Approximate and Improper Integrals

The Fundamental Theorem evaluates ${\int }_a^{b}f(x)\mathbf{d}x$ from an antiderivative of $f$. Two situations put that out of reach. Either $f$ has no elementary antiderivative — as with $e^{x^2}$ or $(\sin x)/x$ — or $f$ is known only as measured data with no formula at all. In both cases the integral is approximated numerically. A separate failure occurs when the interval runs to infinity or $f$ has an infinite discontinuity; there the integral is redefined as a limit, and the question becomes whether that limit exists.

Numerical approximation

The definite integral is a limit of Riemann sums, so any Riemann sum approximates it. Divide $[a,b]$ into $n$ subintervals of width $\Delta x=(b-a)/n$ with endpoints $x_0,x_1,\dots ,x_n$. Sampling the left or right endpoint of each subinterval gives the crude approximations $L_n$ and $R_n$. Two better sampling choices follow.

The Midpoint Rule sums rectangles whose height is read at the middle of each strip; the Trapezoidal Rule replaces each strip's top by the straight chord joining its endpoints, so the area element is a trapezoid. Both are exact for straight-line $f$ and err in proportion to how much $f$ curves.

Simpson's Rule

Fitting a straight chord ignores curvature. Fitting a parabola through three consecutive points captures it. Group the subintervals in pairs (so $n$ must be even) and pass a parabola through each triple of points; the area under one parabola over $[x_{i-1},x_{i+1}]$ works out to $\frac{\Delta x}{3}(y_{i-1}+4y_i+y_{i+1})$. Summing gives the rule.

Error bounds

Each rule comes with a provable bound on the error, the amount that must be added to the approximation to make it exact. The bounds depend on how large a derivative of $f$ can be on $[a,b]$.¹

The Midpoint error bound is half the Trapezoidal, and in practice their errors have opposite signs — Simpson's Rule combines the two as the weighted average $S_{2n}=\frac{1}{3}{T}_n+\frac{2}{3}{M}_n$. Doubling $n$ cuts the trapezoidal and midpoint errors by about $4$ (the $n^2$), but cuts Simpson's error by about $16$ (the $n^4$). Because $f^{(4)}$ vanishes for any cubic, Simpson's Rule is exact on polynomials of degree $3$ or lower.

The rates show clearly when the errors for ${\int }_1^2\frac{1}{x}\mathbf{d}x$ (true value $\ln 2\approx 0.693147$) are tabulated against $n$.

To guarantee accuracy within a tolerance, solve the bound for $n$. For ${\int }_1^2\frac{1}{x}\mathbf{d}x$, here $f^{\prime \prime }(x)=2/x^3$ so $K=2$ on $[1,2]$; the Trapezoidal bound $2/(12n^2)<0.0001$ forces $n\ge 41$, while Simpson's $2/(180n^4)<0.0001$ needs only $n\ge 8$.

The bound also caps the error of a computed approximation before the true value is known.

Improper integrals over infinite intervals

The region under $y=1/x^2$ to the right of $x=1$ extends forever, yet its area is finite. The area out to $x=t$ is

which approaches $1$ as $t\to \infty$. Defining the integral over $[1,\infty )$ as this limit gives it the value $1$.

The nearly identical curve $y=1/x$ behaves oppositely:

so ${\int }_1^{\infty }\frac{1}{x}\mathbf{d}x$ diverges. The two curves look alike for large $x$ and both tend to $0$, but $1/x^2$ decays fast enough for finite area while $1/x$ does not.

The pattern generalizes to a benchmark family, the $p$-integrals, obtained by the same limit computation for a general exponent.

A doubly-infinite integral splits at any finite point and converges only if both halves do.

Improper integrals with a discontinuous integrand

The second kind of impropriety comes from the integrand rather than the interval. If $f$ has a vertical asymptote at an endpoint, integrate up to a movable bound and take the limit toward the bad point.

For $\underset{2}{\overset{5}{\int }}\frac{\mathbf{d}x}{\sqrt{x-2}}$, the integrand blows up at the left endpoint $x=2$, so

a finite value: the region is unbounded in height but has finite area. A logarithmic example converges too: ${\int }_0^1\ln x\mathbf{d}x={\lim }_{t\to 0^{+}}[x\ln x-x{]}_t^1=-1$.

The interior case demands care. The integrand of ${\int }_0^3\frac{\mathbf{d}x}{x-1}$ has an asymptote at $x=1$ inside the interval. Splitting at $c=1$, the piece ${\int }_0^1\frac{\mathbf{d}x}{x-1}$ already diverges, so the whole integral diverges. Treating it as an ordinary integral and writing $\ln |x-1|{|}_0^3=\ln 2$ is wrong: it ignores the discontinuity. Always scan the interval for points where the integrand is undefined before evaluating.

The Comparison Test

When an improper integral has no elementary antiderivative, its value cannot be computed directly, but convergence can still be settled by comparison with a simpler integral of known behavior.

A finite area under the upper curve caps the area under the lower one; an infinite area under the lower curve forces the upper one to be infinite too.

The classic application is the Gaussian integral. The antiderivative of $e^{-x^2}$ is not elementary, yet its convergence is easy. For $x\ge 1$, $x^2\ge x$, so $e^{-x^2}\le e^{-x}$, and

converges. By comparison ${\int }_1^{\infty }{e}^{-x^2}\mathbf{d}x$ converges, and adding the ordinary integral over $[0,1]$ shows ${\int }_0^{\infty }{e}^{-x^2}\mathbf{d}x$ converges. Its exact value $\frac{1}{2}\sqrt{\pi }$ comes later, from multivariable methods, and underlies the normal distribution used in probability. The opposite conclusion follows from a divergent lower bound: since $\frac{1+e^{-x}}{x}\ge \frac{1}{x}$ and ${\int }_1^{\infty }\frac{1}{x}\mathbf{d}x$ diverges, so does ${\int }_1^{\infty }\frac{1+e^{-x}}{x}\mathbf{d}x$.

Between the exact methods and the approximation and limit definitions, every definite integral that arises in practice can be either evaluated exactly, approximated to a stated tolerance, or bounded for convergence.²