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Mathematics Wolfram Language Numerical Computation

Perform a numerical integration?

sahad dubai

Posted 9 years ago

I am getting error while evaluating following integral in Mathematica. Even though I am getting result I am sure it is wrong. ClearAll["GLOBAL'"] {r, a, b, Z, A, B,phi } = {4.087, 1.205, 0.3812, 0, 345.0527, 606741.04395, Pi/3} // Rationalize[#, 0] &; JJ[n_] := r NIntegrate[1/((t Sin[phi])^2 + r^2 + (t Cos[phi] + Z - z)^2 - 2 r t Sin[phi] Cos[eta])^ n, {eta, 0, 2 Pi}, {z, 0, 50}, {t, -5/3, 5/3}, MinRecursion -> 10, MaxRecursion -> 35, WorkingPrecision -> 20]; a b (-A JJ[3] + B JJ[6]) NIntegrate::eincr: The global error of the strategy GlobalAdaptive has increased more than 2000 times. The global error is expected to decrease monotonically after a number of integrand evaluations. Suspect one of the following: the working precision is insufficient for the specified precision goal; the integrand is highly oscillatory or it is not a (piecewise) smooth function; or the true value of the integral is 0. Increasing the value of the GlobalAdaptive option MaxErrorIncreases might lead to a convergent numerical integration. NIntegrate obtained 0.01443218400405474348879705436971438392688666301951432535504827223787189`70. and 1.034077729663448519046495248747417797443269722765919064706177094984896`70.^-10 for the integral and error estimates. >> NIntegrate::eincr: The global error of the strategy GlobalAdaptive has increased more than 2000 times. The global error is expected to decrease monotonically after a number of integrand evaluations. Suspect one of the following: the working precision is insufficient for the specified precision goal; the integrand is highly oscillatory or it is not a (piecewise) smooth function; or the true value of the integral is 0. Increasing the value of the GlobalAdaptive option MaxErrorIncreases might lead to a convergent numerical integration. NIntegrate obtained 6.363900979126543579586998981914238753034987512692844419582498779915941`70.^-6 and 1.243798561618425968533853139468121686264826907454906286861264969523907`70.^-13 for the integral and error estimates. >> -2.100045775865940530 Symbolic integration doesn;t give answer in Mathematica. Is there anyway to speed up while doing symbolic integration in this case/ Attachments:

I am getting error while evaluating following integral in Mathematica. Even though I am getting result I am sure it is wrong.

    ClearAll["GLOBAL'*"]
    {r, a, b, Z, A, B,phi } = {4.087, 1.205, 0.3812, 0, 345.0527, 606741.04395, Pi/3} // Rationalize[#, 0] &;
    JJ[n_] :=  r NIntegrate[1/((t Sin[phi])^2 + r^2 + (t Cos[phi] + Z - z)^2 - 2 r t Sin[phi] Cos[eta])^
    n, {eta, 0, 2 Pi}, {z, 0, 50}, {t, -5/3, 5/3}, MinRecursion -> 10, MaxRecursion -> 35, WorkingPrecision -> 20];
    a b (-A JJ[3] + B JJ[6])

NIntegrate::eincr: The global error of the strategy GlobalAdaptive has increased more than 2000 times. The global error is expected to decrease monotonically after a number of integrand evaluations. Suspect one of the following: the working precision is insufficient for the specified precision goal; the integrand is highly oscillatory or it is not a (piecewise) smooth function; or the true value of the integral is 0. Increasing the value of the GlobalAdaptive option MaxErrorIncreases might lead to a convergent numerical integration. NIntegrate obtained 0.01443218400405474348879705436971438392688666301951432535504827223787189`70. and 1.034077729663448519046495248747417797443269722765919064706177094984896`70.*^-10 for the integral and error estimates. >>

NIntegrate::eincr: The global error of the strategy GlobalAdaptive has increased more than 2000 times. The global error is expected to decrease monotonically after a number of integrand evaluations. Suspect one of the following: the working precision is insufficient for the specified precision goal; the integrand is highly oscillatory or it is not a (piecewise) smooth function; or the true value of the integral is 0. Increasing the value of the GlobalAdaptive option MaxErrorIncreases might lead to a convergent numerical integration. NIntegrate obtained 6.363900979126543579586998981914238753034987512692844419582498779915941`70.*^-6 and 1.243798561618425968533853139468121686264826907454906286861264969523907`70.*^-13 for the integral and error estimates. >>

-2.100045775865940530

Symbolic integration doesn;t give answer in Mathematica. Is there anyway to speed up while doing symbolic integration in this case/

POSTED BY: sahad dubai

11 Replies

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sahad dubai

Posted 9 years ago

POSTED BY: sahad dubai

Neil Singer

Neil Singer, AC Kinetics, Inc.

Posted 9 years ago

POSTED BY: Neil Singer

sahad dubai

Posted 9 years ago

Thank you for your kind cooperation. Is it possible to use symbolic integration using Integrate command. Using one of their two methods I am getting their answer. The one which I have asked here I am not getting their answer that makes me thinking there is some problem with result MMA is giving. The spikes consist of very small region and other space the function goes to zero. So I have a doubt that whether MMA correctly calculating it or Is it missing those spikes.I have read something about such cases.

POSTED BY: sahad dubai

Neil Singer

Neil Singer, AC Kinetics, Inc.

Posted 9 years ago

POSTED BY: Neil Singer

sahad dubai

Posted 9 years ago

Thank you.I tried plotting the integrand for \phi=0 and \phi=pi/2. There is a spike for \phi=pi/2 which is missing in case of \phi=0. Is that spike causing problems. Please see attachment. Attachments:

POSTED BY: sahad dubai

Neil Singer

Neil Singer, AC Kinetics, Inc.

Posted 9 years ago

POSTED BY: Neil Singer

sahad dubai

Posted 9 years ago

I am also thinking the same. But when \phi=0 I get the same value as theirs which can rule out the possibility of a typo. How you came to a conclusion that it is smooth. Did you plot the function by keeping one of the integration variable as a constant. Is it possible to do a symbolic integration rather than numerical integration. I tried it but it takes too long which forced me to kill the job. For \phi=0 the computation time is very less compared for non zero \phi's and without any error messages.

POSTED BY: sahad dubai

Neil Singer

Neil Singer, AC Kinetics, Inc.

Posted 9 years ago

I tried several algorithms (using method->) and get the same answer. I believe you likely have a small discrepancy with your problem source (or they have a typo). I doubt that Mathematica is wrong here because the function is smooth -- there are no numerical instabilities (such as discontinuities). I would be surprised if the answer is wrong.

POSTED BY: Neil Singer

sahad dubai

Posted 9 years ago

Thank you. The answer is -2.00015 which is not matching with answer in the source from which I have taken this problem. The answer from their calculation is -4.4. I dont need 20 digits 5 or 6 is ok. but I am really confused with what is causing such a difference. Is it possible to evaluate it symbolically using Integrate command rather than numerical integration.