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Mathematics Mathematica Calculus Wolfram Language

Different results solving the same integral?

Enrique Rodriguez

Posted 6 years ago

I need to evaluate an integral as in the attached file, for general parameters a, b and c. I obtain a conditional 0 as result using the Integrate function. However, when I evaluate the integral again (numerically and analitically) substituing the parameters a,b and c by numbers (which fulfill the conditions obtained before), the result is not zero. Should I trust the result 0 for general parameters a, b and c, or not? What could be happening? Thanks so much Attachments:

POSTED BY: Enrique Rodriguez

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Alexander Trounev

Alexander Trounev, A&E Trounev IT Consulting

Posted 6 years ago

We need to make an assumption for the integral. For example, In[41]:= Integrate[ 1/Sqrt[(w^2 + a)]/((b - Iw)(c + I*(w + ww))), {w, -Infinity, Infinity}, Assumptions -> {a > 0, c > 0, b > 0, ww > 0}] Out[41]= (I (MeijerG[{{1/2, 1, 1}, {}}, {{1/2, 1/2, 1}, {}}, -(( I Sqrt[a])/b), 1/2] - MeijerG[{{1/2, 1, 1}, {}}, {{1/2, 1/2, 1}, {}}, (I Sqrt[a])/b, 1/ 2]))/(2 Sqrt[a] \[Pi]^(3/2) (b + c + I ww))

We need to make an assumption for the integral. For example,

In[41]:= Integrate[
 1/Sqrt[(w^2 + a)]/((b - I*w)*(c + I*(w + ww))), {w, -Infinity, 
  Infinity}, Assumptions -> {a > 0, c > 0, b > 0, ww > 0}]

Out[41]= (I (MeijerG[{{1/2, 1, 1}, {}}, {{1/2, 1/2, 1}, {}}, -((
     I Sqrt[a])/b), 1/2] - 
   MeijerG[{{1/2, 1, 1}, {}}, {{1/2, 1/2, 1}, {}}, (I Sqrt[a])/b, 1/
    2]))/(2 Sqrt[a] \[Pi]^(3/2) (b + c + I ww))

POSTED BY: Alexander Trounev

Mariusz Iwaniuk

Mariusz Iwaniuk, engineer, math hobbyist.

Posted 6 years ago

Mathematica gives incorrect answer(A bug !!!), is singular,a MMA has problems to calculate. (I (MeijerG[{{1/2, 1, 1}, {}}, {{1/2, 1/2, 1}, {}}, -((I Sqrt[a])/b), 1/2] - MeijerG[{{1/2, 1, 1}, {}}, {{1/2, 1/2, 1}, {}}, (I Sqrt[a])/b, 1/2]))/(2 Sqrt[ a] \[Pi]^(3/2) (b + c + I ww)) /. a -> 1 /. b -> 1 /. c -> 1 /. ww -> 1 // N (* (0.0179587 + 0.0359174 I) ((-11.1367 - 11.1367 I) + MeijerG[{{0.5, 1., 1.}, {}}, {{0.5, 0.5, 1.}, {}}, 0. - 1. I, 0.5])) Output gives no numeric answer. intsymbolic = (I (MeijerG[{{1/2, 1, 1}, {}}, {{1/2, 1/2, 1}, {}}, -((I Sqrt[a])/b), 1/2] - MeijerG[{{1/2, 1, 1}, {}}, {{1/2, 1/2, 1}, {}}, (I Sqrt[a])/b, 1/2]))/(2 Sqrt[a] \[Pi]^(3/2) (b + c + I ww)) /. {a -> 1, b -> b, c -> 1, ww -> 1}; Plot[Evaluate@ReIm[intsymbolic], {b, 9/10, 11/10}, PlotLegends -> "Expressions"] // Quiet As we can see here on the plot gives incorrect result 0.8 - 0.4 I. A correct answer is:1.395872 - 0.983636 I. Solution by Maple 2018.*

Mathematica gives incorrect answer(A bug !!!), is singular,a MMA has problems to calculate.

  (I (MeijerG[{{1/2, 1, 1}, {}}, {{1/2, 1/2, 1}, {}}, -((I Sqrt[a])/b), 
       1/2] - MeijerG[{{1/2, 1, 1}, {}}, {{1/2, 1/2, 
         1}, {}}, (I Sqrt[a])/b, 1/2]))/(2 Sqrt[
     a] \[Pi]^(3/2) (b + c + I ww)) /. a -> 1 /. b -> 1 /. 
  c -> 1 /. ww -> 1 // N

(* (0.0179587 + 0.0359174 I) ((-11.1367 - 11.1367 I) + 
   MeijerG[{{0.5, 1., 1.}, {}}, {{0.5, 0.5, 1.}, {}}, 0. - 1. I, 0.5])*)

Output gives no numeric answer.

   intsymbolic = (I (MeijerG[{{1/2, 1, 1}, {}}, {{1/2, 1/2, 
              1}, {}}, -((I Sqrt[a])/b), 1/2] - 
           MeijerG[{{1/2, 1, 1}, {}}, {{1/2, 1/2, 1}, {}}, (I Sqrt[a])/b,
             1/2]))/(2 Sqrt[a] \[Pi]^(3/2) (b + c + I ww)) /. {a -> 1, 
       b -> b, c -> 1, ww -> 1};
      Plot[Evaluate@ReIm[intsymbolic], {b, 9/10, 11/10}, PlotLegends -> "Expressions"] // Quiet

enter image description here

As we can see here on the plot gives incorrect result 0.8 - 0.4 I. A correct answer is:1.395872 - 0.983636 I.

Solution by Maple 2018. enter image description here

POSTED BY: Mariusz Iwaniuk

Alexander Trounev

Alexander Trounev, A&E Trounev IT Consulting

Posted 6 years ago

In[2]:= (I (MeijerG[{{1/2, 1, 1}, {}}, {{1/2, 1/2, 1}, {}}, -(( I Sqrt[a])/b), 1/2] - MeijerG[{{1/2, 1, 1}, {}}, {{1/2, 1/2, 1}, {}}, (I Sqrt[a])/b, 1/ 2]))/(2 Sqrt[a] \[Pi]^(3/2) (b + c + I ww)) /. {a -> 1.1, b -> 0.5, c -> 0.5, ww -> 1.1} Out[2]= 1.05409 - 1.1595 I

In[2]:= (I (MeijerG[{{1/2, 1, 1}, {}}, {{1/2, 1/2, 1}, {}}, -((
      I Sqrt[a])/b), 1/2] - 
    MeijerG[{{1/2, 1, 1}, {}}, {{1/2, 1/2, 1}, {}}, (I Sqrt[a])/b, 1/
     2]))/(2 Sqrt[a] \[Pi]^(3/2) (b + c + I ww)) /. {a -> 1.1, 
  b -> 0.5, c -> 0.5, ww -> 1.1}

Out[2]= 1.05409 - 1.1595 I

POSTED BY: Alexander Trounev

Gianluca Gorni

Gianluca Gorni, University of Udine

Posted 6 years ago

You may have hit a singular point. With the values `{ww -> 1., a -> 2., b -> 1., c -> 1.}` we get a numerical value.

POSTED BY: Gianluca Gorni

Mariusz Iwaniuk

Mariusz Iwaniuk, engineer, math hobbyist.

Posted 6 years ago

I send e-mail to WRI support with this issue and they gave me the answer: Hello, Thank you for contacting Wolfram Technical Support. I understand that the following is not returning the correct result when compared with the NIntegrate result. Integrate[1/(Sqrt[1+w^2] ((1+I)+w+w^2)),{w,-Infinity,Infinity}] The fundamental problem seems to be with the following term in the Integrate result: MeijerG[{{1/2,1,1},{}},{{1/2,1/2,1},{}},-I,1/2] For some reason, Mathematica is not able to evaluate this numerically. It just returns unevaluated. Also, if you try to evaluate this using arbitrary-precision, you get a bunch of errors. I have filed a report with our developers regarding this behavior. Thank you for bringing this to our attention. Regards, Luke Titus Wolfram Technical Support Wolfram Research Inc. http://support.wolfram.com

I send e-mail to WRI support with this issue and they gave me the answer:

  Hello,

  Thank you for contacting Wolfram Technical Support. I understand that the following is not returning the correct result when 
  compared with the NIntegrate result. 

  Integrate[1/(Sqrt[1+w^2] ((1+I)+w+w^2)),{w,-Infinity,Infinity}]

  The fundamental problem seems to be with the following term in the Integrate result:

  MeijerG[{{1/2,1,1},{}},{{1/2,1/2,1},{}},-I,1/2]

  For some reason, Mathematica is not able to evaluate this numerically. It just returns unevaluated. Also, if you try to evaluate 
  this using arbitrary-precision, you get a bunch of errors. I have filed a report with our developers regarding this behavior. Thank 
  you for bringing this to our attention. 

  Regards,

  Luke Titus
  Wolfram Technical Support
  Wolfram Research Inc.
  http://support.wolfram.com

POSTED BY: Mariusz Iwaniuk