I was looking for the fastest converging method to integrate a family of functions. After some tries, an old-school colleague suggested me a method that he used to use in excel to perform such task.

It relies on a simple procedure:

  1. sample the function at log-spaced sample such that $x_i = x_0 \alpha^i$ for $i=0...N$
  2. compute the values $\hat{y_i} = x_i f(x_i)$
  3. compute the integral as $\int f(x) \approx \log(\alpha)\sum{\hat{y}} $

This "trick" is based on the following equation:

$$ \int_{x_0}^{x_1}f(x)dx=\int_{x_0}^{x_1}xf(x)\frac{dx}{x} = \int_{x_0}^{x_1}xf(x)d(\log(x)) $$ since $x_i$ is logarithmically spaced,

$d(log(x)) = const. = \log(\alpha)$.

Therefore, applying the trivial rectangles rule,

$\int f(x) \approx \sum{x_i f(x_i)\left(\log(x_{i+1})-\log(x_i)\right)} = \log(\alpha)\sum{\hat{y}} $

Now... at a first look this looks like a trick to perform an integration with the rectangle rule with logarithmically spaced point and avoiding the element-wise multiplication imposing $d(log(x)) = const.$

However, when I tested this versus more standard rectangle rules, it's convergence speed was still much faster.

The question therefore is: why is this method so fast and how does it differ from a normal rectangle-rule with log-spaced evaluation points?

To support my claim, here are the results of few test I ran.

With $f(x) = \exp(-x)$ Error with exp(-x)

Since $\exp(-x)$ could be a very special case in conjunction with the logarithmically spaced points, I also tried a less "special" function (that is also of the type that I need to integrate). This is: $f(x) = \frac{1}{\left(1-\left(\frac{x}{4}\right)^2\right)^2 + \left(2\cdot0.05\cdot\frac{x}{4}\right)^2}$. (Some of you might recognize this to be the dynamic amplification factor of a resonant system with natural frequency equal to 4 and damping ratio equal to 5%.)

Mech Adm Mech Adm Error

Despite a less clear and monotonic convergence, the "magic" formula still converges faster than the rectangle rule with the same evaluation points.

Why does this happen?

Note: I initially thought this to be related with this (Integral in log-log space), but I'm now starting to think that this is a quite different topic


Following the requests of @Maxim Umansky and BlaB, here is the convergence plot in loglog scale for the two examples. It looks to me that the four methods converge with a similar power-law, but the "magic" one starts with a much lower error. (NOTE that the evaluation points for the second to fourth lines are the same)

$f(x) = \exp(-x)$

loglog convergence 1

$f(x) = \frac{1}{\left(1-\left(\frac{x}{4}\right)^2\right)^2 + \left(2\cdot0.05\cdot\frac{x}{4}\right)^2}$

loglog convergence 2


Please find below the python 3 code I used to generate these plots

import numpy as np
import matplotlib.pyplot as plt

fun = 'admittance'

if fun == 'exp':
    fun = lambda x: np.exp(-x)
    true_int = 1
elif fun == 'admittance':
    f0 = 4
    xi = 0.05
    fun = lambda x: 1/np.sqrt((1 - (x/f0)**2)**2 + (2*xi*x/f0)**2)
    true_int = 36.546222700939126

N = np.unique(np.round(np.logspace(1, 8, 1000)))

#%% plot function
x = np.linspace(0.001, 100, 200)
y = fun(x)

#%% open new figure

#%% Ractangle w/ linspaced x
err = []
for n in N:
    x = np.linspace(0.001, 100, int(n))
    y = fun(x)
    dx = x[1]-x[0]

    err.append(np.sum(y)*dx - true_int)

plt.loglog(N, np.abs(err), label='Ractangle w/ linspaced x')

#%% Ractangle w/ logspaced x
err_left = []
err_right  = []
for n in N:
    x = np.logspace(-3, 2, int(n))
    y = fun(x)
    dx = np.diff(x)

    err_left.append(np.sum(y[:-1]*dx) - true_int)
    err_right .append(np.sum(y[1: ]*dx) - true_int)

plt.loglog(N, np.abs(err_left), label='Ractangle w/ logspaced x (left)')
plt.loglog(N, np.abs(err_right ), label='Ractangle w/ logspaced x (right)')

#%% ACA
err_ACA = []
for n in N:
    x = np.logspace(-3, 2, int(n))
    y = fun(x)
    xy = x*y
    dlogx = np.log(x[1]) - np.log(x[0])

    err_ACA.append(np.sum(xy) * dlogx - true_int)

plt.loglog(N, np.abs(err_ACA), label=r'$\log(\alpha)\sum{xf(x)}$')

plt.xlabel('Number of integration points')
  • 2
    $\begingroup$ We need to see log(Error) vs. the log of number of points, then we can see the asymptotic rate, that's what matters. If the distribution of integration points is optimized for a particular kind of integrated function that should not change the asymptotic convergence rate. $\endgroup$ – Maxim Umansky Apr 6 at 16:18
  • 1
    $\begingroup$ You really need to show a convergence plot. Is the order of convergence altered ? $\endgroup$ – BlaB Apr 7 at 13:06
  • $\begingroup$ @MaximUmansky I added the loglog plot of the error as you asked. The first one makes me suspect that $f(x) = \exp(-x)$ is basically the optimal case for this method. However the reason for this is not 100% clear to me as the rectangle rule should not match exactly $f(x)=x \exp(-x)$. $\endgroup$ – Luca Apr 7 at 13:20
  • $\begingroup$ @BlaB is this the plot you were asking? Do you mean something else with "convergence plot"? $\endgroup$ – Luca Apr 7 at 13:20
  • 1
    $\begingroup$ @MaximUmansky I totally understand your reasoning, but why does the rectangle rule using the SAME integration points performs much worse? I also tried the trapz rule, but the story doesn't change. To reach the same level of accuracy I need a much larger number of integration points $\endgroup$ – Luca Apr 7 at 15:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.