Use Scipy.Integrate.Quad to Integrate Complex Numbers

Use scipy.integrate.quad to integrate complex numbers

What's wrong with just separating it out into real and imaginary parts? scipy.integrate.quad requires the integrated function return floats (aka real numbers) for the algorithm it uses.

import scipy
from scipy.integrate import quad

def complex_quadrature(func, a, b, **kwargs):
    def real_func(x):
        return scipy.real(func(x))
    def imag_func(x):
        return scipy.imag(func(x))
    real_integral = quad(real_func, a, b, **kwargs)
    imag_integral = quad(imag_func, a, b, **kwargs)
    return (real_integral[0] + 1j*imag_integral[0], real_integral[1:], imag_integral[1:])

E.g.,

>>> complex_quadrature(lambda x: (scipy.exp(1j*x)), 0,scipy.pi/2)
((0.99999999999999989+0.99999999999999989j),
 (1.1102230246251564e-14,),
 (1.1102230246251564e-14,))

which is what you expect to rounding error - integral of exp(i x) from 0, pi/2 is (1/i)(e^i pi/2 - e^0) = -i(i - 1) = 1 + i ~ (0.99999999999999989+0.99999999999999989j).

And for the record in case it isn't 100% clear to everyone, integration is a linear functional, meaning that ∫ { f(x) + k g(x) } dx = ∫ f(x) dx + k ∫ g(x) dx (where k is a constant with respect to x). Or for our specific case ∫ z(x) dx = ∫ Re z(x) dx + i ∫ Im z(x) dx as z(x) = Re z(x) + i Im z(x).

If you are trying to do a integration over a path in the complex plane (other than along the real axis) or region in the complex plane, you'll need a more sophisticated algorithm.

Note: Scipy.integrate will not directly handle complex integration. Why? It does the heavy lifting in the FORTRAN QUADPACK library, specifically in qagse.f which explicitly requires the functions/variables to be real before doing its "global adaptive quadrature based on 21-point Gauss–Kronrod quadrature within each subinterval, with acceleration by Peter Wynn's epsilon algorithm." So unless you want to try and modify the underlying FORTRAN to get it to handle complex numbers, compile it into a new library, you aren't going to get it to work.

If you really want to do the Gauss-Kronrod method with complex numbers in exactly one integration, look at wikipedias page and implement directly as done below (using 15-pt, 7-pt rule). Note, I memoize'd function to repeat common calls to the common variables (assuming function calls are slow as if the function is very complex). Also only did 7-pt and 15-pt rule, since I didn't feel like calculating the nodes/weights myself and those were the ones listed on wikipedia, but getting reasonable errors for test cases (~1e-14)

import scipy
from scipy import array

def quad_routine(func, a, b, x_list, w_list):
    c_1 = (b-a)/2.0
    c_2 = (b+a)/2.0
    eval_points = map(lambda x: c_1*x+c_2, x_list)
    func_evals = map(func, eval_points)
    return c_1 * sum(array(func_evals) * array(w_list))

def quad_gauss_7(func, a, b):
    x_gauss = [-0.949107912342759, -0.741531185599394, -0.405845151377397, 0, 0.405845151377397, 0.741531185599394, 0.949107912342759]
    w_gauss = array([0.129484966168870, 0.279705391489277, 0.381830050505119, 0.417959183673469, 0.381830050505119, 0.279705391489277,0.129484966168870])
    return quad_routine(func,a,b,x_gauss, w_gauss)

def quad_kronrod_15(func, a, b):
    x_kr = [-0.991455371120813,-0.949107912342759, -0.864864423359769, -0.741531185599394, -0.586087235467691,-0.405845151377397, -0.207784955007898, 0.0, 0.207784955007898,0.405845151377397, 0.586087235467691, 0.741531185599394, 0.864864423359769, 0.949107912342759, 0.991455371120813]
    w_kr = [0.022935322010529, 0.063092092629979, 0.104790010322250, 0.140653259715525, 0.169004726639267, 0.190350578064785, 0.204432940075298, 0.209482141084728, 0.204432940075298, 0.190350578064785, 0.169004726639267, 0.140653259715525,  0.104790010322250, 0.063092092629979, 0.022935322010529]
    return quad_routine(func,a,b,x_kr, w_kr)

class Memoize(object):
    def __init__(self, func):
        self.func = func
        self.eval_points = {}
    def __call__(self, *args):
        if args not in self.eval_points:
            self.eval_points[args] = self.func(*args)
        return self.eval_points[args]

def quad(func,a,b):
    ''' Output is the 15 point estimate; and the estimated error '''
    func = Memoize(func) #  Memoize function to skip repeated function calls.
    g7 = quad_gauss_7(func,a,b)
    k15 = quad_kronrod_15(func,a,b)
    # I don't have much faith in this error estimate taken from wikipedia
    # without incorporating how it should scale with changing limits
    return [k15, (200*scipy.absolute(g7-k15))**1.5]

Test case:

>>> quad(lambda x: scipy.exp(1j*x), 0,scipy.pi/2.0)
[(0.99999999999999711+0.99999999999999689j), 9.6120083407040365e-19]

I don't trust the error estimate -- I took something from wiki for recommended error estimate when integrating from [-1 to 1] and the values don't seem reasonable to me. E.g., the error above compared with truth is ~5e-15 not ~1e-19. I'm sure if someone consulted num recipes, you could get a more accurate estimate. (Probably have to multiple by (a-b)/2 to some power or something similar).

Recall, the python version is less accurate than just calling scipy's QUADPACK based integration twice. (You could improve upon it if desired).

Scipy quad integral of imaginary numbers

This is my interpretation of your integral

The answer about how to generalize the integral can be extended to double integral (provided by dblquad function).

import numpy as np
from scipy.integrate import dblquad

def complex_dblquad(func, a, b, g, h, **kwargs):
    def real_func(z, x):
        return np.real(func(z, x))
    def imag_func(z, x):
        return np.imag(func(z, x))
    real_integral = dblquad(real_func, a, b, g, h, **kwargs)
    imag_integral = dblquad(imag_func, a, b, g, h, **kwargs)
    return (real_integral[0] + 1j*imag_integral[0], real_integral[1:], imag_integral[1:])

Then you compute your integral as follows

complex_dblquad(lambda z,x: np.exp(1j*x*z), 0, 1, 0.3, 0.9)

Python's scipy.integrate.quad with complex integration bounds

The integral should be taken over the horizontal half-line going from a to the right. (This will fail if a is a negative real number, but that is a branch cut territory anyway). This halfline is parameterized by a+t where t is real and goes from 0 to infinity. So, integrate exp(-(a+t))/(a+t) from 0 to infinity.

Also, quad from SciPy requires real-valued functions, so separate it into a real and imaginary part. And don't forget that the function must be passed as a callable, like lambda t: np.exp(-t), not just exp(-t)

from scipy import integrate
myfun2_re = integrate.quad(lambda t: np.real(np.exp(-(a+t))/(a+t)), 0, np.inf)[0]
myfun2_im = integrate.quad(lambda t: np.imag(np.exp(-(a+t))/(a+t)), 0, np.inf)[0]
myfun2 = myfun2_re + 1j*myfun2_im
print(myfun2)

This prints

(0.3411120086192922-0.36240971724285814j)

compared to myfun,

array([ 0.34111201-0.36240972j])

By the way, there is no need to wrap myfun into an array if you just want to convert a single number: complex(gammainc(...)) would do.

Scipy: Speeding up calculation of a 2D complex integral

You can gain a factor of about 10 in speed by using Cython, see below:

In [87]: %timeit cythonmodule.doit(lam=lam, y0=y0, zxp=zxp, z=z, k=k, ra=ra)
1 loops, best of 3: 501 ms per loop
In [85]: %timeit doit()
1 loops, best of 3: 4.97 s per loop

This is probably not enough, and the bad news is that this is probably
quite close (maybe factor of 2 at most) to everything-in-C/Fortran speed
--- if using the same algorithm for adaptive integration. (scipy.integrate.quad
itself is already in Fortran.)

To get further, you'd need to consider different ways to do the
integration. This requires some thinking --- can't offer much from
the top of my head now.

Alternatively, you can reduce the tolerance up to which the integral
is evaluated.

# Do in Python
#
# >>> import pyximport; pyximport.install(reload_support=True)
# >>> import cythonmodule

cimport numpy as np
cimport cython

cdef extern from "complex.h":
    double complex csqrt(double complex z) nogil
    double complex cexp(double complex z) nogil
    double creal(double complex z) nogil
    double cimag(double complex z) nogil

from libc.math cimport sqrt

from scipy.integrate import dblquad

cdef class Params:
    cdef public double lam, y0, k, zxp, z, ra

    def __init__(self, lam, y0, k, zxp, z, ra):
        self.lam = lam
        self.y0 = y0
        self.k = k
        self.zxp = zxp
        self.z = z
        self.ra = ra

@cython.cdivision(True)
def integrand_real(double x, double y, Params p):
    R1 = sqrt(x**2 + (y-p.y0)**2 + p.z**2)
    R2 = sqrt(x**2 + y**2 + p.zxp**2)
    return creal(cexp(1j*p.k*(R1-R2)) * (-1j*p.z/p.lam/R2/R1**2) * (1+1j/p.k/R1))

@cython.cdivision(True)
def integrand_imag(double x, double y, Params p):
    R1 = sqrt(x**2 + (y-p.y0)**2 + p.z**2)
    R2 = sqrt(x**2 + y**2 + p.zxp**2)
    return cimag(cexp(1j*p.k*(R1-R2)) * (-1j*p.z/p.lam/R2/R1**2) * (1+1j/p.k/R1))

def ymax(double x, Params p):
    return sqrt(p.ra**2 + x**2)

def doit(lam, y0, k, zxp, z, ra):
    p = Params(lam=lam, y0=y0, k=k, zxp=zxp, z=z, ra=ra)
    rr, err = dblquad(integrand_real, -ra, ra, lambda x: -ymax(x, p), lambda x: ymax(x, p), args=(p,))
    ri, err = dblquad(integrand_imag, -ra, ra, lambda x: -ymax(x, p), lambda x: ymax(x, p), args=(p,))
    return rr + 1j*ri

scipy.integrate.quad precision on big numbers

I believe the issue is due to np.exp(-x) quickly becoming very small as x increases, which results in evaluating as zero due to limited numerical precision. For example, even for x as small as x=10**2*, np.exp(-x) evaluates to 3.72007597602e-44, whereas x values of order 10**3 or above result in 0.

I do not know the implementation specifics of quad, but it probably performs some kind of sampling of the function to be integrated over the given integration range. For a large upper integration limit, most of the samples of np.exp(-x) evaluate to zero, hence the integral value is underestimated. (Note that in these cases the provided absolute error by quad is of the same order as the integral value which is an indicator that the latter is unreliable.)

One approach to avoid this issue is to restrict the integration upper bound to a value above which the numerical function becomes very small (and, hence, contributes marginally to the integral value). From your code snipet, the value of 10**4 appears to be a good choice, however, a value of 10**2 also results in an accurate evaluation of the integral.

Another approach to avoid numerical precision issues is to use a module that performs computation in arbitrary precision arithmetic, such as mpmath. For example, for x=10**5, mpmath evaluates exp(-x) as follows (using the native mpmath exponential function)

import mpmath as mp
print(mp.exp(-10**5))

3.56294956530937e-43430

Note how small this value is. With the standard hardware numerical precision (used by numpy) this value becomes 0.

mpmath offers an integration function (mp.quad), which can provide an accurate estimate of the integral for arbitrary values of the upper integral bound.

import mpmath as mp

print(mp.quad(lambda x : .5 * mp.exp(-.5 * x), [0, mp.inf]))
print(mp.quad(lambda x : .5 * mp.exp(-.5 * x), [0, 10**13]))
print(mp.quad(lambda x : .5 * mp.exp(-.5 * x), [0, 10**8]))
print(mp.quad(lambda x : .5 * mp.exp(-.5 * x), [0, 10**5]))

1.0
0.999999650469474
0.999999999996516
0.999999999999997

We can also obtain even more accurate estimates by increasing the precision to, say, 50 decimal points (from 15 which is the standard precision)

mp.mp.dps = 50; 

print(mp.quad(lambda x : .5 * mp.exp(-.5 * x), [0, mp.inf]))
print(mp.quad(lambda x : .5 * mp.exp(-.5 * x), [0, 10**13]))
print(mp.quad(lambda x : .5 * mp.exp(-.5 * x), [0, 10**8]))
print(mp.quad(lambda x : .5 * mp.exp(-.5 * x), [0, 10**5]))

1.0
0.99999999999999999999999999999999999999999829880262
0.99999999999999999999999999999999999999999999997463
0.99999999999999999999999999999999999999999999999998

In general, the cost for obtaining this accuracy is an increased computation time.

P.S.: It goes without saying that if you are able to evaluate your integral analytically in the first place (e.g., with the help of Sympy) you can forget all the above.

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