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I need to calculate an integral like

B[a_, t_] :=  1/(2 I) (t)^(9/2) NIntegrate[ x^(1/2) (HankelH1[a, x]^3 HankelH2[a, t]^3 -  HankelH2[a, x]^3 HankelH1[a, t]^3), {x, t, 2}, WorkingPrecision -> 40,  MinRecursion -> 10,  MaxRecursion -> 20] // N

$t$ ranges from Exp[-60] to 1. $a$ ranges from 0 to 1.5. Notice that $t$ appears in the lower integration limit, in the argument of Hankel functions and as $t^{9/2}$. Moreover, the integrand is purely imaginary. The integrand diverges in the limit $t\rightarrow 0$. For $a\equiv1.5$, the analytical result predicts a logarithmic divergence for $t\rightarrow 0$. The problem is that 

B[1.5,Exp[-20]]=5.5
B[1.5,Exp[-30]]=-5.89145*10^22

I have tried to increase the precision, but results do not change. I suspect that the Hankel functions are evaluated with too low precision and hence I get wrong results. You can see the (analytical) logarithmic divergence plotting

G[x_, a_, t_] := (t)^(9/2) Im[x^(1/2) (HankelH1[a, x]^3 HankelH2[a, t]^3-HankelH2[a, x]^3 HankelH1[a, t]^3)]
LogLinearPlot[{G[x, 1.5, Exp[-20]], 1/x}, {x, Exp[-10], 2}, PlotRange -> {{0, Exp[-10]}, {0, 100}}, PlotPoints -> 10000]

I am interested in the behavior of the function both for fixed $a$ and for fixed $t$ (but the former is more important).

How can I evaluate this integral? Thanks in advance to those who will try to help me.

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    – Michael E2
    Commented Aug 2, 2016 at 16:42

1 Answer 1

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Clear[B]

Add an optional third argument to control the WorkingPrecision

B[a_?NumericQ, t_?NumericQ, wp_:40] := 
 1/(2 I) (t)^(9/2) NIntegrate[
    x^(1/2) (HankelH1[a, x]^3 HankelH2[a, t]^3 - 
       HankelH2[a, x]^3 HankelH1[a, t]^3), {x, t, 2}, WorkingPrecision -> wp, 
    MinRecursion -> 10, MaxRecursion -> 20] // N

If you provide input values with $MachinePrecision then the calculation will be done with $MachinePrecision. Either input exact numbers

B[3/2, Exp[-20]]

(*  5.46327  *)

or high arbitrary precision numbers (EDIT: see Precision: "Numbers entered in the form digits`p are taken to have precision p.")

B[1.5`45, Exp[-20]]

(*  5.46327  *)

B[3/2, Exp[-30]]

(*  8.0434  *)

B[1.5`60, Exp[-30], 50]

(*  8.0434  *)
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  • $\begingroup$ Thank you very much! This solves my issues. I admit, I'm a newbie. $\endgroup$
    – Xepto
    Commented Aug 2, 2016 at 19:12
  • $\begingroup$ I have a question about your answer. The introduction of "?NumericQ" makes the calculation more efficient? $\endgroup$
    – Xepto
    Commented Aug 2, 2016 at 19:16
  • $\begingroup$ @Xepto - because the definition of B uses a numerical technique (NIntegrate), it can only be successfully evaluated for numeric inputs. The inputs are restricted to being numeric to avoid attempting a symbolic evaluation. $\endgroup$
    – Bob Hanlon
    Commented Aug 2, 2016 at 19:23

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