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Real-valued integral over real variable returns complex result(s)

Jerry
Posted 12 years ago

Why does this real-valued integral over a real variable return a complex-valued result?

fSine[x_] := Piecewise[{{1/(\[Pi]*Sqrt[1 - x^2]), -1 < x < 1}, {0, True}}];

Integrate[fSine[x]*fSine[1/10 - x], {x, -2, 2}];
N[%]
0.444207 - 1.815 I

Sanity check:

Integrate[fSine[x]*fSine[1/10 - x], {x, -2, 2}, Assumptions -> x \[Element] Reals];
N[%]
0.444207 - 1.815 I

Compare to NIntegrate result:

NIntegrate[fSine[x]*fSine[1/10 - x], {x, -2, 2}]
0.444207 - 5.75862*10^-11 I

  • Why is there a non-zero imaginary part for both Integrate and NIntegrate?

  • Why are the two answers different?

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POSTED BY: Jerry
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This might be useful:

 fSine[x_] := HeavisidePi[x/2]/(\[Pi]*Sqrt[1 - x^2]);
    Chop@N[Integrate[fSine[x]*fSine[1/10 - x], {x, -2, 2}], 20]

and 

N[Quiet[NIntegrate[fSine[x]*fSine[1/10 - x], {x, -2, 2}, WorkingPrecision -> 45, MaxRecursion -> 1000]], 20]

both give:

0.44420658153737745178

These work both with the Heaviside definition. We can achieve something similar with the original Piecewise definition. 

fSine[x_] := 
  Piecewise[{{1/(\[Pi]*Sqrt[1 - x^2]), -1 < x < 1}, {0, True}}];

Integrate[fSine[x]*fSine[1/10 - x], {x, -2, 2}, 
  GenerateConditions -> False, PrincipalValue -> True];
Chop@N[%, 20]

which gives:

0.44420658153737745178

,too.

Cheers,

Marco

PS: Oh, yes, and 

SetPrecision[N[Simplify@Integrate[fSine[x]*fSine[1/10 - x], {x, -2, 2}, GenerateConditions -> False, PrincipalValue -> True], 60], Infinity]

suggests that the imaginary part is really zero...
POSTED BY: Marco Thiel
Updating Name
Updating Name
Posted 12 years ago

The symbolic indefinite integral is quite a beast, and includes imaginary terms. This is common: mathematics often forces Mathematica to express a real result in terms of complex numbers. Unfortunately, in this difficult case, Mathematica also seems to misplace a branch cut when computing the definite integral. That's fairly common. You could report it as a bug, and maybe someone will figure out a way to fix this case.

The break points in Piecewise are at the singularities. That's tough for both symbolic and numerical methods. Here, NIntegrate appears by default to do some symbolic processing at the singularities to help out. It apparently has to go complex here, and the poorly cancelled imaginary part results. If you set Method -> {Automatic, "SymbolicProcessing" -> False} you'll get a real result of poor accuracy after complaints of slow convergence.

You can get a better symbolic result by using HeavisidePI instead of Piecewise here:

fSine[x_] := HeavisidePi[x/2]/(\[Pi]*Sqrt[1 - x^2])
Integrate[fSine[x]*fSine[1/10 - x], {x, -2, 2}] // N
NIntegrate[fSine[x]*fSine[1/10 - x], {x, -2, 2}]
(*0.444207 + 0. I*)
(*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 9 recursive bisections in x near {x} = {-0.8985}. NIntegrate obtained 0.4371297669288421` and 0.0061209610092931716` for the integral and error estimates. >>*)
(*0.43713*)

HeavisidePI tends to be more friendly to calculus than Piecewise. You still get the +0. I in the numerical result from the symbolic integral because it involves complex numbers whose imaginary parts cancel. If that's a problem, use Chop or Re when you know the result is real. The NIntegrate result isn't so good with HeavisidePI. It's similar to the result using Piecewise without symbolic processing: real but innaccurate. Some Method setting might work better: there are so many because there's no predicting what method might work best.
POSTED BY: Updating Name
POSTED BY: John Doty
POSTED BY: Nasser M. Abbasi
Jerry
Jerry
Posted 12 years ago
POSTED BY: Jerry
Jerry
Jerry
Posted 12 years ago

Thanks for the tip on preferring HeavisidePi over Piecewise.

Seems like Mathematica should clean up after itself better, wrt the small imaginary part. I’ll send a bug report—at least the symbolic version should get fixed.

FWIW this example comes out of finding the PDF for a sum of random sine variables—convolution. I have a fallback method that doesn’t do the convolutions.
POSTED BY: Jerry
Jerry
Jerry
Posted 12 years ago

But in my example, NIntegrate did not return a non-zero imaginary part. It is small, and one may suspect machine precision problems, but It is not zero. Having a non-zero imaginary part messes up any number of subsequent uses of the result.

I would further maintain that your result, while numerically correct, is at least marginally wrong or inconvenient because it returns any imaginary part. If I perform nearly any other integral such as this, such as

In[6]:= NIntegrate[Sin[x], {x, 0, 1}]

Out[6]= 0.459698

I don’t get “+ 0. I” appended to the end.

I think both of the examples I give, Integrate and NIntegrate, are wrong—buggy. Should I file a bug report?
POSTED BY: Jerry
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