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I wish to understand how to cope with the following error messages:

k = 0.5;
λ = 0.4;
W[t_] := (E^(-t/λ)) ((t)^(k - 1)) GammaRegularized[0.5`, 2.5` t] GammaRegularized[1.5`, 7.5` t]
F[τ_] := NIntegrate[E^((-3 - I 0.16 ) t) W[t] , {t, 0, τ}, MaxRecursion -> 50]
F[∞]

During evaluation of In[113]:= NIntegrate::deodiv: DoubleExponentialOscillatory returns a finite integral estimate, but the integral might be divergent. >>

During evaluation of In[113]:= Power::infy: Infinite expression 1/0. encountered. >>

During evaluation of In[113]:= NIntegrate::deorel: The relative error ComplexInfinity is larger than expected for the integrand -(((0. +1. I) E^(-5.5 (0. +t)) GammaRegularized[0.5,2.5 (0. +t)] GammaRegularized[1.5,7.5 (0. +t)] Sin[0.16 t])/(0. +t)^0.5) over {0,[Infinity]} with DoubleExponentialOscillatory method and tuning parameters, TuningParameters -> {10,5}. >>

During evaluation of In[113]:= NIntegrate::deoncon: DoubleExponentialOscillatory has failed to converge for the integrand -(((0. +1. I) E^(-5.5 (0. +t)) GammaRegularized[0.5,2.5 (0. +t)] GammaRegularized[1.5,7.5 (0. +t)] Sin[0.16 t])/(0. +t)^0.5) over {0,[Infinity]}. DoubleExponentialOscillatory obtained 0. -0.00207921 I and ComplexInfinity for the integral and error estimates. >>

(* 0.408534 - 0.00207501 I*)

While it seems clear that the integral converges quite fast:

Plot[{Re[F[t]], Im[F[t]]}, {t, 0, 4}, PlotLegends -> "Expressions"]

enter image description here

Perhaps I should change the integration method, but I do not have clue on how to proceed...

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The symbolic processing is finding a high frequency oscillatory function. Let's just process the thing numerically:

F[τ_] := NIntegrate[E^((-3 - I 0.16) t) W[t], {t, 0, τ}, 
                    Method -> {"GlobalAdaptive", "SymbolicProcessing" -> 0}]
F[∞]

(* 0.408534 - 0.00207501 I *)
Improve this answer
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3
  • $\begingroup$ Exponential factor should be E^((-3 - I 16/100) $\endgroup$ – Bob Hanlon Oct 22 '15 at 6:15
  • $\begingroup$ Thanks! could you also suggest why exact arithmetic is necessary? $\endgroup$ – altroware Oct 22 '15 at 12:41
  • $\begingroup$ @BobHanlon You're right, thanks!. With that change my old answer stopped working. This new one does work. $\endgroup$ – Dr. belisarius Oct 22 '15 at 13:22

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