nuqad 的用法

$$\begin{align} \int_0^1\mathrm{d}y \int_y^1 \mathrm{d}x \cdot(x^2 + y) = \frac{5}{12} \end{align}$$

figalt

from scipy.integrate import nquad


def func(x, y):
    return x**2 + y


def range_x(y):
    return [0, y]


# 先积 x (y, 1), 再积 y(0, 1)
res = nquad(func, [range_x, [0, 1]])
print(res)
print(5/12)
(0.41666666666666663, 1.473075555508962e-14)
0.4166666666666667

code

定积分换元到质心系

$$\begin{align} \int_0^{2\pi} \mathrm{d}\phi_k \int_0^{2\pi}\mathrm{d}\phi_q \cdot f(\phi_k - \phi_q) = \frac{1}{2}\int_{-2\pi}^{2\pi} \mathrm{d}\phi_- \cdot l(\phi_-) f(\phi_-) \end{align}$$

where

$$\begin{align} \phi_+ =& \phi_k + \phi_q \\ \phi_- =& \phi_k - \phi_q \\ l(\phi_-) =& \left\{\matrix{4\pi - 2\phi_-, \quad \phi_->0\\ 4\pi + 2\phi_-, \quad \phi_->0}\right. \end{align}$$

figalt

import numpy as np
from scipy.integrate import quad
from scipy.integrate import nquad


def l_bound(phi_minus):
    if phi_minus > 0:
        l_bound = 4*np.pi - 2*phi_minus
    else:
        l_bound = 4*np.pi + 2*phi_minus
    return l_bound


def f(phi_minus):
    res = phi_minus + 2*phi_minus**2 + 3*phi_minus**3 - 3**phi_minus
    return res


center = quad(lambda phi_minus: l_bound(phi_minus)*f(phi_minus)/2, -2*np.pi,
              2*np.pi)


normal = nquad(lambda phi_k, phi_q: f(phi_k-phi_q), [[0, 2*np.pi],
                                                     [0, 2*np.pi]])
print(center)
print(normal)
(-303.255886295168, 2.462030579408747e-11)
(-303.2558862951685, 2.2693158156346615e-11)

code

Reference

  • [[https://docs.scipy.org/doc/scipy/reference/generated/scipy.integrate.nquad.html]]
  • Stewart, J. Calculus. (Cengage Learning, 2016). Chap 15.9, Change of Variables in Multiple Integrals