Computational Physics Lectures: Numerical integration, from Newton-Cotes quadrature to Gaussian quadrature

Here we will discuss some of the classical methods for integrating a function. The methods we discuss are

1. Equal step methods like the trapezoidal, rectangular and Simpson's rule, parts of what are called Newton-Cotes quadrature methods.
2. Integration approaches based on Gaussian quadrature.

The integral $$\begin{equation} I=\int_a^bf(x) dx \label{eq:integraldef} \end{equation}$$ has a very simple meaning. The integral is the area enscribed by the function $f(x)$ starting from $x=a$ to $x=b$. It is subdivided in several smaller areas whose evaluation is to be approximated by different techniques. The areas under the curve can for example be approximated by rectangular boxes or trapezoids.

In considering equal step methods, our basic approach is that of approximating a function $f(x)$ with a polynomial of at most degree $N-1$, given $N$ integration points. If our polynomial is of degree $1$, the function will be approximated with $f(x)\approx a_0+a_1x$.

The algorithm for these integration methods is rather simple, and the number of approximations perhaps unlimited!

• Choose a step size $h=(b-a)/N$ where $N$ is the number of steps and $a$ and $b$ the lower and upper limits of integration.
• With a given step length we rewrite the integral as

• The strategy then is to find a reliable polynomial approximation for $f(x)$ in the various intervals. Choosing a given approximation for $f(x)$, we obtain a specific approximation to the integral.
• With this approximation to $f(x)$ we perform the integration by computing the integrals over all subintervals.

One possible strategy then is to find a reliable polynomial expansion for $f(x)$ in the smaller subintervals. Consider for example evaluating $$\begin{equation*} \int_a^{a+2h}f(x)dx, \end{equation*}$$ which we rewrite as $$\begin{equation} \int_a^{a+2h}f(x)dx= \int_{x_0-h}^{x_0+h}f(x)dx. \label{eq:hhint} \end{equation}$$ We have chosen a midpoint $x_0$ and have defined $x_0=a+h$.

Using Lagrange's interpolation formula $$\begin{equation*} P_N(x)=\sum_{i=0}^{N}\prod_{k\ne i} \frac{x-x_k}{x_i-x_k}y_i, \end{equation*}$$ we could attempt to approximate the function $f(x)$ with a first-order polynomial in $x$ in the two sub-intervals $x\in[x_0-h,x_0]$ and $x\in[x_0,x_0+h]$. A first order polynomial means simply that we have for say the interval $x\in[x_0,x_0+h]$ $$\begin{equation*} f(x)\approx P_1(x)=\frac{x-x_0}{(x_0+h)-x_0}f(x_0+h)+\frac{x-(x_0+h)}{x_0-(x_0+h)}f(x_0), \end{equation*}$$ and for the interval $x\in[x_0-h,x_0]$ $$\begin{equation*} f(x)\approx P_1(x)=\frac{x-(x_0-h)}{x_0-(x_0-h)}f(x_0)+\frac{x-x_0}{(x_0-h)-x_0}f(x_0-h). \end{equation*}$$

Having performed this subdivision and polynomial approximation, one from $x_0-h$ to $x_0$ and the other from $x_0$ to $x_0+h$, $$\begin{equation*} \int_a^{a+2h}f(x)dx=\int_{x_0-h}^{x_0}f(x)dx+\int_{x_0}^{x_0+h}f(x)dx, \end{equation*}$$ we can easily calculate for example the second integral as $$\begin{equation*} \int_{x_0}^{x_0+h}f(x)dx\approx \int_{x_0}^{x_0+h}\left(\frac{x-x_0}{(x_0+h)-x_0}f(x_0+h)+\frac{x-(x_0+h)}{x_0-(x_0+h)}f(x_0)\right)dx. \end{equation*}$$

This integral can be simplified to $$\begin{equation*} \int_{x_0}^{x_0+h}f(x)dx\approx \int_{x_0}^{x_0+h}\left(\frac{x-x_0}{h}f(x_0+h)-\frac{x-(x_0+h)}{h}f(x_0)\right)dx, \end{equation*}$$ resulting in $$\begin{equation*} \int_{x_0}^{x_0+h}f(x)dx=\frac{h}{2}\left(f(x_0+h) + f(x_0)\right)+O(h^3). \end{equation*}$$ Here we added the error made in approximating our integral with a polynomial of degree $1$.

The other integral gives $$\begin{equation*} \int_{x_0-h}^{x_0}f(x)dx=\frac{h}{2}\left(f(x_0) + f(x_0-h)\right)+O(h^3), \end{equation*}$$ and adding up we obtain $$\begin{equation} \int_{x_0-h}^{x_0+h}f(x)dx=\frac{h}{2}\left(f(x_0+h) + 2f(x_0) + f(x_0-h)\right)+O(h^3), \label{eq:trapez} \end{equation}$$ which is the well-known trapezoidal rule. Concerning the error in the approximation made, $O(h^3)=O((b-a)^3/N^3)$, you should note that this is the local error. Since we are splitting the integral from $a$ to $b$ in $N$ pieces, we will have to perform approximately $N$ such operations.

This means that the global error goes like $\approx O(h^2)$. The trapezoidal reads then $$\begin{equation} I=\int_a^bf(x) dx=h\left(f(a)/2 + f(a+h) +f(a+2h)+ \dots +f(b-h)+ f_{b}/2\right), \label{eq:trapez1} \end{equation}$$ with a global error which goes like $O(h^2)$.

Hereafter we use the shorthand notations $f_{-h}=f(x_0-h)$, $f_{0}=f(x_0)$ and $f_{h}=f(x_0+h)$.

The correct mathematical expression for the local error for the trapezoidal rule is $$\begin{equation*} \int_a^bf(x)dx -\frac{b-a}{2}\left[f(a)+f(b)\right]=-\frac{h^3}{12}f^{(2)}(\xi), \end{equation*}$$ and the global error reads $$\begin{equation*} \int_a^bf(x)dx -T_h(f)=-\frac{b-a}{12}h^2f^{(2)}(\xi), \end{equation*}$$ where $T_h$ is the trapezoidal result and $\xi \in [a,b]$.

The trapezoidal rule is easy to implement numerically through the following simple algorithm

• Choose the number of mesh points and fix the step length.
• calculate $f(a)$ and $f(b)$ and multiply with $h/2$.
• Perform a loop over $n=1$ to $n-1$ ($f(a)$ and $f(b)$ are known) and sum up the terms $f(a+h) +f(a+2h)+f(a+3h)+\dots +f(b-h)$. Each step in the loop corresponds to a given value $a+nh$.
• Multiply the final result by $h$ and add $hf(a)/2$ and $hf(b)/2$.

A simple function which implements this algorithm is as follows

double TrapezoidalRule(double a, double b, int n, double (*func)(double))
{
      double TrapezSum;
      double fa, fb, x, step;
      int    j;
      step=(b-a)/((double) n);
      fa=(*func)(a)/2. ;
      fb=(*func)(b)/2. ;
      TrapezSum=0.;
      for (j=1; j <= n-1; j++){
         x=j*step+a;
         TrapezSum+=(*func)(x);
      }
      TrapezSum=(TrapezSum+fb+fa)*step;
      return TrapezSum;
}  // end TrapezoidalRule 

The function returns a new value for the specific integral through the variable TrapezSum.

There is one new feature to note here, namely the transfer of a user defined function called func in the definition

  void TrapezoidalRule(double a, double b, int n, double *TrapezSum, double (*func)(double) )  

What happens here is that we are transferring a pointer to the name of a user defined function, which has as input a double precision variable and returns a double precision number. The function TrapezoidalRule is called as

  TrapezoidalRule(a, b, n, &MyFunction )       

in the calling function. We note that a, b and n are called by value, while TrapezSum and the user defined function MyFunction are called by reference.

Python offers an extremely versatile programming environment, allowing for the inclusion of analytical studies in a numerical program. Here we show an example code with the trapezoidal rule using SymPy to evaluate an integral and compute the absolute error with respect to the numerically evaluated one of the integral $4\int_0^1 dx/(1+x^2) = \pi$:

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from math import *

from sympy import *

def Trapez(a,b,f,n):

   h = (b-a)/float(n)

   s = 0

   x = a

   for i in range(1,n,1):

       x = x+h

       s = s+ f(x)

   s = 0.5*(f(a)+f(b)) +s

   return h*s

​

#  function to compute pi

def function(x):

    return 4.0/(1+x*x)

​

a = 0.0;  b = 1.0; n = 100

result = Trapez(a,b,function,n)

print "Trapezoidal rule=", result

# define x as a symbol to be used by sympy

x = Symbol('x')

exact = integrate(function(x), (x, 0.0, 1.0))

print "Sympy integration=", exact

# Find relative error

print "Relative error", abs((exact-result)/exact)

The following extended version of the trapezoidal rule allows you to plot the relative error by comparing with the exact result. By increasing to $10^8$ points one arrives at a region where numerical errors start to accumulate.

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from math import log10

import numpy as np

from sympy import Symbol, integrate

import matplotlib.pyplot as plt

# function for the trapezoidal rule

def Trapez(a,b,f,n):

   h = (b-a)/float(n)

   s = 0

   x = a

   for i in range(1,n,1):

       x = x+h

       s = s+ f(x)

   s = 0.5*(f(a)+f(b)) +s

   return h*s

#  function to compute pi

def function(x):

    return 4.0/(1+x*x)

# define integration limits

a = 0.0;  b = 1.0;

# find result from sympy

# define x as a symbol to be used by sympy

x = Symbol('x')

exact = integrate(function(x), (x, a, b))

# set up the arrays for plotting the relative error

n = np.zeros(9); y = np.zeros(9);

# find the relative error as function of integration points

for i in range(1, 8, 1):

    npts = 10**i

    result = Trapez(a,b,function,npts)

    RelativeError = abs((exact-result)/exact)

    n[i] = log10(npts); y[i] = log10(RelativeError);

plt.plot(n,y, 'ro')

plt.xlabel('n')

plt.ylabel('Relative error')

plt.show()

The last example shows the potential of combining numerical algorithms with symbolic calculations, allowing us thereby to

• Validate and verify our algorithms.
• Including concepts like unit testing, one has the possibility to test and validate several or all parts of the code.
• Validation and verification are then included naturally.
• The above example allows you to test the mathematical error of the algorithm for the trapezoidal rule by changing the number of integration points. You get trained from day one to think error analysis.

Another very simple approach is the so-called midpoint or rectangle method. In this case the integration area is split in a given number of rectangles with length $h$ and height given by the mid-point value of the function. This gives the following simple rule for approximating an integral $$\begin{equation} I=\int_a^bf(x) dx \approx h\sum_{i=1}^N f(x_{i-1/2}), \label{eq:rectangle} \end{equation}$$ where $f(x_{i-1/2})$ is the midpoint value of $f$ for a given rectangle. We will discuss its truncation error below. It is easy to implement this algorithm, as shown here

double RectangleRule(double a, double b, int n, double (*func)(double))
{
      double RectangleSum;
      double fa, fb, x, step;
      int    j;
      step=(b-a)/((double) n);
      RectangleSum=0.;
      for (j = 0; j <= n; j++){
         x = (j+0.5)*step+;   // midpoint of a given rectangle
         RectangleSum+=(*func)(x);   //  add value of function.
      }
      RectangleSum *= step;  //  multiply with step length.
      return RectangleSum;
}  // end RectangleRule 

The correct mathematical expression for the local error for the rectangular rule $R_i(h)$ for element $i$ is $$\begin{equation*} \int_{-h}^hf(x)dx - R_i(h)=-\frac{h^3}{24}f^{(2)}(\xi), \end{equation*}$$ and the global error reads $$\begin{equation*} \int_a^bf(x)dx -R_h(f)=-\frac{b-a}{24}h^2f^{(2)}(\xi), \end{equation*}$$ where $R_h$ is the result obtained with rectangular rule and $\xi \in [a,b]$.

Instead of using the above first-order polynomials approximations for $f$, we attempt at using a second-order polynomials. In this case we need three points in order to define a second-order polynomial approximation $$\begin{equation*} f(x) \approx P_2(x)=a_0+a_1x+a_2x^2. \end{equation*}$$ Using again Lagrange's interpolation formula we have $$\begin{equation*} P_2(x)=\frac{(x-x_0)(x-x_1)}{(x_2-x_0)(x_2-x_1)}y_2+ \frac{(x-x_0)(x-x_2)}{(x_1-x_0)(x_1-x_2)}y_1+ \frac{(x-x_1)(x-x_2)}{(x_0-x_1)(x_0-x_2)}y_0. \end{equation*}$$ Inserting this formula in the integral of Eq. $\eqref{eq:hhint}$ we obtain $$\begin{equation*} \int_{-h}^{+h}f(x)dx=\frac{h}{3}\left(f_h + 4f_0 + f_{-h}\right)+O(h^5), \end{equation*}$$ which is Simpson's rule.

Note that the improved accuracy in the evaluation of the derivatives gives a better error approximation, $O(h^5)$ vs.\ $O(h^3)$ . But this is again the local error approximation. Using Simpson's rule we can easily compute the integral of Eq. $\eqref{eq:integraldef}$ to be $$\begin{equation} I=\int_a^bf(x) dx=\frac{h}{3}\left(f(a) + 4f(a+h) +2f(a+2h)+ \dots +4f(b-h)+ f_{b}\right), \label{eq:simpson} \end{equation}$$ with a global error which goes like $O(h^4)$.

More formal expressions for the local and global errors are for the local error $$\begin{equation*} \int_a^bf(x)dx -\frac{b-a}{6}\left[f(a)+4f((a+b)/2)+f(b)\right]=-\frac{h^5}{90}f^{(4)}(\xi), \end{equation*}$$ and for the global error $$\begin{equation*} \int_a^bf(x)dx -S_h(f)=-\frac{b-a}{180}h^4f^{(4)}(\xi). \end{equation*}$$ with $\xi\in[a,b]$ and $S_h$ the results obtained with Simpson's method.

The method can easily be implemented numerically through the following simple algorithm

• Choose the number of mesh points and fix the step.
• calculate $f(a)$ and $f(b)$
• Perform a loop over $n=1$ to $n-1$ ($f(a)$ and $f(b)$ are known) and sum up the terms $4f(a+h) +2f(a+2h)+4f(a+3h)+\dots +4f(b-h)$. Each step in the loop corresponds to a given value $a+nh$. Odd values of $n$ give $4$ as factor while even values yield $2$ as factor.
• Multiply the final result by $\frac{h}{3}$.

In more general terms, what we have done here is to approximate a given function $f(x)$ with a polynomial of a certain degree. One can show that given $n+1$ distinct points $x_0,\dots, x_n\in[a,b]$ and $n+1$ values $y_0,\dots,y_n$ there exists a unique polynomial $P_n(x)$ with the property $$\begin{equation*} P_n(x_j) = y_j\hspace{0.5cm} j=0,\dots,n \end{equation*}$$

In the Lagrange representation the interpolating polynomial is given by $$\begin{equation*} P_n = \sum_{k=0}^nl_ky_k, \end{equation*}$$ with the Lagrange factors $$\begin{equation*} l_k(x) = \prod_{\begin{array}{c}i=0 \\ i\ne k\end{array}}^n\frac{x-x_i}{x_k-x_i}\hspace{0.2cm} k=0,\dots,n. \end{equation*}$$

If we for example set $n=1$, we obtain $$\begin{equation*} P_1(x) = y_0\frac{x-x_1}{x_0-x_1}+y_1\frac{x-x_0}{x_1-x_0}=\frac{y_1-y_0}{x_1-x_0}x-\frac{y_1x_0+y_0x_1}{x_1-x_0}, \end{equation*}$$ which we recognize as the equation for a straight line.

The polynomial interpolatory quadrature of order $n$ with equidistant quadrature points $x_k=a+kh$ and step $h=(b-a)/n$ is called the Newton-Cotes quadrature formula of order $n$.

The methods we have presented hitherto are tailored to problems where the mesh points $x_i$ are equidistantly spaced, $x_i$ differing from $x_{i+1}$ by the step $h$.

The basic idea behind all integration methods is to approximate the integral $$\begin{equation*} I=\int_a^bf(x)dx \approx \sum_{i=1}^N\omega_if(x_i), \end{equation*}$$ where $\omega$ and $x$ are the weights and the chosen mesh points, respectively. In our previous discussion, these mesh points were fixed at the beginning, by choosing a given number of points $N$. The weigths $\omega$ resulted then from the integration method we applied. Simpson's rule, see Eq. $\eqref{eq:simpson}$ would give $$\begin{equation*} \omega : \left\{h/3,4h/3,2h/3,4h/3,\dots,4h/3,h/3\right\}, \end{equation*}$$ for the weights, while the trapezoidal rule resulted in $$\begin{equation*} \omega : \left\{h/2,h,h,\dots,h,h/2\right\}. \end{equation*}$$

In general, an integration formula which is based on a Taylor series using $N$ points, will integrate exactly a polynomial $P$ of degree $N-1$. That is, the $N$ weights $\omega_n$ can be chosen to satisfy $N$ linear equations, see chapter 3 of Ref.\ [3]. A greater precision for a given amount of numerical work can be achieved if we are willing to give up the requirement of equally spaced integration points. In Gaussian quadrature (hereafter GQ), both the mesh points and the weights are to be determined. The points will not be equally spaced.

The theory behind GQ is to obtain an arbitrary weight $\omega$ through the use of so-called orthogonal polynomials. These polynomials are orthogonal in some interval say e.g., [-1,1]. Our points $x_i$ are chosen in some optimal sense subject only to the constraint that they should lie in this interval. Together with the weights we have then $2N$ ($N$ the number of points) parameters at our disposal.

Even though the integrand is not smooth, we could render it smooth by extracting from it the weight function of an orthogonal polynomial, i.e., we are rewriting $$\begin{equation} I= \int_a^b f(x)dx =\int_a^b W(x)g(x)dx \approx \sum_{i=1}^N\omega_ig(x_i), \label{eq:generalint} \end{equation}$$ where $g$ is smooth and $W$ is the weight function, which is to be associated with a given orthogonal polynomial. Note that with a given weight function we end up evaluating the integrand for the function $g(x_i)$.

The weight function $W$ is non-negative in the integration interval $x\in [a,b]$ such that for any $n \ge 0$, the integral $\int_a^b |x|^n W(x) dx$ is integrable. The naming weight function arises from the fact that it may be used to give more emphasis to one part of the interval than another. A quadrature formula $$\begin{equation} \int_a^b W(x)f(x)dx \approx \sum_{i=1}^N\omega_if(x_i), \label{_auto1} \end{equation}$$ with $N$ distinct quadrature points (mesh points) is a called a Gaussian quadrature formula if it integrates all polynomials $p\in P_{2N-1}$ exactly, that is $$\begin{equation} \int_a^bW(x)p(x)dx =\sum_{i=1}^N\omega_ip(x_i), \label{_auto2} \end{equation}$$ It is assumed that $W(x)$ is continuous and positive and that the integral $$\begin{equation*} \int_a^bW(x)dx \end{equation*}$$ exists. Note that the replacement of $f\rightarrow Wg$ is normally a better approximation due to the fact that we may isolate possible singularities of $W$ and its derivatives at the endpoints of the interval.

The quadrature weights or just weights (not to be confused with the weight function) are positive and the sequence of Gaussian quadrature formulae is convergent if the sequence $Q_N$ of quadrature formulae $$\begin{equation*} Q_N(f)\rightarrow Q(f)=\int_a^bf(x)dx, \end{equation*}$$ in the limit $N\rightarrow \infty$.

Then we say that the sequence $$\begin{equation*} Q_N(f) = \sum_{i=1}^N\omega_i^{(N)}f(x_i^{(N)}), \end{equation*}$$ is convergent for all polynomials $p$, that is $$\begin{equation*} Q_N(p) = Q(p) \end{equation*}$$ if there exits a constant $C$ such that $$\begin{equation*} \sum_{i=1}^N|\omega_i^{(N)}| \le C, \end{equation*}$$ for all $N$ which are natural numbers.

The error for the Gaussian quadrature formulae of order $N$ is given by $$\begin{equation*} \int_a^bW(x)f(x)dx-\sum_{k=1}^Nw_kf(x_k)=\frac{f^{2N}(\xi)}{(2N)!}\int_a^bW(x)[q_{N}(x)]^2dx \end{equation*}$$ where $q_{N}$ is the chosen orthogonal polynomial and $\xi$ is a number in the interval $[a,b]$. We have assumed that $f\in C^{2N}[a,b]$, viz. the space of all real or complex $2N$ times continuously differentiable functions.

In science there are several important orthogonal polynomials which arise from the solution of differential equations. Well-known examples are the Legendre, Hermite, Laguerre and Chebyshev polynomials. They have the following weight functions

The importance of the use of orthogonal polynomials in the evaluation of integrals can be summarized as follows.

Methods based on Taylor series using $N$ points will integrate exactly a polynomial $P$ of degree $N-1$. If a function $f(x)$ can be approximated with a polynomial of degree $N-1$ $$\begin{equation*} f(x)\approx P_{N-1}(x), \end{equation*}$$ with $N$ mesh points we should be able to integrate exactly the polynomial $P_{N-1}$.

Gaussian quadrature methods promise more than this. We can get a better polynomial approximation with order greater than $N$ to $f(x)$ and still get away with only $N$ mesh points. More precisely, we approximate $$\begin{equation*} f(x) \approx P_{2N-1}(x), \end{equation*}$$ and with only $N$ mesh points these methods promise that $$\begin{equation*} \int f(x)dx \approx \int P_{2N-1}(x)dx=\sum_{i=0}^{N-1} P_{2N-1}(x_i)\omega_i, \end{equation*}$$

The reason why we can represent a function $f(x)$ with a polynomial of degree $2N-1$ is due to the fact that we have $2N$ equations, $N$ for the mesh points and $N$ for the weights.

The mesh points are the zeros of the chosen orthogonal polynomial of order $N$, and the weights are determined from the inverse of a matrix. An orthogonal polynomials of degree $N$ defined in an interval $[a,b]$ has precisely $N$ distinct zeros on the open interval $(a,b)$.

Before we detail how to obtain mesh points and weights with orthogonal polynomials, let us revisit some features of orthogonal polynomials by specializing to Legendre polynomials. In the text below, we reserve hereafter the labelling $L_N$ for a Legendre polynomial of order $N$, while $P_N$ is an arbitrary polynomial of order $N$. These polynomials form then the basis for the Gauss-Legendre method.

The following python code allows you to run interactively either in a browser or using ipython notebook. It compares the trapezoidal rule and Gaussian quadrature with the exact result from symbolic python SYMPY up to 1000 integration points for the integral $$I = 2 = \int_0^{\infty} x^2 \exp{-x} dx.$$ For the trapezoidal rule the results will vary strongly depending on how the infinity limit is approximated. Try to run the code below for different finite approximations to $\infty$.

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from math import exp

import numpy as np

from sympy import Symbol, integrate, exp, oo

​

# function for the trapezoidal rule

def TrapezoidalRule(a,b,f,n):

   h = (b-a)/float(n)

   s = 0

   x = a

   for i in range(1,n,1):

       x = x+h

       s = s+ f(x)

   s = 0.5*(f(a)+f(b)) +s

   return h*s

​

# function for the Gaussian quadrature with Laguerre polynomials

def GaussLaguerreRule(n):

   s = 0

   xgauleg, wgauleg = np.polynomial.laguerre.laggauss(n)

   for i in range(1,n,1):

       s = s+ xgauleg[i]*xgauleg[i]*wgauleg[i]

   return s

​

#  function to compute

def function(x):

    return x*x*exp(-x)

​

​

# Integration limits for the Trapezoidal rule

a = 0.0; b = 10000.0

# define x as a symbol to be used by sympy

x = Symbol('x')

# find result from sympy

exact = integrate(function(x), (x, a, oo))

# set up the arrays for plotting the relative error

n = np.zeros(40); Trapez = np.zeros(4); LagGauss = np.zeros(4);

# find the relative error as function of integration points

for i in range(1, 3, 1):

    npts = 10**i

    n[i] = npts

    Trapez[i] = abs((TrapezoidalRule(a,b,function,npts)-exact)/exact)

    LagGauss[i] = abs((GaussLaguerreRule(npts)-exact)/exact)

print "Integration points=", n[1], n[2]

print "Trapezoidal relative error=", Trapez[1], Trapez[2]

print "LagGuass relative error=", LagGauss[1], LagGauss[2]

So-called principal value (PV) integrals are often employed in physics, from Green's functions for scattering to dispersion relations. Dispersion relations are often related to measurable quantities and provide important consistency checks in atomic, nuclear and particle physics. A PV integral is defined as $$\begin{equation*} I(x)={\cal P}\int_a^bdt\frac{f(t)}{t-x}=\lim_{\epsilon\rightarrow 0^+} \left[\int_a^{x-\epsilon}dt\frac{f(t)}{t-x}+\int_{x+\epsilon}^bdt\frac{f(t)}{t-x}\right], \end{equation*}$$ and arises in applications of Cauchy's residue theorem when the pole $x$ lies on the real axis within the interval of integration $[a,b]$. Here ${\cal P}$ stands for the principal value. An important assumption is that the function $f(t)$ is continuous on the interval of integration.

In case $f(t)$ is a closed form expression or it has an analytic continuation in the complex plane, it may be possible to obtain an expression on closed form for the above integral.

However, the situation which we are often confronted with is that $f(t)$ is only known at some points $t_i$ with corresponding values $f(t_i)$. In order to obtain $I(x)$ we need to resort to a numerical evaluation.

To evaluate such an integral, let us first rewrite it as $$\begin{equation*} {\cal P}\int_a^bdt\frac{f(t)}{t-x}= \int_a^{x-\Delta}dt\frac{f(t)}{t-x}+\int_{x+\Delta}^bdt\frac{f(t)}{t-x}+ {\cal P}\int_{x-\Delta}^{x+\Delta}dt\frac{f(t)}{t-x}, \end{equation*}$$ where we have isolated the principal value part in the last integral.

Defining a new variable $u=t-x$, we can rewrite the principal value integral as $$\begin{equation} I_{\Delta}(x)={\cal P}\int_{-\Delta}^{+\Delta}du\frac{f(u+x)}{u}. \label{eq:deltaint} \end{equation}$$ One possibility is to Taylor expand $f(u+x)$ around $u=0$, and compute derivatives to a certain order as we did for the Trapezoidal rule or Simpson's rule. Since all terms with even powers of $u$ in the Taylor expansion dissapear, we have that $$\begin{equation*} I_{\Delta}(x)\approx \sum_{n=0}^{N_{max}}f^{(2n+1)}(x) \frac{\Delta^{2n+1}}{(2n+1)(2n+1)!}. \end{equation*}$$

To evaluate higher-order derivatives may be both time consuming and delicate from a numerical point of view, since there is always the risk of loosing precision when calculating derivatives numerically. Unless we have an analytic expression for $f(u+x)$ and can evaluate the derivatives in a closed form, the above approach is not the preferred one.

Rather, we show here how to use the Gauss-Legendre method to compute Eq. $\eqref{eq:deltaint}$. Let us first introduce a new variable $s=u/\Delta$ and rewrite Eq. $\eqref{eq:deltaint}$ as $$\begin{equation} I_{\Delta}(x)={\cal P}\int_{-1}^{+1}ds\frac{f(\Delta s+x)}{s}. \label{eq:deltaint2} \end{equation}$$

The integration limits are now from $-1$ to $1$, as for the Legendre polynomials. The principal value in Eq. $\eqref{eq:deltaint2}$ is however rather tricky to evaluate numerically, mainly since computers have limited precision. We will here use a subtraction trick often used when dealing with singular integrals in numerical calculations. We introduce first the calculus relation $$\begin{equation*} \int_{-1}^{+1} \frac{ds}{s} =0. \end{equation*}$$ It means that the curve $1/(s)$ has equal and opposite areas on both sides of the singular point $s=0$.

If we then note that $f(x)$ is just a constant, we have also $$\begin{equation*} f(x)\int_{-1}^{+1} \frac{ds}{s}=\int_{-1}^{+1}f(x) \frac{ds}{s} =0. \end{equation*}$$

Subtracting this equation from Eq. $\eqref{eq:deltaint2}$ yields $$\begin{equation} I_{\Delta}(x)={\cal P}\int_{-1}^{+1}ds\frac{f(\Delta s+x)}{s}=\int_{-1}^{+1}ds\frac{f(\Delta s+x)-f(x)}{s}, \label{eq:deltaint3} \end{equation}$$ and the integrand is no longer singular since we have that $\lim_{s \rightarrow 0} (f(s+x) -f(x))=0$ and for the particular case $s=0$ the integrand is now finite.

Eq. $\eqref{eq:deltaint3}$ is now rewritten using the Gauss-Legendre method resulting in $$\begin{equation} \int_{-1}^{+1}ds\frac{f(\Delta s+x)-f(x)}{s}=\sum_{i=1}^{N}\omega_i\frac{f(\Delta s_i+x)-f(x)}{s_i}, \label{eq:deltaint4} \end{equation}$$ where $s_i$ are the mesh points ($N$ in total) and $\omega_i$ are the weights.

In the selection of mesh points for a PV integral, it is important to use an even number of points, since an odd number of mesh points always picks $s_i=0$ as one of the mesh points. The sum in Eq. $\eqref{eq:deltaint4}$ will then diverge.

Let us apply this method to the integral $$\begin{equation} I(x)={\cal P}\int_{-1}^{+1}dt\frac{e^t}{t}. \label{eq:deltaint5} \end{equation}$$ The integrand diverges at $x=t=0$. We rewrite it using Eq. $\eqref{eq:deltaint3}$ as $$\begin{equation} {\cal P}\int_{-1}^{+1}dt\frac{e^t}{t}=\int_{-1}^{+1}\frac{e^t-1}{t}, \label{eq:deltaint6} \end{equation}$$ since $e^x=e^0=1$. With Eq. $\eqref{eq:deltaint4}$ we have then $$\begin{equation} \int_{-1}^{+1}\frac{e^t-1}{t}\approx \sum_{i=1}^{N}\omega_i\frac{e^{t_i}-1}{t_i}. \label{eq:deltaint7} \end{equation}$$

The exact results is $2.11450175075....$. With just two mesh points we recall from the previous subsection that $\omega_1=\omega_2=1$ and that the mesh points are the zeros of $L_2(x)$, namely $x_1=-1/\sqrt{3}$ and $x_2=1/\sqrt{3}$. Setting $N=2$ and inserting these values in the last equation gives $$\begin{equation*} I_2(x=0)=\sqrt{3}\left(e^{1/\sqrt{3}}-e^{-1/\sqrt{3}}\right)=2.1129772845. \end{equation*}$$ With six mesh points we get even the exact result to the tenth digit $$\begin{equation*} I_6(x=0)=2.11450175075! \end{equation*}$$

We can repeat the above subtraction trick for more complicated integrands. First we modify the integration limits to $\pm \infty$ and use the fact that $$\begin{equation*} \int_{-\infty}^{\infty} \frac{dk}{k-k_0}= \int_{-\infty}^{0} \frac{dk}{k-k_0}+ \int_{0}^{\infty} \frac{dk}{k-k_0} =0. \end{equation*}$$ A change of variable $u=-k$ in the integral with limits from $-\infty$ to $0$ gives $$\begin{equation*} \int_{-\infty}^{\infty} \frac{dk}{k-k_0}= \int_{\infty}^{0} \frac{-du}{-u-k_0}+ \int_{0}^{\infty} \frac{dk}{k-k_0}= \int_{0}^{\infty} \frac{dk}{-k-k_0}+ \int_{0}^{\infty} \frac{dk}{k-k_0}=0. \end{equation*}$$

It means that the curve $1/(k-k_0)$ has equal and opposite areas on both sides of the singular point $k_0$. If we break the integral into one over positive $k$ and one over negative $k$, a change of variable $k\rightarrow -k$ allows us to rewrite the last equation as $$\begin{equation*} \int_{0}^{\infty} \frac{dk}{k^2-k_0^2} =0. \end{equation*}$$

We can use this to express a principal values integral as $$\begin{equation} {\cal P}\int_{0}^{\infty} \frac{f(k)dk}{k^2-k_0^2} = \int_{0}^{\infty} \frac{(f(k)-f(k_0))dk}{k^2-k_0^2}, \label{eq:trick_pintegral} \end{equation}$$ where the right-hand side is no longer singular at $k=k_0$, it is proportional to the derivative $df/dk$, and can be evaluated numerically as any other integral.

Such a trick is often used when evaluating integral equations.

Here we show an example of a multidimensional integral which appears in quantum mechanical calculations.

The ansatz for the wave function for two electrons is given by the product of two $1s$ wave functions as $$\Psi({\bf r}_1,{\bf r}_2) = \exp{-(\alpha (r_1+r_2))}.$$ The integral we need to solve is the quantum mechanical expectation value of the correlation energy between two electrons, namely $$I = \int_{-\infty}^{\infty} d{\bf r}_1d{\bf r}_2 \exp{-2(\alpha (r_1+r_2))}\frac{1}{|{\bf r}_1-{\bf r}_2|}.$$ The integral has an exact solution $5\pi^2/16 = 0.19277$.

If we use Gaussian quadrature with Legendre polynomials (without rewriting the integral), we have

     double *x = new double [N];
     double *w = new double [N];
//   set up the mesh points and weights
     GaussLegendrePoints(a,b,x,w, N);

//   evaluate the integral with the Gauss-Legendre method
//   Note that we initialize the sum
     double int_gauss = 0.;
//   six-double loops
     for (int i=0;i<N;i++){
	     for (int j = 0;j<N;j++){
	     for (int k = 0;k<N;k++){
	     for (int l = 0;l<N;l++){
	     for (int m = 0;m<N;m++){
	     for (int n = 0;n<N;n++){
        int_gauss+=w[i]*w[j]*w[k]*w[l]*w[m]*w[n]
       *int_function(x[i],x[j],x[k],x[l],x[m],x[n]);
     		}}}}}
	}

Using Legendre polynomials for the Gaussian quadrature is not very efficient. There are several reasons for this:

• You can easily end up in situations where the integrand diverges
• The limits $\pm \infty$ have to be approximated with a finite number

This means that our integral becomes $$I=\int_0^{\infty} r_1^2dr_1 \int_0^{\infty}r_2^2dr_2 \int_0^{\pi}dcos(\theta_1)\int_0^{\pi}dcos(\theta_2)\int_0^{2\pi}d\phi_1\int_0^{2\pi}d\phi_2 \frac{\exp{-2\alpha (r_1+r_2)}}{r_{12}}$$ where we have defined $$\frac{1}{r_{12}}= \frac{1}{\sqrt{r_1^2+r_2^2-2r_1r_2cos(\beta)}}$$ with $$\cos(\beta) = \cos(\theta_1)\cos(\theta_2)+\sin(\theta_1)\sin(\theta_2)\cos(\phi_1-\phi_2))$$

Our integral is now given by $$I=\int_0^{\infty} r_1^2dr_1 \int_0^{\infty}r_2^2dr_2 \int_0^{\pi}dcos(\theta_1)\int_0^{\pi}dcos(\theta_2)\int_0^{2\pi}d\phi_1\int_0^{2\pi}d\phi_2 \frac{\exp{-2\alpha (r_1+r_2)}}{r_{12}}$$ For the angles we need to perform the integrations over $\theta_i\in [0,\pi]$ and $\phi_i \in [0,2\pi]$. However, for the radial part we can now either use

• Gauss-Legendre wth an appropriate mapping or
• Gauss-Laguerre taking properly care of the integrands involving the $r_i^2 \exp{-(2\alpha r_i)}$ terms.