think. so just modified.
    $endgroup$
    – Xminer
    2 days ago











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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



enter image description here



One can further a few other methods which may eventually enable a more accurate estimate of the integral.



Have fun!






share|improve this answer









$endgroup$









  • 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 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










  • $begingroup$
    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


















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



enter image description here



One can further a few other methods which may eventually enable a more accurate estimate of the integral.



Have fun!






share|improve this answer









$endgroup$









  • 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 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










  • $begingroup$
    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
















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



enter image description here



One can further a few other methods which may eventually enable a more accurate estimate of the integral.



Have fun!






share|improve this answer









$endgroup$



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



enter image description here



One can further a few other methods which may eventually enable a more accurate estimate of the integral.



Have fun!







share|improve this answer












share|improve this answer



share|improve this answer










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 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










  • $begingroup$
    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




    $begingroup$
    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










  • $begingroup$
    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













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.






share|improve this answer











$endgroup$













  • $begingroup$
    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 returns Nothing when the condition met,I think. so just modified.
    $endgroup$
    – Xminer
    2 days ago
















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.






share|improve this answer











$endgroup$













  • $begingroup$
    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 returns Nothing when the condition met,I think. so just modified.
    $endgroup$
    – Xminer
    2 days ago














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.






share|improve this answer











$endgroup$



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.







share|improve this answer














share|improve this answer



share|improve this answer








edited 2 days ago

























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 returns Nothing when 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












  • $begingroup$
    @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
















$begingroup$
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










Javier Alaminos is a new contributor. Be nice, and check out our Code of Conduct.










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