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Math Analysis 1

Improper integrals: definition, reference integrals and convergence tests

Integrals over unbounded intervals and of unbounded functions, the reference integrals, comparison and limit comparison, the integral test for series.

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The definite integral was built for closed bounded intervals and bounded functions. Many useful integrals break these conditions: the energy of a signal that lasts forever, \int_0^{+\infty} e^{-t}dt, or the area under \frac{1}{\sqrt x} near 0. Improper integrals define them as limits and tell you when they have a finite value.

Two kinds of improper integral

Unbounded interval. \displaystyle\int_a^{+\infty} f(x)\,dx = \lim_{M\to+\infty}\int_a^M f(x)\,dx.

Function unbounded near an endpoint. If f blows up at a: \displaystyle\int_a^b f(x)\,dx = \lim_{\varepsilon\to0^+}\int_{a+\varepsilon}^b f(x)\,dx.

If the limit is finite the integral converges, if it is infinite it diverges, otherwise it does not exist. If there is more than one problem (for instance on all of \mathbb R), split the integral so that each piece has only one, and the integral converges if every piece converges.

The reference integrals

Everything rests on two results, checked by computing the antiderivative \frac{x^{1-\alpha}}{1-\alpha} (or \ln x for \alpha = 1):

    \[ \int_1^{+\infty}\frac{dx}{x^\alpha} \ \text{converges} \iff \alpha > 1, \qquad \int_0^{1}\frac{dx}{x^\alpha} \ \text{converges} \iff \alpha < 1. \]

The case \alpha = 1 diverges in both: \int_1^M\frac{dx}{x} = \ln M \to \infty. Examples: \int_1^{+\infty}\frac{dx}{x^2} = 1, \int_0^1 \frac{dx}{\sqrt x} = 2. For the exponential: \int_0^{+\infty} e^{-ax}dx = \frac1a for a > 0.

Convergence tests

Often the antiderivative cannot be found, but you want to know whether the integral converges. For positive functions:

  • Comparison: if 0 \le f \le g and \int g converges, so does \int f; if \int f diverges, so does \int g.
  • Limit comparison: if f \sim g (the ratio tends to a positive number), the two integrals behave the same way. This is the most used test: replace the function with the power that approximates it.
  • Absolute convergence: if \int |f| converges, so does \int f.

Examples.

  • \int_1^{+\infty}\frac{x + 1}{x^3 + 2}\,dx: at infinity the function is \sim \frac{1}{x^2}, it converges.
  • \int_0^1\frac{dx}{\sin x}: near 0, \sin x \sim x, so it behaves like \int_0^1\frac{dx}{x} and diverges.
  • \int_1^{+\infty}\frac{\sin x}{x^2}\,dx: \left|\frac{\sin x}{x^2}\right| \le \frac{1}{x^2}, it converges absolutely.
  • \int_0^{+\infty} e^{-x^2}dx: for x \ge 1, e^{-x^2} \le e^{-x}, so it converges. Its value is \frac{\sqrt\pi}{2}, a result proved in multivariable calculus that underpins all the statistics of the normal distribution.

The link with series

The integral test: if f is positive and decreasing, the series \sum f(n) and the integral \int_1^{+\infty} f(x)\,dx converge or diverge together. This gives the behaviour of the generalised harmonic series \sum\frac{1}{n^\alpha}, which converges only for \alpha > 1.

Why engineers care

The energy of a signal is \int_{-\infty}^{+\infty} |x(t)|^2 dt: an exponential pulse e^{-t/\tau} for t \ge 0 has energy \frac{\tau}{2}, which is finite. The Fourier and Laplace transforms are improper integrals, and whether they exist depends on exactly these tests: \int_0^{+\infty} f(t)e^{-st}dt converges when the decaying exponential “beats” the growth of f.

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