numerical_integral (GSL C library) does not work with complex integrands?

asked 2018-03-12 16:30:59 +0100

joaoff
220 ●1 ●6 ●21

I am unable to use numerical_integral (GSL C library) with a complex integrand. I haven't found anything about this issue in the documentation. Is this correct? nintegral works correctly.

e.g.

f(x) = (exp(i*x) + exp(-i*x))/2
f(x).nintegral(x,0,1)
numerical_integral(f,0,1)

the above code outputs

(0.8414709848078965, 9.34220461887732e-15, 21, 0)
Error in lines 3-3
...
TypeError: unable to coerce to a real number

Tested with SageMath version 7.5.1 and Cocalc.

Cordially

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answered 2018-03-12 19:14:58 +0100

dan_fulea
5537 ●5 ●43 ●95

In the documentation of numerical_integral there is no explicit statement that the function should be (defined on some proper or improper interval) with values in $\mathbb R$, but the following sentences in the doc string strongly suggest this is the case:

Note the coercion to the real field RR, which prevents underflow:

      sage: f = exp(-x**2)
      sage: numerical_integral(f, -Infinity, +Infinity)  # random output
      (1.7724538509060035, 3.4295192165889879e-08)

and

One can integrate any real-valued callable function:

      sage: numerical_integral(lambda x: abs(zeta(x)), [1.1,1.5])  # random output
      (1.8488570602160455, 2.052643677492633e-14)

In our case the solution / a work-around is simple, we just isolate the imaginary and the real parts of the function to be integrated.

sage: f(x) = (exp(i*x) + exp(-i*x))/2
sage: numerical_integral( real(f), (0,1) )
(0.8414709848078964, 9.342204618877319e-15)
sage: numerical_integral( imag(f), (0,1) )
(0.0, 0.0)

which matches

sage: numerical_integral( cos(x), (0,1) )
(0.8414709848078964, 9.342204618877319e-15)

Further examples with different, but similar data:

sage: numerical_integral( lambda x: exp(i*x).real(), (0, pi/6) )
(0.4999999999999999, 5.551115123125781e-15)
sage: numerical_integral( lambda x: exp(i*x).imag(), (0, pi/6) )
(0.13397459621556138, 1.487416814333746e-15)

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In general, the integral of a complex function depends on the integration path if the function has any singularities ; see this for a smalll introduction to a vast subject.

This answer assumes that the integration path is the segment $[0~1]$ of $\mathbb{R}$. What does justify this assumption ?

Emmanuel Charpentier ( 2018-03-12 19:46:33 +0100 )edit

Thank you for the replies. In fact, I was trying to optimize my code finding a replacer for nintegral(), that works for me, but seems to be much slower than the equivalent function in Mathematica.

joaoff ( 2018-03-13 10:45:17 +0100 )edit

@Emmanuel Charpentier The question also assumes the integration path is the segment $[0,1]$ in $\mathbb R$. Moreover, the integrand is entire, so given this example and still the question, the only point of the question could have been the functionality of numerical_integral. If we really need to understand monodromy, we can still reduce the numerical calculus to the above:

def myint(f, P=0, eps=0.1):
    var( 't' );
    path(t)  = P + eps*exp(i*t)
    g(t) = f( path(t) ) * diff( path(t), t )
    return ( numerical_integral( lambda u: real(g(u)), (0, 2*pi) )[0]
             +
             numerical_integral( lambda u: imag(g(u)), (0, 2*pi) )[0] * i  )

Then myint(log), myint(log,0,1) shows "it".

dan_fulea ( 2018-03-14 00:47:37 +0100 )edit

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