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Improper Integrals



What Makes an Integral Improper?
Infinite Limits of Integration
Improper Integral Notation
Discontinuous Integrands
Convergence vs Divergence
The p-Test
Comparison Test
Limit Comparison Test
Summary: Tests for Improper Integral Convergence
Common Pitfalls
Improper Integrals FAQ



Integrating Beyond Bounds


Standard definite integrals require finite intervals and bounded integrands. Improper integrals remove these restrictions, extending integration to infinite intervals and functions with vertical asymptotes.

The key idea: replace the problematic bound or point with a variable, compute the resulting proper integral, then take a limit. If the limit exists and is finite, the improper integral converges to that value. If not, it diverges.

Some infinite regions have finite area. The region under 1/x21/x^2 from 11 to ∞\infty has area exactly 11. Other infinite regions have infinite area—the region under 1/x1/x from 11 to ∞\infty diverges. The distinction matters throughout mathematics and physics.

Key Terms

Improper Integral— infinite interval or unbounded integrand, evaluated as a limit
Definite Integral— improper integrals extend the definite integral framework
Bounds of Integration— one or both bounds may be ±∞\pm\infty
Limit— convergence or divergence determined by whether the limit exists

See All Calculus Definitions →


What Makes an Integral Improper?

How much area lies under y = 1/x² from x = 1 onward, with no end?
The region under the curve gets thinner and thinner but never stops. It is a fair question whether its total area is some finite number or simply infinite. Written as an integral, the question is
What Makes an Integral Improper?
∫1∞1x2 dx\int_1^{\infty} \frac{1}{x^2}\, dx
y = 1/x^2 from x = 1 onward never reaches the axisxy123456789101x = 1y = 1/x²the region runs on to the right foreverEver thinner, but it never ends
The region starts at x=1x = 1 and continues past every number on the axis.
The definite integral, as defined so far, cannot answer it.
A definite integral runs between two numbers, and ∞\infty is not a number: there is no right-hand endpoint to evaluate an antiderivative at, and no finite interval to cut into rectangles. The same trouble appears when the interval is finite but the function is not — here the integrand blows up at x=0x = 0:
What Makes an Integral Improper?
∫011x dx\int_0^1 \frac{1}{\sqrt{x}}\, dx
Both conditions can occur simultaneously. The integral
What Makes an Integral Improper?
∫0∞1x(1+x) dx\int_0^{\infty} \frac{1}{\sqrt{x}(1 + x)}\, dx
has an infinite upper limit and an unbounded integrand at x=0x = 0.
-4 -3 -2 -1 1 2 3 4 -4 -3 -2 -1 1 2 3 4 x y f(x) = 1/x²
1/x² near x = 0: the integrand is unbounded

Both probes, at −0.5 and 0.5, read 4, and moving them inward sends the values to infinity: the curve has a vertical asymptote at 0, so no rectangle of finite height can cover the region beside it. An integral across this point must stop short at t and let t approach 0 in a limit. Squeeze the probes toward the asymptote on the limit explorer.

Stopping at a finite point, and moving that point outward, gets close to an answer.
Replace ∞\infty by a number bb. From 11 to bb the integral is an ordinary one: ∫1b1x2 dx=1−1b\int_1^b \frac{1}{x^2}\, dx = 1 - \frac{1}{b}. Now push bb out.
stop at barea from 1 to b: 1 − 1/bstill missing
20.50.5
50.80.2
100.90.1
1000.990.01
10000.9990.001
The area so far keeps growing, but by less and less, and what is still missing shrinks toward 00. The total heads for 11.
Both kinds of impropriety are handled the same way: replace the bad endpoint by a variable and take a limit.
The definition
An improper integral is an integral over an infinite interval, or of a function that is unbounded on the interval. Its value is defined as the limit of ordinary definite integrals, for example ∫a∞f(x) dx=lim⁡b→∞∫abf(x) dx\int_a^{\infty} f(x)\, dx = \lim_{b \to \infty} \int_a^b f(x)\, dx.
It converges when that limit is a finite number and diverges otherwise. Here ∫1∞1x2 dx=1\int_1^{\infty} \frac{1}{x^2}\, dx = 1.
An improper integral is a number when it converges, and a verdict of divergence when it does not.
When the limit exists, the improper integral is one number — an infinitely long region with finite area, as here. When the limit does not exist, there is no number at all, only the statement that the integral diverges.
A table of values suggests an answer but does not prove it.
Compare 1x\frac{1}{x}, a curve that looks almost the same.
stop at bunder 1/x²: 1 − 1/bunder 1/x: ln b
100.92.30
1000.994.61
10000.9996.91
10⁶0.99999913.82
The second column also grows by less and less, and from a table alone it could pass for levelling off. It does not: ln⁡b\ln b grows without bound, so ∫1∞1x dx\int_1^{\infty} \frac{1}{x}\, dx diverges. What settles the first column is algebra, not the table — 1−1b1 - \frac{1}{b} really does tend to 11. Deciding which case you are in, without a table, is what the rest of this page is about.
The rest of the page defines each type precisely — infinite limits of integration and unbounded integrands — then gives the notation, the p-integral test, worked convergence and divergence, and the comparison tests that decide convergence without computing the integral.

Infinite Limits of Integration


Replace the infinite limit with a finite variable and take a limit.

Type 1: Upper limit infinite

Improper Integral (Infinite Limits)
∫a∞f(x) dx=lim⁡b→∞∫abf(x) dx\int_a^{\infty} f(x)\, dx = \lim_{b \to \infty} \int_a^b f(x)\, dx
Learn more about this formula: Improper Integral (Infinite Limits) →


Type 2: Lower limit infinite

Infinite Limits of Integration
∫−∞bf(x) dx=lim⁡a→−∞∫abf(x) dx\int_{-\infty}^b f(x)\, dx = \lim_{a \to -\infty} \int_a^b f(x)\, dx

Type 3: Both limits infinite

Infinite Limits of Integration
∫−∞∞f(x) dx=∫−∞cf(x) dx+∫c∞f(x) dx\int_{-\infty}^{\infty} f(x)\, dx = \int_{-\infty}^c f(x)\, dx + \int_c^{\infty} f(x)\, dx

Pitfall: Running both ends out together
Using one variable for both ends gives lim⁡R→∞∫−RR2x1+x2 dx=0\lim_{R \to \infty} \int_{-R}^{R} \frac{2x}{1 + x^2}\, dx = 0. Split at c=0c = 0 instead: ∫0b2x1+x2 dx=ln⁡(1+b2)→∞\int_0^b \frac{2x}{1 + x^2}\, dx = \ln(1 + b^2) \to \infty, so the integral diverges.

Both integrals must converge independently. The choice of cc is arbitrary—any finite value works.

Splitting turns one integral with two infinite ends into two integrals with one infinite end each: a Type 2 integral on (−∞,c](-\infty, c] and a Type 1 integral on [c,∞)[c, \infty). Moving cc shifts area from one piece to the other, which changes the two pieces but not whether they converge, and not what they add up to.

Improper Integral Notation

Notation

Improper Integral Notation

A symbol that quietly means a limit, the scaffold that makes the limit visible, and the verdict vocabulary that comes with it.
∫ab\int_a^b and its parts — definite integrals; ∞\infty and lim⁡=∞\lim = \infty — limits at infinity; one-sided arrows — one-sided limits.
∫a∞f(x) dx\int_a^{\infty} f(x)\, dx
The integral from a to infinity
∞\infty in the bound is not a point — the whole symbol is defined as lim⁡b→∞∫abf(x) dx\lim_{b \to \infty} \int_a^b f(x)\,dx, the scaffold of Infinite Limits of Integration above. The compact form hides the limit; the definition is the limit.
CasesLower bound −∞-\infty mirrors it. Both bounds infinite — split at an arbitrary cc, and each half must converge on its own.
Also written∫a+∞\int_a^{+\infty}, with the sign spelled out — matching the signed-infinity tradition of limits at infinity.
Do not confuseThe symmetric limit lim⁡R→∞∫−RR\lim_{R \to \infty} \int_{-R}^{R} — Cauchy's principal value, written PV ⁣∫\mathrm{PV}\!\int. Not the same object: ∫−∞∞x dx\int_{-\infty}^{\infty} x\,dx diverges, while its principal value is 00. Tying the two ends to one variable changes the definition.
lim⁡t→b−∫atf(x) dx\lim\limits_{t \to b^{-}} \int_a^t f(x)\, dx
The limit, as t approaches b from below, of the integral from a to t
The scaffold for an unbounded integrand: the bad endpoint is replaced by a variable that approaches it from inside the interval — the one-sided superscript carries the direction. Nothing in ∫011x dx\int_0^1 \frac{1}{\sqrt{x}}\,dx announces impropriety; the integrand has to be inspected, then the scaffold written.
CasesAsymptote at the left endpoint — t→a+t \to a^{+}; at the right — t→b−t \to b^{-}; at an interior cc — two scaffolds with independent letters tt and ss, as in Discontinuous Integrands below.
Also writtenε\varepsilon-style: lim⁡ε→0+∫ab−ε\lim_{\varepsilon \to 0^{+}} \int_a^{b-\varepsilon}, common in analysis texts, where the small excision is named directly.
Do not confuseOne letter for both halves of an interior split. That ties the two limits together — a hidden principal value again — and can declare a divergent integral finite.
converges · diverges · ∫<∞\int < \infty
The integral converges; the integral diverges
Verdicts about the defining limit, as in Convergence vs Divergence below. Only after “converges” is the equation form ∫1∞1x2 dx=1\int_1^{\infty} \frac{1}{x^2}\,dx = 1 legitimate — the value is the limit.
CasesDivergence comes in two species: =∞= \infty, borrowing the equals-abuse of infinite limits, or by oscillation — ∫0∞sin⁡x dx\int_0^{\infty} \sin x\,dx — where no verdict value exists at all.
Also written∫a∞∣f∣<∞\int_a^{\infty} |f| < \infty — the analyst's shorthand for “converges absolutely”, an inequality doing the work of a sentence.
Do not confuse“Diverges” with “equals infinity”. Every =∞= \infty diverges; not every divergence is an ∞\infty — the oscillating case fails without ever growing.

Discontinuous Integrands


When ff has a vertical asymptote at cc within [a,b][a, b], split the integral and use limits.

Asymptote at left endpoint:

Discontinuous Integrands
∫abf(x) dx=lim⁡t→a+∫tbf(x) dx\int_a^b f(x)\, dx = \lim_{t \to a^+} \int_t^b f(x)\, dx

Asymptote at right endpoint:

Discontinuous Integrands
∫abf(x) dx=lim⁡t→b−∫atf(x) dx\int_a^b f(x)\, dx = \lim_{t \to b^-} \int_a^t f(x)\, dx

Asymptote at interior point:

Discontinuous Integrands
∫abf(x) dx=lim⁡t→c−∫atf(x) dx+lim⁡s→c+∫sbf(x) dx\int_a^b f(x)\, dx = \lim_{t \to c^-} \int_a^t f(x)\, dx + \lim_{s \to c^+} \int_s^b f(x)\, dx

Improper Integral (Discontinuous Integrand)
∫abf(x) dx=lim⁡t→b−∫atf(x) dxwhen f has an asymptote at b\int_a^b f(x)\, dx = \lim_{t \to b^-} \int_a^t f(x)\, dx \quad \text{when } f \text{ has an asymptote at } b
Learn more about this formula: Improper Integral (Discontinuous Integrand) →


Pitfall: Missing an asymptote inside the interval
Evaluating ∫−111x2 dx\int_{-1}^{1} \frac{1}{x^2}\, dx straight through gives [−1x]−11=−2\left[-\frac{1}{x}\right]_{-1}^{1} = -2, a negative value for a positive integrand. Split at 00: ∫s11x2 dx=1s−1→∞\int_s^1 \frac{1}{x^2}\, dx = \frac{1}{s} - 1 \to \infty as s→0+s \to 0^+, so the integral diverges.

Both limits must exist independently.

The interior case is the one the notation does not announce: both endpoints are ordinary numbers, and the asymptote shows only when the integrand is checked at every point between them. Once it is found at cc, cutting the interval there turns each piece into an endpoint case, with its own variable and its own one-sided limit.
Improper feature Setup as a limit Convergence requirement
Upper limit = ∞ ∫a∞ f(x) dx = limb → ∞ ∫ab f(x) dx single limit must exist and be finite
Lower limit = −∞ ∫−∞b f(x) dx = lima → −∞ ∫ab f(x) dx single limit must exist and be finite
Both limits infinite ∫−∞∞ f(x) dx = ∫−∞c f + ∫c∞ f  (any finite c) both pieces must converge independently
Asymptote at left endpoint a ∫ab f(x) dx = limt → a⁺ ∫tb f(x) dx single one-sided limit must exist and be finite
Asymptote at right endpoint b ∫ab f(x) dx = limt → b⁻ ∫at f(x) dx single one-sided limit must exist and be finite
Asymptote at interior point c split at c: limt → c⁻ ∫at f  +  lims → c⁺ ∫sb f both one-sided limits must converge independently

Convergence vs Divergence


An improper integral converges if the defining limit exists and is finite. It diverges if the limit is infinite or fails to exist.

Convergent example:

Convergence vs Divergence
∫1∞1x2 dx=lim⁡b→∞[−1x]1b=lim⁡b→∞(−1b+1)=1\int_1^{\infty} \frac{1}{x^2}\, dx = \lim_{b \to \infty} \left[-\frac{1}{x}\right]_1^b = \lim_{b \to \infty} \left(-\frac{1}{b} + 1\right) = 1

Divergent example:

Convergence vs Divergence
∫1∞1x dx=lim⁡b→∞[ln⁡x]1b=lim⁡b→∞ln⁡b=∞\int_1^{\infty} \frac{1}{x}\, dx = \lim_{b \to \infty} [\ln x]_1^b = \lim_{b \to \infty} \ln b = \infty

Pitfall: Taking a vanishing integrand as enough
Concluding convergence from f(x)→0f(x) \to 0 fails for 1x\frac{1}{\sqrt{x}}: it tends to 00, but ∫1b1x dx=2b−2→∞\int_1^b \frac{1}{\sqrt{x}}\, dx = 2\sqrt{b} - 2 \to \infty. So ∫1∞1x dx\int_1^{\infty} \frac{1}{\sqrt{x}}\, dx diverges.

Convergence depends on how fast the integrand decays. The 1/x21/x^2 decays fast enough; 1/x1/x does not.

How fast is measured against powers of xx. At x=100x = 100, 1x2=0.0001\frac{1}{x^2} = 0.0001 while 1x=0.01\frac{1}{x} = 0.01, a hundred times larger, and the ratio between them is xx itself, so the gap keeps widening as xx grows. The p-test in the next section turns this comparison into a rule.
Area under 1/x^2 from 1 to b is 1 - 1/b and levels off at 1; area under 1/x is ln b and keeps growingby2468101214161820123limit 1under 1/x²: 1 − 1/bunder 1/x: ln barea from 1 to b, as b grows1/x² totals 1; 1/x totals ln b, which never stops growing
Area so far, as the upper limit grows

For each upper limit b, the curves show the area from 1 to b. Under 1/x² the area is 1 − 1/b: it climbs quickly and then flattens against the dashed level 1, so ∫₁^∞ dx/x² converges to 1. Under 1/x the area is ln b: it slows down but never levels off, so ∫₁^∞ dx/x diverges. The integrands differ only in how fast they decay.

1/x is the boundary case p = 1 of the p-test: any power above 1 levels off.

The p-Test


The integrals of 1/xp1/x^p serve as benchmarks.

At infinity:

p-Test for Improper Integrals
∫1∞1xp dx{convergesp>1divergesp≤1\int_1^{\infty} \frac{1}{x^p}\, dx \quad \begin{cases} \text{converges} & p > 1 \\ \text{diverges} & p \leq 1 \end{cases}
Learn more about this formula: p-Test for Improper Integrals →


At zero:

The p-Test
∫011xp dx{convergesp<1divergesp≥1\int_0^1 \frac{1}{x^p}\, dx \quad \begin{cases} \text{converges} & p < 1 \\ \text{diverges} & p \geq 1 \end{cases}

Pitfall: Using the rule for infinity near zero
Calling ∫011x3/2 dx\int_0^1 \frac{1}{x^{3/2}}\, dx convergent because p=32>1p = \frac{3}{2} > 1 uses the wrong half of the test. ∫t1x−3/2 dx=2t−2→∞\int_t^1 x^{-3/2}\, dx = \frac{2}{\sqrt{t}} - 2 \to \infty as t→0+t \to 0^+, so it diverges.

The boundary case p=1p = 1 always diverges—∫1/x dx=ln⁡∣x∣\int 1/x\, dx = \ln|x|, which is unbounded both as x→∞x \to \infty and as x→0+x \to 0^+.

The two halves of the test point in opposite directions because the trouble sits at opposite ends. On [1,∞)[1, \infty) a larger pp makes 1xp\frac{1}{x^p} smaller, so the tail is thinner. On (0,1](0, 1] a larger pp makes 1xp\frac{1}{x^p} larger, so the spike near 00 is taller.
Form p > 1 p = 1 p < 1
∫1∞ (1 / xp) dx converges diverges diverges
∫01 (1 / xp) dx diverges diverges converges

Comparison Test


Compare an unknown integral to one with known behavior.

Direct comparison: For f(x)≥0f(x) \geq 0 and g(x)≥0g(x) \geq 0:

If f(x)≤g(x)f(x) \leq g(x) and ∫g\int g converges, then ∫f\int f converges.

If f(x)≥g(x)f(x) \geq g(x) and ∫g\int g diverges, then ∫f\int f diverges.

Example
Does ∫1∞1x2+1 dx\int_1^{\infty} \frac{1}{x^2 + 1}\, dx converge?
Since 1x2+1<1x2\dfrac{1}{x^2 + 1} < \dfrac{1}{x^2} and ∫1∞1x2 dx\int_1^{\infty} \dfrac{1}{x^2}\, dx converges, the given integral converges by comparison.
1/(x^2 + 1) lies below 1/x^2 on [1, infinity); the larger area is 1, so the smaller area is finitexy12345611/x²: total area 11/(x² + 1)smaller integrand, area at most 1Trapped under a convergent integral, the smaller one converges too
Comparison: trapped below a finite area

The section’s 1/(x² + 1) on [1, ∞) lies below 1/x² at every x. The shaded region under the smaller curve is therefore contained in the region under the larger one, whose total area is 1. A region inside a finite area is finite, so ∫₁^∞ dx/(x² + 1) converges; its value is π/4.

The test only works one way: lying below a divergent integral proves nothing.

Limit Comparison Test


When direct comparison is awkward, use limits.

For f(x)>0f(x) > 0 and g(x)>0g(x) > 0, if:

Limit Comparison Test
lim⁡x→∞f(x)g(x)=Lwhere 0<L<∞\lim_{x \to \infty} \frac{f(x)}{g(x)} = L \quad \text{where } 0 < L < \infty

then ∫f\int f and ∫g\int g both converge or both diverge.

Example
Does ∫1∞xx3+5 dx\int_1^{\infty} \frac{x}{x^3 + 5}\, dx converge?
Compare to g(x)=1/x2g(x) = 1/x^2:
lim⁡x→∞x/(x3+5)1/x2=lim⁡x→∞x3x3+5=1\lim_{x \to \infty} \frac{x/(x^3 + 5)}{1/x^2} = \lim_{x \to \infty} \frac{x^3}{x^3 + 5} = 1
Since ∫1∞1/x2 dx\int_1^{\infty} 1/x^2\, dx converges, so does the given integral.

Summary: Tests for Improper Integral Convergence


Determining whether an improper integral converges or diverges typically uses one of a small set of standard tests. The table below collects them in one place, pairing each test with the kind of integrand it suits, the conclusion it produces, and a worked benchmark from the sections above. Read the "When to use" column first when scanning an unfamiliar integral; the right row points directly to the test that settles convergence.
Test When to use Conclusion Example or benchmark
Direct evaluation an antiderivative is available exact value if the limit is finite; otherwise diverges ∫1∞ 1/x² dx = 1
p-test at ∞ integrand behaves like 1/xp as x → ∞ converges iff p > 1 ∫1∞ 1/xp dx (benchmark)
p-test near 0 integrand behaves like 1/xp as x → 0⁺ converges iff p < 1 ∫01 1/√x dx (p = 1/2)
Direct comparison f ≥ 0 and bounded above by convergent g, or below by divergent g f inherits the convergence behavior of the comparator 1/(x² + 1) < 1/x²  ⇒  ∫ 1/(x² + 1) converges
Limit comparison f, g > 0 and limx → ∞ f/g = L with 0 < L < ∞ ∫ f and ∫ g both converge or both diverge x/(x³ + 5) ~ 1/x²  ⇒  ∫ x/(x³ + 5) converges
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Before you read
Three tests for improper integrals
Most improper integrals cannot be evaluated directly. These three tests decide convergence by comparing with integrals whose behaviour is known.
Convergence testsfor improper integrals
p-test∫1∞1xp dx\int_1^\infty \frac{1}{x^p}\,dx
Verdictconverges exactly when p>1p > 1
Comparison
Howbound by a known integral

works in one direction only

Limit comparison
Howratio limit finite and positive
Verdictboth integrals behave the same
Takeaway
Start from the p-test
The p-test supplies the benchmarks; comparison and limit comparison carry a benchmark's verdict over to the integral you have.
Convergence tests for improper integrals·/calculus/integrals/improperLearn Math Class

Common Pitfalls


Five mistakes cause most wrong verdicts on improper integrals. Four of them are marked where they happen in the sections above; the fifth is the one-way rule that the section on the comparison test warns about. Here each is worked through in full, with the wrong and the right working side by side.

Running both ends out together

An integral over the whole real line is evaluated with one variable at both ends, ∫−RR\int_{-R}^{R}, and RR is sent to ∞\infty. For an odd integrand the two halves cancel at every RR, and the result looks like a convergent value.

Wrong working: ∫−RR2x1+x2 dx=[ln⁡(1+x2)]−RR=ln⁡(1+R2)−ln⁡(1+R2)=0\int_{-R}^{R} \dfrac{2x}{1 + x^2}\, dx = \left[\ln(1 + x^2)\right]_{-R}^{R} = \ln(1 + R^2) - \ln(1 + R^2) = 0, so ∫−∞∞2x1+x2 dx=0\int_{-\infty}^{\infty} \dfrac{2x}{1 + x^2}\, dx = 0.

Right working: split at c=0c = 0. ∫0∞2x1+x2 dx=lim⁡b→∞ln⁡(1+b2)=∞\int_0^{\infty} \dfrac{2x}{1 + x^2}\, dx = \lim_{b \to \infty} \ln(1 + b^2) = \infty. One piece diverges, so the whole integral diverges, whatever the other piece does.

Why it happens: the symmetric limit is shorter and gives a tidy answer, and the rule that an odd function integrates to 00 over [−a,a][-a, a] makes 00 look right. That rule is for finite intervals; on the whole line, the definition asks for two separate limits.

How to check: let the two ends move at different speeds. ∫−R2R2x1+x2 dx=ln⁡1+4R21+R2→ln⁡4≈1.39\int_{-R}^{2R} \dfrac{2x}{1 + x^2}\, dx = \ln\dfrac{1 + 4R^2}{1 + R^2} \to \ln 4 \approx 1.39, not 00. A value that depends on how the ends move is not a limit of the integral.

Missing an asymptote inside the interval

Both endpoints are ordinary numbers, but the integrand blows up between them. The antiderivative is evaluated at the endpoints as if the integral were proper.

Wrong working: ∫−111x2 dx=[−1x]−11=−1−1=−2\int_{-1}^{1} \dfrac{1}{x^2}\, dx = \left[-\dfrac{1}{x}\right]_{-1}^{1} = -1 - 1 = -2.

Right working: split at 00. ∫−111x2 dx=lim⁡t→0−∫−1t1x2 dx+lim⁡s→0+∫s11x2 dx\int_{-1}^{1} \dfrac{1}{x^2}\, dx = \lim_{t \to 0^-} \int_{-1}^{t} \dfrac{1}{x^2}\, dx + \lim_{s \to 0^+} \int_{s}^{1} \dfrac{1}{x^2}\, dx. The second piece is lim⁡s→0+(1s−1)=∞\lim_{s \to 0^+} \left(\dfrac{1}{s} - 1\right) = \infty, so the integral diverges.

Why it happens: the Fundamental Theorem of Calculus needs a continuous integrand on [a,b][a, b], and that condition is easy to skip because the formula still produces a number.

How to check: look at the sign. 1x2\dfrac{1}{x^2} is positive at every x≠0x \neq 0, so a negative result is impossible. Then check the integrand at every point of the interval: its denominator is 00 at x=0x = 0, inside [−1,1][-1, 1].

Taking a vanishing integrand as enough

The integrand tends to 00 as x→∞x \to \infty, and the area under it is taken to be finite for that reason alone.

Wrong working: 1x→0\dfrac{1}{\sqrt{x}} \to 0 as x→∞x \to \infty, so ∫1∞1x dx\int_1^{\infty} \dfrac{1}{\sqrt{x}}\, dx converges.

Right working: ∫1bx−1/2 dx=[2x]1b=2b−2→∞\int_1^b x^{-1/2}\, dx = \left[2\sqrt{x}\right]_1^b = 2\sqrt{b} - 2 \to \infty, so the integral diverges. The p-test gives the same verdict, since p=12≤1p = \frac{1}{2} \leq 1.

Why it happens: every convergent example on [1,∞)[1, \infty) has an integrand that shrinks to 00, so shrinking looks like the reason for convergence. It is only part of it; the integrand must shrink fast enough.

How to check: compute the area up to a large bb. At b=10000b = 10000 it is already 2⋅100−2=1982 \cdot 100 - 2 = 198 and still growing.

Using the rule for infinity near zero

The p-test condition for [1,∞)[1, \infty), p>1p > 1, is applied to an integral on [0,1][0, 1], where the condition is reversed.

Wrong working: ∫011x3/2 dx\int_0^1 \dfrac{1}{x^{3/2}}\, dx converges because p=32>1p = \frac{3}{2} > 1.

Right working: ∫t1x−3/2 dx=[−2x−1/2]t1=−2+2t→∞\int_t^1 x^{-3/2}\, dx = \left[-2x^{-1/2}\right]_t^1 = -2 + \dfrac{2}{\sqrt{t}} \to \infty as t→0+t \to 0^+, so the integral diverges. Near zero, ∫011xp dx\int_0^1 \dfrac{1}{x^p}\, dx converges only for p<1p < 1.

Why it happens: "converges for p>1p > 1" is the version met first and remembered, often without the interval it belongs to.

How to check: ask where the trouble is. At infinity the integrand must be small, and a larger pp makes it smaller. Near 00 the integrand is large, and a larger pp makes it larger: at x=0.01x = 0.01, x−3/2=1000x^{-3/2} = 1000 while x−1/2=10x^{-1/2} = 10.

Comparing in the wrong direction

The integrand lies below a divergent integrand, and the integral is declared divergent. Direct comparison gives no verdict in that case.

Wrong working: on [1,∞)[1, \infty), 1x2+1<1x\dfrac{1}{x^2 + 1} < \dfrac{1}{x}, and ∫1∞1x dx\int_1^{\infty} \dfrac{1}{x}\, dx diverges, so ∫1∞1x2+1 dx\int_1^{\infty} \dfrac{1}{x^2 + 1}\, dx diverges.

Right working: compare with 1x2\dfrac{1}{x^2} instead. 1x2+1<1x2\dfrac{1}{x^2 + 1} < \dfrac{1}{x^2} and ∫1∞1x2 dx\int_1^{\infty} \dfrac{1}{x^2}\, dx converges, so the integral converges. Its value is lim⁡b→∞(arctan⁡b−arctan⁡1)=π2−π4=π4\lim_{b \to \infty} (\arctan b - \arctan 1) = \dfrac{\pi}{2} - \dfrac{\pi}{4} = \dfrac{\pi}{4}.

Why it happens: the test has two halves, and being smaller than something infinite sounds as if it should say something.

How to check: before concluding, read the inequality. To prove convergence, ff must lie below a convergent integrand; to prove divergence, above a divergent one. Below a divergent integrand, or above a convergent one, decides nothing.

Improper Integrals FAQ

How do you know an integral is improper if it looks normal?

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Inspect the integrand, because nothing in the notation warns you. ∫₀¹ (1/√x) dx looks like an ordinary definite integral, but the integrand blows up at x = 0, so it is improper and needs a limit. Applying the Fundamental Theorem straight to it happens to work here; on a divergent example it would produce a confident, wrong number.Read more →

Why must an integral from −∞ to ∞ be split into two separate limits?

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Because tying both ends to one variable changes the definition. Split at any finite c and require each half to converge on its own. The symmetric version, letting −R and R run out together, is Cauchy's principal value — a different object. The integral of x over the whole line diverges, yet its principal value is 0.Read more →

Does an integral that diverges always equal infinity?

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No. Divergence comes in two species. One is growth without bound, written = ∞ by the same equals-sign abuse used for infinite limits. The other is oscillation: ∫₀^∞ sin x dx never settles and never grows, so no value exists at all. Every integral equal to infinity diverges, but not every divergent integral is infinite.Read more →

Why do the two sides of an interior asymptote need separate limits?

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Because one variable for both sides secretly cancels the two infinities against each other. With an asymptote at c inside [a, b], write the left limit as t → c⁻ and the right as s → c⁺, using independent letters. Both must converge on their own; a shared letter can declare a divergent integral finite.Read more →

Why does the p-test flip between infinity and zero?

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Because the danger sits at opposite ends. Out at infinity, ∫₁^∞ (1/xᵖ) dx converges when p > 1 — the integrand has to decay fast enough. Near zero, ∫₀¹ (1/xᵖ) dx converges when p < 1 — the blow-up has to be mild enough. Larger p helps at infinity and hurts at zero. At p = 1 both diverge.Read more →