The
definite integral
\int_a^b f(x)\,dx was built for a finite interval
and a bounded integrand. An improper integral relaxes one of
those: the interval may run off to \infty, or the integrand may blow
up at a point. We give it meaning the only honest way — as a
limit.
Over an infinite interval, integrate up to a finite cut-off and then let the
cut-off run away:
\int_a^{\infty} f(x)\,dx = \lim_{b \to \infty} \int_a^{b} f(x)\,dx.
Where the integrand has a vertical asymptote at an endpoint — say
f is unbounded near a — creep up to the
bad point from the safe side:
\int_a^{b} f(x)\,dx = \lim_{t \to a^{+}} \int_t^{b} f(x)\,dx.
If the limit is a finite number, the integral converges to that number. If the
limit is infinite or fails to exist, the integral diverges. (A two-sided
improper integral, infinite at both ends or with a bad point in the middle, is split into pieces
and each piece must converge on its own.)
The benchmark: \displaystyle\int_1^{\infty} x^{-p}\,dx
One family settles almost every example you will meet by comparison, so we evaluate it once and
for all. The behaviour turns entirely on whether the exponent p is
bigger than 1.
Step 1 — replace \infty by a finite cut-off
b. By definition the improper integral is the limit of an
ordinary one:
\int_1^{\infty} x^{-p}\,dx = \lim_{b \to \infty} \int_1^{b} x^{-p}\,dx.
Step 2 — antidifferentiate, assuming p \neq 1. The
power rule for integration adds one to the exponent and divides by the new exponent
(1 - p):
\int_1^{b} x^{-p}\,dx = \left[\frac{x^{1-p}}{1-p}\right]_{1}^{b} = \frac{b^{1-p} - 1}{1-p}.
Step 3 — take the limit, and watch the sign of the exponent
1 - p. Everything hinges on what
b^{1-p} does as b \to \infty.
Case p > 1. Then 1 - p < 0,
so b^{1-p} = 1/b^{\,p-1} \to 0. The lone survivor is the
-1:
\lim_{b \to \infty} \frac{b^{1-p} - 1}{1-p} = \frac{0 - 1}{1 - p} = \frac{1}{p - 1}. \qquad \textbf{Converges.}
Case p < 1. Then 1 - p > 0,
so b^{1-p} \to \infty and the whole expression blows up:
\lim_{b \to \infty} \frac{b^{1-p} - 1}{1-p} = +\infty. \qquad \textbf{Diverges.}
Case p = 1. The power rule breaks (it would divide by
1 - p = 0), and the antiderivative is the logarithm instead:
\int_1^{b} \frac{dx}{x} = \big[\ln x\big]_1^{b} = \ln b \;\xrightarrow{\;b \to \infty\;}\; +\infty. \qquad \textbf{Diverges.}
So the integral converges precisely when p > 1 — and at the borderline
p = 1 the logarithm's famously lazy growth still wins for infinity. The
area under 1/x out to infinity is infinite; under
1/x^2 it is exactly 1.
For the benchmark improper integral over [1, \infty):
-
If p > 1, the integral converges, with value
\displaystyle \int_1^{\infty} x^{-p}\,dx = \frac{1}{p-1}.
-
If p \le 1 (including p = 1, the
logarithm), the integral diverges to +\infty.
-
The companion near the origin flips the inequality:
\displaystyle \int_0^{1} x^{-p}\,dx converges iff
p < 1 — a small power tames the spike, a large one does not.
Spin the curve y = 1/x for x \ge 1 about
the x-axis. The solid it sweeps out — Gabriel's horn — has a
finite volume but an infinite surface area, a clash that
delighted (and unsettled) seventeenth-century mathematicians.
The volume uses the
disk method,
an integrand of 1/x^2 — a p = 2 > 1 case,
so it converges:
V = \pi \int_1^{\infty} \left(\frac{1}{x}\right)^{2} dx = \pi \int_1^{\infty} x^{-2}\,dx = \pi \cdot \frac{1}{2-1} = \pi.
The surface area carries an extra factor of \sqrt{1 + f'(x)^2} \ge 1,
which drags the integrand down toward 1/x — a
p = 1 case, which diverges:
S = 2\pi \int_1^{\infty} \frac{1}{x}\sqrt{1 + \tfrac{1}{x^4}}\,dx \;>\; 2\pi \int_1^{\infty} \frac{dx}{x} = +\infty.
You could fill the horn with \pi cubic units of paint, yet never
have enough to coat its inside. The paradox dissolves once you notice that "coat" assumes a
layer of fixed thickness — which the throat, shrinking forever, cannot accommodate. The
mathematics is innocent; the intuition about paint is the liar.