These notes are based on Chapter 11 of Jones, O., Maillardet, R., & Robinson, A. (2009). Introduction to scientific programming and simulation using R. CRC Press.

Introduction

A definite integral $$\int_a^b f(x)dx$$ is easy to compute if we can find an antiderivative $$F$$ such that $$F'(x) = f(x)$$ because $$\int_a^b f(x)dx = F(b)-F(a)$$.

Unfortunately, for many functions, it is impossible to write down an antiderivative in a “closed form” so that we can compute definite integrals without numerical methods.

Numerical integration could provide good approximation of definite integrals. For example, the cumulative distribution function (CDF) of the standard normal distribution $\Phi(z) = \int_{-\infty}^z\frac{1}{\sqrt{2\pi}}e^{-x^2/2}dx$ is commonly computed via numerical integration. R has pnorm for $$\Phi$$ and, for example,

pnorm(-1)
## [1] 0.1586553

is an approximation of $$\Phi(-1)$$.

We cover three numerical integration techniques:

• Trapezoidal rule,
• Simpson’s rule,
• Adaptive quadrature.

These are deterministic methods for numerical integration. In contrast, Monte Carlo integration is a class of stochastic methods.

A test problem is to integrate $$f(x) = x\sin(x)+5$$ over $$[0,3\pi]$$, which can be analytically done using integration by parts. \begin{align} \int_0^{3\pi} x\sin(x)+5 dx &= \int_0^{3\pi} x\sin(x) dx + \int_0^{3\pi} 5 dx\\ &= \left( -x\cos(x)\Bigr\rvert_0^{3\pi} + \int_0^{3\pi}\cos(x)dx \right) + 15\pi\\ &= 18\pi\\ &\approx 56.5486677646 \end{align}

test1 <- function(x) x*sin(x)+5
integrate(test1, 0, 3*pi) # R built-in function
## 56.54867 with absolute error < 8e-12

Trapezoidal rule

Given a finite set of subintervals, the trapezoidal rule approximates the area under $$y=f(x)$$ by the finite sum of trapezoid areas, which are easy to calculate.

The following is an example with $$n=6$$ intervals.

curve(x*sin(x)+5, xaxt="n", lwd=1.5, from=-.5, to=9.9, ylab="y")
abline(h=0)
title(expression(y==x*sin(x)+5))
x <- seq(0, 3*pi, .5*pi)
# How can we generate these expressions using a loop?
letters <- c(expression(x[0]==0), expression(x[1]), expression(x[2]),
expression(x[3]), expression(x[4]), expression(x[5]),
expression(x[6]==3*pi))
axis(1, at=x, labels=letters)
from_x <- c(0, .5*pi, pi, 1.5*pi, 2*pi, 2.5*pi, 3*pi)
from_y <- rep(0,7)
to_x <- from_x
to_y <- sapply(from_x, function(x) x*sin(x)+5)
segments(from_x, from_y, to_x, to_y, lty=3)

from_x <- from_x[-length(from_x)]
from_y <- to_y[-length(to_y)]
to_x <- to_x[-1]
to_y <- to_y[-1]
segments(from_x, from_y, to_x, to_y, lty=3)

In general, suppose we try to approximate $$\int_a^b f(x)dx$$ using $$n$$ subintervals. Let the length of each interval be denoted by $$h = (b-a)/n$$. Then, the sum of all trapezoid areas $$(T)$$ is given by: \begin{align} T &= \frac{h}{2}(f(x_0)+f(x_1))+\frac{h}{2}(f(x_1)+f(x_2))+\dots+\frac{h}{2}(f(x_{n-1})+f(x_n))\\ &= \frac{h}{2}(f(x_0)+2f(x_1)+2f(x_2)+\dots+2f(x_{n-1})+f(x_n)) \end{align}

Since it is a type of Riemann sum, we expect that the approximation improves with larger $$n$$.

trapezoid <- function(f, a, b, n=100) {
h <- (b-a)/n
x <- seq(a+h, b-h, by=h)
T <- h/2*(f(a) + 2*sum(f(x)) + f(b))

return(T)
}
trapezoid(test1, 0, 3*pi)
## [1] 56.54169

Simpson’s rule

As implied by the use of trapezoids, the trapezoidal rule approximates $$f$$ by a piece-wise linear function. Simpson’s rule uses, instead, a piece-wise quadratic function for approximation of $$f$$.

Since a quadratic function $$p(x) = ax^2 + bx + c$$ has three parameters $$(a,b,c)$$, three non-collinear points $$(u,f(u))$$, $$(v,f(v))$$, and $$(w,f(w))$$ pin down the function through a system of three linear equations. (Without loss of generality, assume $$u<v<w$$.)

Collinearity means $$(v,f(v)) = t(u,f(u))+(1-t)(w,f(w))$$ for $$t\in(0,1)$$. Can you see its implication for $$p(x) = ax^2 + bx + c$$?

$\begin{cases} au^2 + bu + c = f(u)\\ av^2 + bv + c = f(v)\\ aw^2 + bw + c = f(w) \end{cases}$

Solving the system of equations, we get $p(x) = f(u)\frac{(x-v)(x-w)}{(u-v)(u-w)} + f(v)\frac{(x-u)(x-w)}{(v-u)(v-w)} \\+ f(w)\frac{(x-u)(x-v)}{(w-u)(w-v)} .$

Plug in each of three points and see this make sense.

If $$u=v-h$$ and $$w=v+h$$, then we get the following integral $\int_u^w p(x)dx = \frac{h}{3}(f(u)+4f(v)+f(w)) .$

I have no trick of doing this integral. I laboured twice and got the same answer. So, it is likely a correct one.

Now, consider $$n$$ number of subintervals where $$n$$ is even. Then, we can carry out the above integration $$n/2$$ times and sum them up to get an approximation $$S$$.

\begin{align} S &= \frac{h}{3}(f(x_0)+4f(x_1)+f(x_2)) + \frac{h}{3}(f(x_2)+4f(x_3)+f(x_4)) + \dots \\ &\qquad + \frac{h}{3}(f(x_{n-2})+4f(x_{n-1})+f(x_{n}))\\ &= \frac{h}{3}(f(x_0)+4f(x_1)+2f(x_2)+4f(x_3)+\dots+4f(x_{n-1})+f(x_{n})) \end{align}

Notice that the function values for even points (except $$0$$ and $$n$$) get $$2$$ as each of them is used for two adjacent subintervals.

Let’s apply Simpson’s rule to the test problem.

simpson_n <- function(f, a, b, n=50) {
n <- max(c(2*(n%/%2), 2)) # %/% (integer division) for even n >= 2
h <- (b-a)/n
y.a <- f(a)
y.b <- f(b)
y <- f(seq(a+h, b-h, by=h))
coeff <- c(rep(c(4,2), n/2-1), 4) # coefficients 4,2,...,4,2,4
S <- h/3*(y.a + sum(coeff*y) + y.b) # coeff*y: element-wise product

return(S)
}
simpson_n(test1, 0, 3*pi, 50)
## [1] 56.54873

Simpson with $$n=50$$ works better than Trapezoid with $$n=100$$.

How large should $$n$$ be?

When using Simpson’s rule in practice, we need some rule for choosing $$n$$ which results in reasonably accurate approximation.

Suppose that $$f$$ is Riemann integrable or continuous almost everywhere. Let $$I = \int_a^b f(x)dx$$. Also, let $$S(n)$$ be the approximate integral via Simpson’s rule using $$n$$ subintervals. Then, we have $$|S(n) - I|\to0$$. Furthermore, by the triangle inequality, \begin{align} |S(2n)-S(n)| &= |S(2n)-I-S(n)+I|\\ &= |S(2n)-I+(-(S(n)-I))|\\ &\leq |S(2n)-I|+|-(S(n)-I)|\\ &= |S(2n)-I|+|S(n)-I|\\ &\to 0 \end{align} $$|S(2n)-S(n)|$$ can be seen as an improvement after doubling the number of subintervals. Thus, given some small $$\epsilon$$, we can use $$|S(2n)-S(n)|\leq\epsilon$$ as a stopping criterion as we keep doubling $$n$$.

Now, modify the above Simpson’s rule so that, given $$\epsilon$$, we can search for a large enough $$n$$. Observe that we increase $$n$$ by a factor of $$2$$ each time, so that we can reuse the previous function evaluations.

In practice, evaluating $$f$$ is the most expensive operation when doing numerical integration. If we increased $$n$$ by an arbitrary even number, then two sets of evaluation points before and after the increase will likely be misaligned (except $$a$$ and $$b$$). So, we will have to re-calculate $$f$$ at all points. In contrast, if we double $$n$$, we can reuse all the existing $$n$$ evaluations at the original set of points, and we will simply evaluate $$n$$ more points in each subinterval.

simpson <- function(f, a, b, tol=1e-5) {
# While this block is the same as S <- simpson_n(f,a,b,2),
# we want y.a, y.b, and y to be computed here.
n <- 2
h <- (b-a)/n
y.a <- f(a)
y.b <- f(b)
y <- f(seq(a+h, b-h, by=h))
coeff <- c(rep(c(4,2), n/2-1), 4)
S <- h/3*(y.a + sum(coeff*y) + y.b)

S.diff <- 2*tol # to ensure to loop at least once
while (S.diff > tol) {
S.old <- S
n <- 2*n
h <- h/2
y[seq(2, n-2, by=2)] <- y # reuse
y[seq(1, n-1, by=2)] <- f(seq(a+h, b-h, by=2*h)) # new values
coeff <- c(rep(c(4,2), n/2-1), 4)
S <- h/3*(y.a + sum(coeff*y) + y.b)

S.diff <- abs(S - S.old)
}

return(list("S"=S, "Number of intervals"=n))
}
simpson(test1, 0, 3*pi, 1e-4)
## $S ## [1] 56.54867 ## ##$Number of intervals
## [1] 128

Adaptive quadrature

Evenly dividing the integration range is intuitive and will likely work well if we have computational capacity for large $$n$$. If not, we can divide the range adaptively and achieve the same level of performance with smaller $$n$$.

A typical situation in which adaptivity makes difference is that the function to integrate has varying curvature, i.e., consisting of flat and steep parts. In steep portions, the function changes rapidly, so we need fine subdivision for good approximation.

Adaptive quadrature is Simpson’s rule with subinterval length $$h$$ locally adapted. The key idea is that:

• Subinterval division occurs only where needed.
• Need is determined locally by examining whether an improvement $$|S(4)-S(2)|$$ is still large for each subinterval.

Here is the algorithm. Do the following for each subinterval that is not yet marked “terminated”.

1. Compute $$S(2)$$ using Simpson’s rule with $$n=2$$.
2. Compute $$S(4)$$ using Simpson’s rule with $$n=4$$.
3. If an improvement $$|S(4)-S(2)|$$ is still large, subdivide it in half; otherwise, terminate subdivision of that subinterval.

Continue until all subintervals are marked terminated.

Since a size of improvement depends on the shape of the function over a subinterval, subdivision in one region where the function is steep may stop only for a short subinterval while subdivision in another region where the function is flat may stop even for a long subinterval. This results in varying lengths of subintervals.

This type of algorithms is efficiently and naturally implemented using a recursive function because, compared to linear processes implemented in loops, the halving process is more “organic” where a single subinterval spawns two subintervals, each of which spawns two… The price to pay is difficulty in following the flow of the algorithm; it is hard to keep track of the order of subdivisions and terminations.

Let’s apply adaptive quadrature to $$f(x) = -4x\log(x)$$ over $$[0,1]$$, which is steep near the origin as $$f'(x) = -4(\log(x)+1) \to \infty$$.

Using integration by parts with $$t = \log(x)$$, we can get $$\int_0^1f(x)dx=1$$.

curve(-4*x*log(x), ylab="y", lwd=1.5, n=1000,
main=expression(y==-4*x*log(x)))