Using scipy.quad with iε trick: Bad results

Question

In order to circumvent the cauchy principle value, I tried to integrate an integral using a small shift iε into the complex plane to evade the pole. However, as can be inferred from the figure below, the result is pretty bad. The code for this result is shown below. Do you have ideas how to improve this method? Why is it not working? I already tried changing ε or the limit in the integral.

Edit: I included the method "cauchy" with the principle value, which seems not to work at all.

import matplotlib.pyplot as plt
from scipy.integrate import quad
import numpy as np 

def cquad(func, a, b, **kwargs):
    real_integral = quad(lambda x: np.real(func(x)), a, b, limit = 200,**kwargs)
    imag_integral = quad(lambda x: np.imag(func(x)), a, b, limit = 200,**kwargs)
    return (real_integral[0] + 1j*imag_integral[0], real_integral[1:], imag_integral[1:])

def k_(a):
    ϵ = 1e-32
    return (cquad(lambda x: np.exp(-1j*x)/(x**2 - a**2 - 1j*ϵ),-np.inf,np.inf)[0])

def k2_(a):
    return (cquad(lambda x: np.exp(-1j*x)/(x**2 - a**2),-1e6,1e6, weight='cauchy', wvar = a)[0])

k  = np.vectorize(k_)
k2 = np.vectorize(k2_)

fig, ax = plt.subplots()

a = np.linspace(-10,10,300)
ax.plot(a,np.real(k(a)),".-",label = "numerical result")
ax.plot(a,np.real(k2(a)),".-",label = "numerical result (cauchy)")
ax.plot(a, - np.pi*np.sin(a)/a,"-",label="analytical result")
ax.set_ylim(-5,5)
ax.set_ylabel("f(x)")
ax.set_xlabel("x")
ax.set_title(r"$\mathcal{P}\int_{-\infty}^{\infty} \frac{e^{-i y}}{y^2 - x^2}\mathrm{d}y = -\frac{\pi\sin(x)}{x}$")
plt.legend()
plt.savefig("./bad_result.png")
plt.show()

Answer 1

The main problem is that the integrand has poles at both x=a and x=-a. ev-br's post show how to deal with a pole at x=a. All that's needed then is to find a way to massage the integral into a form that avoids integrating through the other pole at x=-a. Taking advantage of evenness allows us to "fold the integral over", so instead of having two poles we just need to deal with one pole at x=a.

---

The real part of

np.exp(-1j*x) / (x**2 - a**2) = (np.cos(x) - 1j * np.sin(x)) / (x**2 - a**2)

is an even function of x so integrating the real part from x = -infinity to infinity would equal twice the integral from x = 0 to infinity. The imaginary part of the integrand is an odd function of x. The integral from x = -infinity to infinity equals the integral from x = -infinity to 0, plus the integral from x = 0 to infinity. These two parts cancel each other out since the (imaginary) integrand is odd. So the integral of the imaginary part equals 0.

Finally, using ev-br's suggestion, since

1 / (x**2 - a**2) = 1 / ((x - a)(x + a))

using weight='cauchy', wvar=a implicitly weights the integrand by 1 / (x - a) thus allowing us to reduce the explicit integrand to

np.cos(x) / (x + a)

Since the integrand is an even function of a, we can assume without loss of generality that a is positive:

a = abs(a)

Now integrating from x = 0 to infinity avoids the pole at x = -a.

---

import matplotlib.pyplot as plt
from scipy.integrate import quad
import numpy as np


def cquad(func, a, b, **kwargs):
    real_integral = quad(lambda x: np.real(func(x)), a, b, limit=200, **kwargs)
    imag_integral = quad(lambda x: np.imag(func(x)), a, b, limit=200, **kwargs)
    return (real_integral[0] + 1j*imag_integral[0], real_integral[1:], imag_integral[1:])


def k2_(a):
    a = abs(a)
    # return 2*(cquad(lambda x: np.exp(-1j*x)/(x + a), 0, 1e6, weight='cauchy', wvar=a)[0]) # also works
    # return 2*(cquad(lambda x: np.cos(x)/(x + a), 0, 1e6, weight='cauchy', wvar=a)[0]) # also works, but not necessary
    return 2*quad(lambda x: np.cos(x)/(x + a), 0, 1e6, limit=200, weight='cauchy', wvar=a)[0]


k2 = np.vectorize(k2_)

fig, ax = plt.subplots()

a = np.linspace(-10, 10, 300)
ax.plot(a, np.real(k2(a)), ".-", label="numerical result (cauchy)")
ax.plot(a, - np.pi*np.sin(a)/a, "-", label="analytical result")
ax.set_ylim(-5, 5)
ax.set_ylabel("f(x)")
ax.set_xlabel("x")
ax.set_title(
    r"$\mathcal{P}\int_{-\infty}^{\infty} \frac{e^{-i y}}{y^2 - x^2}\mathrm{d}y = -\frac{\pi\sin(x)}{x}$")
plt.legend()
plt.show()

Answer 2

You can use instead the weight="cauchy" argument to quad. https://docs.scipy.org/doc/scipy/reference/generated/scipy.integrate.quad.html