3
$\begingroup$

I have a function that is real when u is real and positive.

And L[u,s,t] is an even function in s and t, and it is non-negative with above conditions.

L[u_, s_, t_] := 
Simplify[(1/(16*Pi^2*u^(1/2)*t^3))*(((
  4 t (I s + t^2) - (s - I (-2 + t) t) (s - 
      I t (2 + t)) (Log[-2 + (I s)/t + t] - 
      Log[2 + (I s)/t + t]))/(8 t^2)) + ((
  4 t (-I s + t^2) - (s + I (-2 + t) t) (s + 
      I t (2 + t)) (Log[-2 - (I s)/t + t] - 
      Log[2 - (I s)/t + t]))/(8 t^2))), 
Assumptions -> {u > 0, s \[Element] Reals, t \[Element] Reals}]

As one can check,

Plot3D does show up (means it is real valued) using u=1

u = 1
Plot3D[L[u, s, t], {s, -10^3 - 1, 10^3 - 1}, {t, 0, 10^3}]

However, this Integral

a[u_?NumericQ] := NIntegrate[
u^(5/2)/2*(Log[1 + L[u,s,t]] - L[u,s,t]/(1 + L[u,s,t])), {t, 0, 
10^3}, {s, -10^3, 10^3}, Exclusions -> {0, 0}] // Chop

And seems like this integral,returns me complex.

For instance,

a[1.]

gives

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 4.84092 +2.25059 I and 
1.8852748239554364` for the integral and error estimates.

With result of

4.84092 + 2.25059 I

where I is imaginary unit.

Why is this happening? Any help will be greatly thanked!

Thank you!

--------------------------ADDED-----------------------------------

What I really meant to do above was addition of complex conjugates

L[u_, s_, 
t_] := (1/(16*Pi^2*u^(1/2)*t^3))*(((
 4 t (I s + t^2) - (s - I (-2 + t) t) (s - 
     I t (2 + t)) (Log[-2 + (I s)/t + t] - 
     Log[2 + (I s)/t + t]))/(8 t^2)) + 
Conjugate[((
  4 t (I s + t^2) - (s - I (-2 + t) t) (s - 
      I t (2 + t)) (Log[-2 + (I s)/t + t] - 
      Log[2 + (I s)/t + t]))/(8 t^2))])

Maybe I took complex conjugate wrong before. This does give me real answers. (Maybe I missed something again?).

But the integral written above still provides me complex answer (with the error above)

Another issue I encountered was that if I perform Rationalize and FullSimplify on L[u,s,t], the plot actually looks different from the original version of the equation. But for now, I will use the original form and not worry about this.

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9
  • $\begingroup$ L[1, 0, 1] is (4 - 3 I π + Log[27])/(64 π^2) $\endgroup$
    – bbgodfrey
    Nov 26, 2017 at 16:00
  • $\begingroup$ @bbgodfrey Thank you. I think when s=0, t cannot be less or equal to 2 because of log expression has $log(-2+t)$ in it. Now I want to exclude this $s=0, t\leq 2$ line from my integration (s=0 only). How should I modify the code? I tried to search exclusion option for it, but perhaps I am doing something wrong. $\endgroup$
    – Duke Smith
    Nov 26, 2017 at 16:12
  • $\begingroup$ You also could try a second definition for L for s == 0 && t < 2. $\endgroup$
    – bbgodfrey
    Nov 26, 2017 at 16:30
  • $\begingroup$ Please let me know how you resolve this matter. There is more than meets the eye, I believe. $\endgroup$
    – bbgodfrey
    Nov 26, 2017 at 17:46
  • $\begingroup$ @bbgodfrey I took your advice above and used Piecewise option. But I did notice that Maybe I had the equation wrong. Please take a look at my added part of my question above. $\endgroup$
    – Duke Smith
    Nov 26, 2017 at 18:19

1 Answer 1

2
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I get (in V11.2.0):

a[1]

NIntegrate::slwcon: Numerical integration converging too slowly; suspect one of the following: singularity, value of the integration is 0, highly oscillatory integrand, or WorkingPrecision too small.

NIntegrate::ncvb: NIntegrate failed to converge to prescribed accuracy after 18 recursive bisections in s near {t,s} = {9.86073,465.576}. NIntegrate obtained 4.7853 +2.25347 I and 1.8445652241078074` for the integral and error estimates.

4.7853 + 2.25347 I

If I follow the advice and increase WorkingPrecision, I get an error-free result:

a[u_?NumericQ] := 
 NIntegrate[
  u^(5/2)/2*(Log[1 + L[u, s, t]] - L[u, s, t]/(1 + L[u, s, t])),
  {t, 0, 10^3}, {s, -10^3, 10^3}, Exclusions -> {0, 0}, 
  PrecisionGoal -> 8, WorkingPrecision -> 32]

a[1]
(*  0.0099144575004650534254476105632829  *)
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7
  • $\begingroup$ And yet it does not help with everything, for example, NIntegrate[1/(Exp[I*s] (0.45 Cos[s]+1)), {s, 0, 2 Pi}] $\endgroup$ Jan 5, 2022 at 8:06
  • $\begingroup$ @ВалерийЗаподовников “…every unhappy family is unhappy in its own way”: NIntegrate[1/(Exp[I*s](0.45 Cos[s] + 1)), {s, 0, 2 Pi}, Method -> “Trapezoidal”] — Is there a relationship between your integral and the one in the question? $\endgroup$
    – Michael E2
    Jan 5, 2022 at 17:08
  • $\begingroup$ You can also use Integrate on your example. — Wait, I just tried your integral. There’s no problem, with the original or with raising the working precision. What did you mean? $\endgroup$
    – Michael E2
    Jan 5, 2022 at 17:11
  • $\begingroup$ Here both print very small +I just like alpha. But yes, Integrate with doing 45/100 hack does print (40*(1 - 20/Sqrt[319])*Pi)/9, which is precise. $\endgroup$ Jan 5, 2022 at 17:45
  • $\begingroup$ What is strange is that some integrals no longer need this hack! mathematica.stackexchange.com/questions/171707/… $\endgroup$ Jan 5, 2022 at 19:23

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