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Solve a Cauchy integral if the singular node is an endpoint?

GROUPS:

I have one definite integral that needs to be evaluated from -1 to 1 in the Cauchy principal value sense. Math programs only let me put the singular points between the integration limits , but my singular point is -1.

The integral is complex,it has a strong singularity on -1, but how can I integrate from -1 to 1?

Here is the function for simple copy and paste:

-(((1 - xi)*xi*Sqrt[(0.056249999999999994*(1 - xi) + 0.10625000000000001*xi - 0.049999999999999996*(1 + xi))^2]*
    (0. + ((0.056249999999999994*(1 - xi) + 0.10625000000000001*xi - 0.049999999999999996*(1 + xi))*(-0.1 - 0.05*(1 - xi)*xi + 0.10625*(1 - xi)*(1 + xi) + 0.05625*xi*(1 + xi))*
       (1.3 + 4*(BesselK[0, 628.318530717959*Sqrt[0. + (-0.1 - 0.05*(1 - xi)*xi + 0.10625*(1 - xi)*(1 + xi) + 0.05625*xi*(1 + xi))^2]] + 
          (0.003183098861837905*(-(0.0015915494309189525/Sqrt[0. + (-0.1 - 0.05*(1 - xi)*xi + 0.10625*(1 - xi)*(1 + xi) + 0.05625*xi*(1 + xi))^2]) + 
             BesselK[1, 628.318530717959*Sqrt[0. + (-0.1 - 0.05*(1 - xi)*xi + 0.10625*(1 - xi)*(1 + xi) + 0.05625*xi*(1 + xi))^2]]))/
           Sqrt[0. + (-0.1 - 0.05*(1 - xi)*xi + 0.10625*(1 - xi)*(1 + xi) + 0.05625*xi*(1 + xi))^2])))/(Sqrt[(0.056249999999999994*(1 - xi)
+ 0.10625000000000001*xi - 0.049999999999999996*(1 + xi))^2]*
       Sqrt[0. + (-0.1 - 0.05*(1 - xi)*xi + 0.10625*(1 - xi)*(1 + xi) + 0.05625*xi*(1 + xi))^2]) + (0.*(-0.1 - 0.05*(1 - xi)*xi + 0.10625*(1 - xi)*(1 + xi) + 0.05625*xi*(1 + xi))*
       (0.7 + 1256.637061435918*Sqrt[0. + (-0.1 - 0.05*(1 - xi)*xi + 0.10625*(1 - xi)*(1 + xi) + 0.05625*xi*(1 + xi))^2]*
         BesselK[1, 628.318530717959*Sqrt[0. + (-0.1 - 0.05*(1 - xi)*xi + 0.10625*(1 - xi)*(1 + xi) + 0.05625*xi*(1 + xi))^2]] + 
        8*(BesselK[0, 628.318530717959*Sqrt[0. + (-0.1 - 0.05*(1 - xi)*xi + 0.10625*(1 - xi)*(1 + xi) + 0.05625*xi*(1 + xi))^2]] + 
          (0.003183098861837905*(-(0.0015915494309189525/Sqrt[0. + (-0.1 - 0.05*(1 - xi)*xi + 0.10625*(1 - xi)*(1 + xi) + 0.05625*xi*(1 + xi))^2]) + 
             BesselK[1, 628.318530717959*Sqrt[0. + (-0.1 - 0.05*(1 - xi)*xi + 0.10625*(1 - xi)*(1 + xi) + 0.05625*xi*(1 + xi))^2]]))/
           Sqrt[0. + (-0.1 - 0.05*(1 - xi)*xi + 0.10625*(1 - xi)*(1 + xi) + 0.05625*xi*(1 + xi))^2])))/(Sqrt[(0.056249999999999994*(1 - xi)
+ 0.10625000000000001*xi - 0.049999999999999996*(1 + xi))^2]*
       Sqrt[0. + (-0.1 - 0.05*(1 - xi)*xi + 0.10625*(1 - xi)*(1 + xi) + 0.05625*xi*(1 + xi))^2])))/(8*Pi*Sqrt[0. + (-0.1 - 0.05*(1 - xi)*xi + 0.10625*(1 - xi)*(1 + xi) + 0.05625*xi*(1 + xi))^2]))

Could mathematica solve this integral since the singular point is on the endpoint of the integration?

POSTED BY: Romildo Junior
Answer
4 months ago

Does anyone have a hint on how to solve the above integral ?

POSTED BY: Romildo Junior
Answer
4 months ago

How do you define this integral? That is to say, what normalization do you have in mind to remove the singularity.

POSTED BY: Daniel Lichtblau
Answer
4 months ago

Thanks for the reply, It is a strong singularity of type (1/r) , no adaptivity technique seems to be working, and cauchy could only be evaluated if the singular point is in the interval . I tried to solve it analytically like and got the input back! This integral is getting on my nerves !

POSTED BY: Romildo Junior
Answer
4 months ago

You say the singularity must lie within the interval. A perhaps stupid question: what about calculating the integral from -1+x to 1 for several small x-es and then fitting a curve through the { 1 - x , intergralvalue} points , for example a polynomial (perhaps of second order ) . Then let x = 0 in this function?

POSTED BY: Hans Dolhaine
Answer
4 months ago

Thanks Hans, that gave me a very high value like -3521, I think that might not be the answer. I have a fortran code that uses the singularity subtraction to calculate and it gives a value of like 0.1000 for the integral from -1 to 1, I wanted to know if Mathematica could solve a Integral like that without a external code or technique. I am starting to think that Mathematica maybe cannot solve this integral. I wanted to upgrade my code to Mathematica by using its functions to solve the integral more easily. If anyone got a way to go please help

POSTED BY: Romildo Junior
Answer
4 months ago

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