Improper integrals and convergence

An endless region can have a finite total

Start with the positive function f(x)=1/x² for x≥1. From 1 to 2 its integral is 1/2. Extending the upper endpoint to 4 gives 3/4; extending it to 8 gives 7/8. Every extension adds something, but the totals get closer to 1. The question is whether all sufficiently long finite pieces approach that same number.

For a finite cutoff R≥1, the antiderivative −1/x gives A(R)=∫₁ᴿ1/x² dx=1−1/R. As R grows without bound, 1/R tends to zero. We define the improper integral ∫₁^∞1/x² dx to be this limiting value, 1. Infinity describes a limiting direction; it is not a real input at which to evaluate the function.

The omitted tail after R has integral 1/R. At R=8 it is 1/8, even though the drawing must stop somewhere. A graph can show the finite piece; the formula establishes what remains beyond every drawing.

The function is 1/x² on x≥1. Predict the accumulated integral from 1 to R and the contribution still beyond R.

Solid blue: the original function. Shading: the selected finite integral. Dashed orange: its cutoff.

The finite integral is 1−1/R and the remaining tail is 1/R. The improper total is 1. The drawing displays only a finite window; the exact tail continues beyond it.

The positive examples need limits and finite integral rules. The final optional proof also uses integration by parts; its Cauchy argument is explained here.

You need signed finite integrals, antiderivatives and limits. Use the preparation links beside the opening graph when a rule is unfamiliar. On a first reading, follow the positive power examples, split-point examples and comparisons. Return to the optional signed-oscillation section after reviewing integration by parts and radian trigonometric derivatives.

Replace the troublesome endpoint before integrating

An ordinary Riemann integral uses a bounded function on a finite closed interval. Continuity there is a sufficient condition for its existence. To handle an endless interval, first assume f is Riemann integrable on every finite interval [a,R] with R>a. Continuity on each such interval is a convenient sufficient condition. Define ∫ₐ^∞f(x)dx=lim as R→∞ of ∫ₐᴿf(x)dx when that limit is a finite real number.

If the limit is finite, the improper integral converges. If it grows to +∞ or −∞, or fails to approach any limit, it diverges. Writing “diverges to +∞” records a direction of growth; it does not turn infinity into a finite integral value. For example, ∫₁ᴿcos x dx=sin R−sin 1 remains bounded but oscillates, so it has no limit as R→∞.

For lettered finite cutoffs, write ∫[A,B] f for the integral of f from A to B, with A<B. For a left infinite end, use ∫_{−∞}ᵇf=lim as L→−∞ of ∫[L,b] f, with L<b and the lower cutoff moving left. A convergent integral of a negative function can have a negative value; signed accumulation is not always geometric area.

Evaluate an antiderivative only on each permitted finite interval, then take the limit of its endpoint difference. A formal substitution of an undefined or infinite endpoint does not check existence. A bounded jump discontinuity need not make an integral improper: many bounded discontinuous functions have ordinary Riemann integrals. The issue here is an unbounded interval or an unbounded endpoint behaviour.

An unbounded height can still contribute a finite amount

The function 1/√x is defined for x>0 and grows without bound as x approaches 0 from the right. For ε>0, integrate over [ε,1]: ∫_ε¹1/√x dx=[2√x]_ε¹=2−2√ε. Letting ε→0⁺ gives 2. The integral converges although the function has no finite height at zero.

This is a one-sided limit of finite integrals. More generally, if f is integrable on every [t,b] with a<t<b, define ∫ₐᵇf=lim as t→a⁺ of ∫ₜᵇf when that limit is finite. An unbounded right endpoint instead uses ∫ₐᵗf with t→b⁻. The cutoff stays inside the original domain.

For the reciprocal-root example, ε=1/4, 1/16 and 1/64 give finite totals 1, 3/2 and 7/4. The missing pieces are 1, 1/2 and 1/4. By contrast, ∫_ε¹1/x dx=−ln ε grows without bound. Being close to the endpoint does not by itself make the remaining contribution small.

The function is 1/√x on x>0. Predict the integral from ε to 1 as ε decreases towards zero.

Solid blue: the original function. Shading: the selected finite integral. Dashed orange: its cutoff.

The finite integral is 2−2√ε and the missing contribution near zero is 2√ε. The limiting total is 2 even though the height is unbounded.

The same power has opposite thresholds at zero and infinity

For a real exponent p and positive x, write x^(−p)=1/xᵖ. If p≠1, its antiderivative is x^(1−p)/(1−p). Hence ∫₁ᴿx^(−p)dx=(R^(1−p)−1)/(1−p). As R→∞, this has the finite value 1/(p−1) exactly when p>1. For p<1 it grows without bound; at p=1 the separate formula ln R also grows without bound.

Near zero, the finite expression is ∫_ε¹x^(−p)dx=(1−ε^(1−p))/(1−p). Now ε→0⁺, so it converges exactly when p<1, to 1/(1−p). For p≥1 it diverges, with the logarithm again handling p=1. When p≤0 the function has a bounded continuous extension at zero, so the limiting result agrees with an ordinary integral of that extension.

The threshold changes because one limit sends x to infinity while the other sends x to zero. In particular, x^(−1/2) is integrable near zero but not over an infinite positive tail; x^(−2) has the opposite behaviour. No single x^(−p) has a convergent integral over all of (0,∞): split at 1 and both conditions would have to hold.

Moving a finite starting point from 1 to another positive a changes a tail's value, not its convergence, provided the intervening finite integral exists. Likewise, changing the length of a neighbourhood around a single endpoint does not change its local convergence when the added finite piece is integrable.

Each unbounded end or interior singularity needs its own limit

Suppose f becomes unbounded at an interior point c of [a,b]. Split the expression into ∫ₐᶜf and ∫꜀ᵇf. The first uses a cutoff approaching c from the left; the second uses a separate cutoff approaching c from the right. Both limits must exist as finite real numbers before their values can be added. One divergent piece makes the original improper integral divergent.

Similarly, ∫_{−∞}^∞f means ∫_{−∞}ᶜf+∫꜀^∞f for any fixed finite split point c at which these pieces make sense, with both tails converging independently. If a tail itself contains a singularity, split it again. For several troublesome points, isolate each one. A valid finite change of the split point does not change a convergent total because finite integrals are additive.

For ∫₋₁¹1/√|x| dx, the left side becomes ∫₀¹1/√u du after u=−x, and is 2. The right side is also 2, so the improper integral is 4. Its two unbounded heights are acceptable because each side has a finite accumulated limit. Replacing the function by 1/|x| makes each side diverge instead.

Matching two cutoffs can hide two divergent pieces

Consider 1/x on [−1,1], excluding zero. With a left gap α>0, ∫₋₁^(−α)1/x dx=ln α, which tends to −∞ as α→0⁺. With a right gap β>0, ∫_β¹1/x dx=−ln β, which tends to +∞. Neither one-sided integral converges, so the ordinary improper integral diverges.

If we force α=β=ε before taking a limit, the two finite terms add to zero for every ε. This linked-cutoff limit is called a Cauchy principal value; here it is 0. It answers a different question from the ordinary improper integral. If the right gap is instead 2ε while the left is ε, the finite sum is −ln 2 for every sufficiently small ε. Other fixed ratios give other answers.

The function is 1/x. The left finite interval is [−1,−α]; the right is [β,1]. Predict the separate signs and the combined finite value.

Solid blue: the left contribution. Dashed orange: the right contribution. Dotted green outline: their finite sum. A hollow circle marks an exact zero.

These are finite contributions, not values of the separate improper integrals. The left limit is −∞ and the right limit is +∞. Matching gaps can make their finite sum zero; it cannot make the ordinary improper integral converge.

The function is 1/x. The left finite interval is [−1,−α]; the right is [β,1]. Predict the separate signs and the combined finite value.

Solid blue: the left contribution. Dashed orange: the right contribution. Dotted green outline: their finite sum. A hollow circle marks an exact zero.

These are finite contributions, not values of the separate improper integrals. The left limit is −∞ and the right limit is +∞. Matching gaps can make their finite sum zero; it cannot make the ordinary improper integral converge.

The general finite sum is ln(α/β). A value obtained by tying the two gaps together cannot repair the failed independent limits. Oddness alone therefore does not justify a zero improper integral across a singularity or two infinite ends. First establish convergence of the separate pieces.

A larger finite total can control a smaller one

Suppose 0≤f(x)≤g(x) beyond a finite starting point, and both functions are integrable on each finite interval there. If ∫g converges over that tail, then ∫f converges too. The truncated integrals of f increase as the endpoint moves right and are bounded above by a finite total for g; an increasing bounded accumulation has a finite limit. Any earlier finite piece must also be integrable.

For divergence, reverse the useful comparison direction: if a nonnegative lower function has a divergent integral, every larger nonnegative function also has a divergent integral. An upper function that diverges gives no conclusion about the smaller one. Likewise, a convergent lower function does not decide what a larger function does.

For x≥1, 1/x²≤(2+sin x)/x²≤3/x². The upper integral is 3, so the middle integral converges; its total lies between 1 and 3. More usefully, after any R≥1 its remaining positive tail is at most 3/R. The comparison proves a limit and a bound without supplying an exact formula for the middle integral.

Solid blue: (2+sin x)/x². Dashed orange: 1/x². Dotted green: 3/x².

The exact inequalities give a total between 1 and 3 and a remaining positive tail at most 3/R. The finite drawing is an illustration of those inequalities, not a numerical evaluation of the total.

For x≥1, x²+1≤2x², so 1/√(x²+1)≥1/(√2 x). The lower reciprocal integral diverges, so the larger integral diverges too. The same comparison principles work near a finite endpoint, using the appropriate one-sided cutoffs. Check the direction of the inequality on the actual domain.

A positive finite ratio gives the same convergence behaviour

Suppose f and g are nonnegative and locally integrable on a tail, g is positive there, and f(x)/g(x) tends to a finite number L>0 as x→∞. For all sufficiently large x, the ratio lies between L/2 and 3L/2. Thus (L/2)g≤f≤(3L/2)g. Positive comparison in both directions proves that the two tail integrals either both converge or both diverge.

For f(x)=(3x+1)/(x³+4) on x≥1, compare with g(x)=1/x². Their ratio is (3x³+x²)/(x³+4), which tends to 3. The reciprocal-square integral converges, so the integral of f converges. The ratio describes their eventual scale; it does not assert equality of the functions or of their integral values.

A ratio tending to zero or infinity does not meet this two-sided theorem's hypotheses. Return to an actual one-sided inequality. For example, both 1/x and 1/x² have ratio tending to zero when compared with g(x)=1, yet the first integral diverges and the second converges. A limiting ratio also says nothing about an unchecked singularity in an earlier finite interval.

An absolute bound controls signed cancellation

Assume the signed real function f is Riemann integrable on every finite interval in the tail under discussion. Then |f| is also integrable on each such interval. Absolute convergence means that the improper integral of |f| converges, and it implies convergence of the integral of f. On every finite tail [A,B], |∫[A,B] f|≤∫[A,B] |f|. A small remaining absolute contribution therefore bounds every sufficiently late signed change.

This is the Cauchy tail condition: for every tolerance η>0, there must be a cutoff M such that |∫[A,B] f|<η whenever M≤A<B. For a function integrable on every finite piece, this condition is equivalent to convergence of its accumulated integrals. It tests every late cutoff pair, not just selected samples.

Here is the real-number fact behind that equivalence. Let F(R) be the accumulated finite integral from a fixed start to R. Its values at successive whole-number cutoffs eventually differ from one another by less than any prescribed tolerance; these values form a Cauchy sequence. Completeness of the real numbers says that such values approach a real number S. The tail condition also makes F(R) close to a late whole-number-cutoff value for every intervening real R, so all cutoffs approach the same S. Conversely, if F(R) approaches S, the triangle inequality bounds |F(B)−F(A)| by |F(B)−S|+|F(A)−S|; make each term less than η/2. No series test is needed.

For f(x)=sin x/x² on x≥1, |f(x)|≤1/x². Thus the integral converges absolutely, and the remaining signed tail after any R≥1 has magnitude at most 1/R. This is an error bound, not an exact value of that error. The same absolute comparison can be applied separately at each troublesome endpoint.

A signed limit can exist even when absolute accumulation diverges

This optional section uses integration by parts and the radian derivative rules (sin x)′=cos x and (cos x)′=−sin x. On 1≤A<B, integration by parts gives ∫[A,B] (sin x/x) dx=cos A/A−cos B/B−∫[A,B] (cos x/x²) dx. Taking magnitudes and using |cos x|≤1 yields |∫[A,B] (sin x/x) dx|≤1/A+1/B+(1/A−1/B)=2/A.

Given any η>0, choose M>max(1,2/η). Every B>A≥M then has |∫[A,B] (sin x/x) dx|<η. The Cauchy condition proves that ∫₁^∞sin x/x dx converges, and its remaining signed tail after any R≥1 has magnitude at most 2/R. This argument establishes convergence and an error bound, without claiming an exact value of the whole integral.

Its absolute integral diverges. On each interval [nπ,(n+1)π] for an integer n≥1, 1/x≥1/((n+1)π) and ∫|sin x|dx over that half-period is 2. Therefore its absolute contribution is at least 2/((n+1)π). Group the indices n=2ᵏ through 2ᵏ⁺¹−1: there are 2ᵏ such intervals and each contributes at least 2/(2ᵏ⁺¹π), so every group contributes at least 1/π. Infinitely many positive groups cannot have a finite total.

Convergence without absolute convergence is called conditional convergence. Here cancellation occurs within a single ordered tail whose every late cutoff is controlled by the 2/A estimate. It is distinct from imposing matching cutoffs on two tails that individually diverge. The sine curve's decreasing envelope suggests the mechanism; the estimates establish it.

Solid blue: sin x/x. Dashed orange: ±1/x. The vertical scale is function height; it is not accumulated integral.

For every B>A≥1, the signed contribution from A to B has magnitude at most 2/A. This estimate proves convergence even though the integral of |sin x|/x diverges.

Combine a finite expression before taking its limit

A valid algebraic rewrite on a finite interval remains useful, even when its separate terms would diverge after extension to infinity. For R>2, ∫₂ᴿ1/[x(x+1)]dx=∫₂ᴿ(1/x−1/(x+1))dx=ln(R/(R+1))+ln(3/2). The combined finite expression tends to ln(3/2) as R→∞.

It would be invalid to turn this into a difference between the two separate infinite integrals: both ∫₂^∞1/x and ∫₂^∞1/(x+1) diverge. The finite identity is valid for every common R, and its combined limit exists. There is no interior singularity on [2,R] and no addition of independently undefined improper values. Linearity of improper integrals requires the relevant separate convergence before their limiting values are combined.

For an actual approximation, distinguish truncation error from error in calculating the finite integral. A proved tail bound controls only the first. For the reciprocal-square example, using R=1000 leaves exactly 0.001 beyond the cutoff; a requirement of strictly less than 0.001 needs R>1000. For a comparison tail bounded by 3/R, choosing R>3000 gives that strict guarantee.

Check every troublesome place before reporting a total

Inspect the domain first. Mark every infinite end and every unbounded interior or endpoint behaviour. Split until each limiting piece has only one troublesome end. Write its finite cutoff integral and approach from the permitted side. If a checked primitive is available, take the limit of its full endpoint difference.

If an exact primitive is unhelpful, choose a positive comparison, a finite positive limiting ratio, an absolute bound or a signed tail estimate with stated hypotheses. Record whether you established a finite value, convergence alone or a numerical error bound. Small values of the integrand, a smooth-looking graph and agreement at a few cutoffs do not prove convergence.

When an integral represents a physical accumulation, the quantity, sign, units and model over the whole interval still need justification. Extending a model to endless time or arbitrarily close to a singular point is an additional modelling decision. Convergence is a mathematical property of that stated function and domain; it does not establish that the model remains physically valid there.

Worked examples

Try it yourself

1. For which real p does ∫₀¹x^(−p)dx converge, and for which p does ∫₁^∞x^(−p)dx converge?

   Show the answer and reasoning

   Near zero: p<1. At infinity: p>1.

   Use ε→0⁺ in the first finite power expression and R→∞ in the second. At p=1 each logarithm diverges.

2. Is there a real p for which ∫₀^∞x^(−p)dx converges?

   Show the answer and reasoning

   No.

   Splitting at 1 requires p<1 and p>1 simultaneously. Convergence of just one end is insufficient.

3. The finite integrals ∫₀ᴿcos x dx stay between −1 and 1. Does the improper integral converge?

   Show the answer and reasoning

   No.

   The finite total is sin R, which has no single limit. Boundedness alone controls no convergence for signed accumulations.

4. For 1/x on [−1,1], exclude a left gap ε and a right gap 2ε. What is the combined finite integral, and does it establish an ordinary improper value?

   Show the answer and reasoning

   −ln 2; it does not establish an ordinary improper value.

   The finite sum is ln(ε/(2ε)). The separate one-sided limits diverge to opposite infinities, so the ordinary improper integral diverges.

5. Since 0≤1/x²≤1/x for x≥1 and the integral of 1/x diverges, must the integral of 1/x² diverge?

   Show the answer and reasoning

   No.

   A divergent upper comparison is inconclusive. The smaller reciprocal-square integral converges to 1.

6. If |f(x)|≤4/x² for x≥2 and f is integrable on every finite interval there, what follows about ∫₂^∞f and its tail after R≥2?

   Show the answer and reasoning

   It converges absolutely, and the remaining signed tail has magnitude at most 4/R.

   The positive bound has a finite integral. Its tail controls ∫|f| and hence the magnitude of the signed tail; this is a bound, not a claimed exact value.

7. A positive tail is at most 3/R. Does R=300 guarantee an error strictly less than 0.01?

   Show the answer and reasoning

   No. Choose R>300 for that guarantee.

   At R=300 the bound is 0.01, so it only guarantees error at most 0.01. A strict inequality requires a strictly smaller bound.

8. If one side of an interior singularity converges to 2 and the other diverges to +∞, does the full improper integral converge?

   Show the answer and reasoning

   No.

   Every one-sided piece must converge to a finite value. One successful piece cannot repair a divergent one.

9. Does convergence of ∫₁^∞sin x/x dx imply convergence of ∫₁^∞|sin x|/x dx?

   Show the answer and reasoning

   No; this example is conditionally convergent.

   The signed tail obeys the 2/A estimate, while positive half-period contributions of the absolute value have an unbounded total. Absolute convergence would imply signed convergence, but the converse fails.

10. For f(x)=x/(1+x²), the integral over [−R,R] is zero for every R>0. Does ∫_{−∞}^∞f converge?

    Show the answer and reasoning

    No.

    On [0,R] the integral is (1/2)ln(1+R²), which diverges to +∞. The left tail diverges to −∞. Their symmetric cancellation is a principal-value calculation, not two convergent tails.

11. May −1/x simply be evaluated at x=∞ to justify ∫₁^∞1/x² dx?

    Show the answer and reasoning

    Use a limit instead.

    For every finite R≥1 the integral is 1−1/R; that expression tends to 1. Infinity is a direction for R, not a real function input.

What to carry forward

• An improper integral is a limit of valid finite integrals; convergence requires a finite real limit.
• Check every unbounded end and every singular side independently before adding values.
• The power threshold is p>1 at infinity and p<1 near zero; p=1 needs its logarithm.
• Positive comparison and finite positive ratio comparison need verified inequalities, domains and local integrability.
• Absolute convergence controls signed convergence and tail error; conditional convergence needs its own signed argument.
• A Cauchy principal value uses linked cutoffs and can exist when the ordinary improper integral diverges.
• Keep cancellations inside a valid finite expression until its combined limit has been checked; numerical samples do not prove convergence.