A Fourier series only ever integrates a periodic function over one of its own periods. That's a perfectly ordinary, finite integral. The Fourier transform, several chapters ahead, integrates over all of (-\infty, \infty) instead. Deciding whether a signal even qualifies for a transform means deciding whether that infinite integral converges. This section is a fast pass over the two ways an integral turns improper, and the one test that decides convergence without an exact antiderivative.
Picture the area under a curve as water filling a container that stretches out infinitely far in one direction. Maybe it runs off to the horizon, or narrows down into a crack without ever fully closing. "Does the improper integral converge" is really asking one thing. Is there a finite amount of water in that infinitely long container, or does it need an infinite amount to fill it? A container that narrows fast enough holds a finite amount even though it never ends. That's convergence. One that stays wide, or gets wider, never finishes filling. That's divergence.
The comparison test judges one container by a nearby one you already know the answer for, without measuring either exactly. Suppose a wider container next to yours is already known to hold a finite amount. Then your narrower one, sitting entirely inside it, can't possibly hold more, so it converges too. Suppose instead a narrower container next to yours already needs infinite water. Then your wider one, containing it, needs at least that much, so it diverges too. The container being compared has to actually sit on the correct side, bigger or smaller, of the one in question. Otherwise the comparison says nothing at all.
An integral is improper in one of two ways. Either the interval is unbounded (a type 1 improper integral, \int_a^{\infty} f(x)\,dx or similar), or the integrand blows up at some point in the interval, most often an endpoint (a type 2 improper integral). Both are defined the same way, as a limit of an ordinary, proper integral, with the trouble spot pushed toward but never reached: \int_a^{\infty} f(x)\,dx = \lim_{t\to\infty}\int_a^{t} f(x)\,dx, \qquad \int_a^{b} f(x)\,dx = \lim_{t\to b^-}\int_a^{t} f(x)\,dx \text{ (if } f \text{ blows up at } b\text{)}. The improper integral converges if that limit exists and is finite. It diverges otherwise, whether by running off to \pm\infty or by simply failing to settle.
Not every improper integral needs an antiderivative to decide. The comparison test settles convergence from an inequality alone. For 0 \le f(x) \le g(x) on [a,\infty): if \int_a^{\infty} g converges, then \int_a^{\infty} f converges too, squeezed underneath a finite number. If \int_a^{\infty} f diverges, then \int_a^{\infty} g diverges too, dragged up from below by something already infinite. The direction of the inequality is what makes the argument valid. Comparing against a smaller convergent integral proves nothing about a larger one.
Theorem
Comparison Test for Improper Integrals
Recalled from Calculus II — Integral Calculus & Infinite Series · Comparison tests for improper integrals — first taught there.
Suppose f and g are continuous on [a,\infty) with 0 \le f(x) \le g(x) for all x\ge a.
- If \displaystyle\int_a^\infty g(x)\,dx converges, then \displaystyle\int_a^\infty f(x)\,dx converges too.
- If \displaystyle\int_a^\infty f(x)\,dx diverges, then \displaystyle\int_a^\infty g(x)\,dx diverges too.
(An identical statement holds for an improper integral caused by a finite discontinuity instead of an infinite bound.)
Let F(t)=\int_a^t f(x)\,dx. Since f\ge0, F is nondecreasing in t, so F(t) either grows without bound or approaches a finite limit — there is no other option for a nondecreasing function. If \int_a^\infty g\,dx converges to some value L, then for every t\ge a, F(t)=\int_a^t f\,dx \le \int_a^t g\,dx \le L, since f\le g pointwise and g\ge 0 makes \int_a^t g\,dx itself nondecreasing in t with limit L. A nondecreasing function bounded above by L must converge to some limit at most L — it cannot grow without bound — so \int_a^\infty f\,dx converges. The second bullet is the direct contrapositive: if f's integral grew without bound, the same inequality F(t)\le\int_a^t g\,dx would force g's integral to grow without bound as well.
How fast the tail has to shrink
Drag p from below 1 to above 1. Watch how much area survives out past x=1.
Loading interactive…
A type-1 integral settling toward a number
Drag t out toward infinity. Watch the running value level off instead of climbing forever.
Loading interactive…
Recipe Test an improper integral for convergence Use when: the integral has an infinite limit of integration, or the integrand is undefined (blows up) somewhere on the interval Goal: either the exact value the integral converges to, or a converges-or-diverges verdict backed by a valid comparison
Quick recipe
-
1
Decide whether the trouble is an unbounded interval or a point where the integrand blows up, then rewrite the integral with that trouble spot replaced by a variable t.
Example
\displaystyle\int_1^{\infty}\frac{1}{x^3}\,dx = \lim_{t\to\infty}\int_1^t x^{-3}\,dx, an unbounded interval.
- the interval itself is unbounded, such as a to \infty — rewrite as \lim_{t\to\infty}\int_a^t f(x)\,dx.
- the integrand blows up at a point c in the interval, often an endpoint — rewrite as a limit of t toward c instead.
Why: an improper integral is only ever defined as this limit; treating infinity as a number you can plug in skips the definition entirely.
-
2
Integrate the function normally up to t. Then take the limit as t moves toward the trouble spot; a finite limit means convergence, an infinite or non-existent one means divergence.
Example
\int_1^t x^{-3}\,dx = \frac12-\frac{1}{2t^2}, and as t\to\infty that goes to \frac12, so \int_1^{\infty}x^{-3}\,dx=\frac12, a genuine convergent value.
- an antiderivative is available and the limit settles to a finite number — the integral converges to that number; done.
- an antiderivative is available but the limit runs off to infinity or fails to exist — the integral diverges; done.
- no elementary antiderivative can be found for the integrand — skip evaluating and continue to Step 3, the comparison test.
Why: the limit's own behavior, finite versus infinite or non-existent, is literally what convergence and divergence mean.
-
3
Choose a comparison function g(x), matching the inequality's direction to what needs proving.
Example
Since -1\le\cos x\le1, 2\le 3+\cos x\le4, so 0\le\frac{3+\cos x}{x^3}\le\frac{4}{x^3} for every x\ge1.
- the goal is proving convergence — find g already known convergent, with f(x)\le g(x) on the whole interval.
- the goal is proving divergence — find g already known divergent, with f(x)\ge g(x) on the whole interval.
Why: the direction of the inequality is what makes the argument valid; comparing against a smaller convergent integral, or a larger divergent one, proves nothing.
-
4
Once the comparison function's own status is settled, the original integral inherits the same converges-or-diverges verdict, with no exact value attached.
Example
4\int_1^{\infty}x^{-3}\,dx = 4\left(\frac12\right)=2, a finite number, so \int_1^{\infty}\frac{3+\cos x}{x^3}\,dx converges too, even though its own exact value stays unknown.
Why: squeezing f underneath a finite g, or above an infinite one, only pins down converges-or-diverges, since an inequality is not an equality.
Common mistakes
- Plugging infinity directly into the antiderivative instead of taking a genuine limit.
- Comparing against a smaller convergent integral, or a larger divergent one, the wrong-direction pairing that proves nothing.
- Reporting an exact value from a comparison test, when it only ever proves converges or diverges.
- Assuming the integrand shrinking to zero is enough for convergence, when the rate it shrinks at is what actually matters.
Worked example
Evaluating a convergent type-1 integral
-
1
Evaluate \displaystyle\int_1^{\infty} \frac{1}{x^2}\,dx, following Steps 1 and 2 of the recipe above: rewrite as a limit, then evaluate the antiderivative directly.
-
2the trouble spot is pushed to t, never evaluated exactly at infinity
Rewrite as a limit: \displaystyle\lim_{t\to\infty}\int_1^t x^{-2}\,dx.
the trouble spot is pushed to t, never evaluated exactly at infinity
-
3\int_1^t x^{-2}\,dx = \left[-\frac{1}{x}\right]_1^t = -\frac{1}{t}+1.
As t\to\infty, -\frac1t \to 0.
-
4\boxed{\int_1^{\infty}\frac{1}{x^2}\,dx = 1}
Finite. The tail's area really does settle at exactly 1.
Worked example
A type-1 integral that only looks similar
-
1
Evaluate \displaystyle\int_1^{\infty} \frac{1}{x}\,dx.
Common trap. \frac1x \to 0 just as \frac{1}{x^2} does. It's tempting to assume "shrinks to zero" is enough for the area to be finite, but Steps 1 and 2 of the recipe above still have to be run before trusting that instinct.
-
2\int_1^t x^{-1}\,dx = \big[\ln x\big]_1^t = \ln t - \ln 1 = \ln t.
-
3
As t\to\infty, \ln t \to \infty. The antiderivative itself has no ceiling, even though the integrand's height is shrinking to 0.
-
4
\boxed{\text{Diverges}}. Shrinking to zero isn't the same as shrinking fast enough. \frac{1}{x^2} converged and \frac1x diverges precisely because p=2>1 and p=1 sits right at the boundary the next example names explicitly.
Loading interactive…
A type-2 integral that never settles
Drag t in toward the discontinuity at x=0. Watch the running value climb with no ceiling.
Loading interactive…
Worked example
Comparison without an antiderivative
-
1
Does \displaystyle\int_1^{\infty} \frac{2+\sin x}{x^2}\,dx converge?
Recognition. 2+\sin x has no clean antiderivative to pair with x^{-2}, so integrating directly is the wrong first move. The shape, a bounded numerator over x^2, is what should trigger Steps 3 and 4 of the recipe above, the comparison test, instead.
-
2
Since -1\le\sin x\le 1, the numerator satisfies 1\le 2+\sin x\le 3, so 0 \le \frac{2+\sin x}{x^2} \le \frac{3}{x^2} \quad \text{for all } x\ge 1.
-
3
\int_1^{\infty}\frac{3}{x^2}\,dx = 3\int_1^{\infty}\frac{1}{x^2}\,dx = 3(1)=3 converges. It's a constant multiple of the convergent integral from the first example. So the smaller integral is squeezed under a finite number too.
-
4
\boxed{\text{Converges}}, without ever finding an antiderivative for 2+\sin x over x^2.
Where this goes next
The Fourier transform, chapter 7
The Fourier transform of a signal f(t) is itself an improper integral, \int_{-\infty}^{\infty} f(t)e^{-i\omega t}\,dt. A signal only has a well-behaved transform when that integral converges. In practice, that's usually checked through exactly the comparison-test reasoning above, rather than by finding a closed-form antiderivative for an arbitrary signal.
Check your understanding (counts toward your progress)
Does \displaystyle\int_1^{\infty}\frac{1}{\sqrt{x}}\,dx converge or diverge?
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
\frac{1}{\sqrt{x}} = x^{-1/2} — this is the p=\frac12 case, and p\le 1 is exactly the boundary where the tail no longer shrinks fast enough.
Why
This is \int_1^\infty x^{-1/2}\,dx, a p-integral with p=\frac12 \le 1, which diverges — the same threshold that made \int_1^\infty \frac1x\,dx diverge in the lesson.
Rewrite \frac{1}{\sqrt{x}} as x^{-1/2} and compare its exponent to the p=1 threshold.
Does \displaystyle\int_1^{\infty}\frac{1+\cos x}{x^3}\,dx converge?
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
\cos x is bounded, between -1 and 1 — it never makes the integrand undefined.
A comparison settles this without ever finding an antiderivative for the numerator.
Why
Since 0\le 1+\cos x\le 2, the integrand is squeezed under \frac{2}{x^3}, whose integral converges (p=3>1), so the original integral converges too.
Bound 1+\cos x above by a constant, then compare the resulting integral to a p-integral.
Evaluate \displaystyle\int_1^{\infty}\frac{1}{x^3}\,dx.
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Any equivalent form counts — 2x+3, 3 + 2·x, … Use the keypad below the field for fractions, roots, and other symbols.
Why
\int_1^t x^{-3}\,dx = \left[-\frac{1}{2x^2}\right]_1^t \to \frac12 as t\to\infty.
Correct answer
Also accepted:
Why
Practice: Type 1, type 2, or compare — decide which (4 problems · doesn’t affect your progress)
Determine the value of the improper integral \displaystyle\int_1^{\infty}\frac{1}{x^4}\,dx.
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Any equivalent form counts — 2x+3, 3 + 2·x, … Use the keypad below the field for fractions, roots, and other symbols.
Why
\int_1^t x^{-4}\,dx = \left[-\frac{1}{3x^3}\right]_1^t \to \frac13 as t\to\infty.
Correct answer
Also accepted:
Why
Does \displaystyle\int_0^{1}\frac{1}{\sqrt{x}}\,dx converge or diverge?
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
\frac{1}{\sqrt x}=x^{-1/2} blows up at x=0, but a type-2 integral \int_0^1 x^{-p}\,dx converges whenever p<1 — here p=\frac12<1.
Why
This is a type-2 improper integral with p=\frac12<1, which converges — the opposite threshold from a type-1 tail, where p>1 is what's needed.
For a type-2 integral \int_0^1 x^{-p}\,dx, convergence needs p<1 — a different threshold than the type-1 case.
Converges
Determine whether \displaystyle\int_1^{\infty}\frac{2+\cos x}{x^3}\,dx converges.
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
\cos x is bounded between -1 and 1 — it never makes the integrand undefined.
A comparison settles convergence without ever finding an antiderivative for the numerator.
Why
Since 0\le 2+\cos x\le 3, the integrand is squeezed under \frac{3}{x^3}, whose integral converges (p=3>1), so the original integral converges too.
Yes, by comparison to \int_1^\infty \frac{3}{x^3}\,dx
Determine whether \displaystyle\int_2^{\infty}\frac{1}{x}\,dx converges.
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
\int_2^t x^{-1}\,dx = \ln t - \ln 2, and \ln t\to\infty as t\to\infty — the antiderivative has no ceiling.
Why
\frac1x sits right at the p=1 boundary: it shrinks to 0, but not fast enough for the tail's area to settle. Moving the lower limit from 1 to 2 changes nothing about that.
Diverges