The Central Idea
Improper integrals are definite integrals interpreted through limits.
The two most common causes of impropriety are:
- the interval of integration is infinite;
- the integrand becomes unbounded somewhere on the interval.
In either case, replace the problematic endpoint or singular point with a variable and take a limit.
\[ \boxed{\text{improper integral}=\text{limit of ordinary definite integrals}}. \]Why They Are Called Improper
The usual definite integral
\[ \int_a^b f(x)\,dx \]assumes finite endpoints and sufficiently well-behaved values on the interval. Infinite endpoints or unbounded behavior require a limiting interpretation.
Type I: Infinite Intervals
If the upper endpoint extends to infinity, define
\[ \boxed{\int_a^\infty f(x)\,dx=\lim_{b\to\infty}\int_a^b f(x)\,dx}. \]The integral converges only if the limit exists and is finite.
Infinite Lower Endpoint
\[ \boxed{\int_{-\infty}^{b}f(x)\,dx=\lim_{a\to-\infty}\int_a^b f(x)\,dx}. \]Both Endpoints Infinite
Choose any finite splitting point \(c\):
\[ \boxed{\int_{-\infty}^{\infty}f(x)\,dx=\int_{-\infty}^{c}f(x)\,dx+\int_c^\infty f(x)\,dx}. \]Both improper integrals must converge separately.
Do Not Use One Symmetric Limit as the Definition
The symmetric expression
\[ \lim_{R\to\infty}\int_{-R}^{R}f(x)\,dx \]can exist even when the ordinary improper integral diverges. That symmetric limit is related to a Cauchy principal value, not the ordinary definition.
Type II: Unbounded Integrands
An integral is improper when the function becomes unbounded at an endpoint or at a point inside the interval.
For example,
\[ \int_0^1\frac{1}{\sqrt{x}}\,dx \]is improper because \(1/\sqrt{x}\to\infty\) as \(x\to0^+\).
Unbounded at the Left Endpoint
\[ \boxed{\int_a^b f(x)\,dx=\lim_{t\to a^+}\int_t^b f(x)\,dx}. \]Unbounded at the Right Endpoint
\[ \boxed{\int_a^b f(x)\,dx=\lim_{t\to b^-}\int_a^t f(x)\,dx}. \]Unbounded at an Interior Point
If \(a Each piece is interpreted by its own one-sided limit.
Both Sides Must Converge
At an interior singularity, each side must converge independently. If either side diverges, the entire improper integral diverges.
Convergence and Divergence
An improper integral converges if its defining limit exists and is finite.
It diverges if the limit is infinite, fails to exist, or any required piece diverges.
Example: A Convergent Infinite-Interval Integral
\[ \int_1^\infty\frac{1}{x^2}\,dx =\lim_{b\to\infty}\int_1^b x^{-2}\,dx. \] \[ =\lim_{b\to\infty}\left[-\frac1x\right]_1^b =\lim_{b\to\infty}\left(1-\frac1b\right). \]Therefore,
\[ \boxed{\int_1^\infty\frac1{x^2}\,dx=1}. \]Example: A Divergent Infinite-Interval Integral
\[ \int_1^\infty\frac1x\,dx =\lim_{b\to\infty}[\ln x]_1^b =\lim_{b\to\infty}\ln b. \]Since \(\ln b\to\infty\),
\[ \boxed{\int_1^\infty\frac1x\,dx\text{ diverges}}. \]The \(p\)-Integral on \([1,\infty)\)
\[ \int_1^\infty\frac1{x^p}\,dx \]converges exactly when
\[ \boxed{p>1} \]and diverges when \(p\le1\).
The \(p\)-Integral Near Zero
\[ \int_0^1\frac1{x^p}\,dx \]converges exactly when
\[ \boxed{p<1} \]and diverges when \(p\ge1\).
Do Not Mix Up the Two \(p\)-Integral Rules
\[ \boxed{\int_1^\infty x^{-p}\,dx\text{ converges if }p>1} \]but
\[ \boxed{\int_0^1 x^{-p}\,dx\text{ converges if }p<1}. \]Example: A Convergent Endpoint Singularity
\[ \int_0^1\frac1{\sqrt{x}}\,dx =\lim_{t\to0^+}\int_t^1x^{-1/2}\,dx. \] \[ =\lim_{t\to0^+}[2\sqrt{x}]_t^1 =\lim_{t\to0^+}(2-2\sqrt{t}). \]Hence,
\[ \boxed{2}. \]Example: A Divergent Endpoint Singularity
\[ \int_0^1\frac1x\,dx =\lim_{t\to0^+}[\ln x]_t^1 =\lim_{t\to0^+}(-\ln t). \]Since \(-\ln t\to\infty\), the integral diverges.
Example: Interior Singularity
Consider
\[ \int_{-1}^{1}\frac1{x^2}\,dx. \]The integrand is unbounded at \(x=0\), so split:
\[ \int_{-1}^{0}\frac1{x^2}\,dx+\int_0^1\frac1{x^2}\,dx. \]Each side diverges. Therefore,
\[ \boxed{\int_{-1}^{1}\frac1{x^2}\,dx\text{ diverges}}. \]The Cancellation Trap
Although \(1/x\) is odd,
\[ \int_{-1}^{1}\frac1x\,dx \]does not equal zero as an ordinary improper integral. The singularity at \(0\) forces two separate improper integrals, and both diverge.
\[ \boxed{\int_{-1}^{1}\frac1x\,dx\text{ diverges}}. \]Example: Exponential Decay
\[ \int_0^\infty e^{-x}\,dx =\lim_{b\to\infty}[-e^{-x}]_0^b =\lim_{b\to\infty}(1-e^{-b}). \]Therefore,
\[ \boxed{1}. \]Example: Exponential Decay with a Parameter
If \(k>0\), then
\[ \boxed{\int_0^\infty e^{-kx}\,dx=\frac1k}. \]Example: Rational Decay
\[ \int_0^\infty\frac1{1+x^2}\,dx =\lim_{b\to\infty}[\arctan x]_0^b. \]Since \(\arctan b\to\pi/2\),
\[ \boxed{\frac{\pi}{2}}. \]Example: Integral over the Entire Real Line
\[ \int_{-\infty}^{\infty}\frac1{1+x^2}\,dx =\int_{-\infty}^{0}\frac1{1+x^2}\,dx+\int_0^\infty\frac1{1+x^2}\,dx. \]Each side equals \(\pi/2\), so
\[ \boxed{\pi}. \]Comparison as a Convergence Tool
Sometimes exact evaluation is difficult, but convergence can be decided by comparing the integrand with a simpler known function.
Direct Comparison for Nonnegative Functions
Suppose \(0\le f(x)\le g(x)\) for sufficiently large \(x\).
- If \(\int_a^\infty g(x)\,dx\) converges, then \(\int_a^\infty f(x)\,dx\) converges.
- If \(\int_a^\infty f(x)\,dx\) diverges, then \(\int_a^\infty g(x)\,dx\) diverges.
Example: Comparison with a \(p\)-Integral
For \(x\ge1\),
\[ 0<\frac1{x^2+1}\le\frac1{x^2}. \]Since \(\int_1^\infty 1/x^2\,dx\) converges,
\[ \boxed{\int_1^\infty\frac1{x^2+1}\,dx\text{ converges}}. \]Limit Comparison
If positive functions satisfy
\[ \lim_{x\to\infty}\frac{f(x)}{g(x)}=L, \qquad 0Example: Leading-Term Comparison
\[ \frac{3x+2}{x^3+4}\sim\frac{3}{x^2} \]for large \(x\). Since the comparison behaves like a \(p\)-integral with \(p=2\),
\[ \boxed{\int_1^\infty\frac{3x+2}{x^3+4}\,dx\text{ converges}}. \]Decay Rate Matters
A function can approach zero and still have a divergent improper integral. The classic example is \(1/x\):
\[ \frac1x\to0 \]but
\[ \int_1^\infty\frac1x\,dx \]diverges.
Infinite Width Can Still Produce Finite Area
The graph of \(1/x^2\) extends forever, yet
\[ \int_1^\infty\frac1{x^2}\,dx=1. \]The height decreases quickly enough for the total accumulation to remain finite.
Infinite Height Can Still Produce Finite Area
The function \(1/\sqrt{x}\) becomes unbounded as \(x\to0^+\), yet
\[ \int_0^1\frac1{\sqrt{x}}\,dx=2. \]The singularity is mild enough for the total area to remain finite.
Absolute Convergence
An improper integral converges absolutely if
\[ \int_a^\infty|f(x)|\,dx \]converges. Absolute convergence guarantees convergence of \(\int_a^\infty f(x)\,dx\).
Conditional Convergence
Some sign-changing improper integrals converge even though the corresponding absolute-value integral diverges. Such convergence is called conditional convergence.
A Reliable Procedure for Infinite Intervals
- Identify which endpoint is infinite.
- Replace infinity by a finite variable.
- Evaluate the resulting proper definite integral.
- Take the appropriate limit.
- State whether the limit is finite.
- Conclude convergence or divergence.
A Reliable Procedure for Unbounded Integrands
- Locate every singular point in the interval.
- Split at each interior singularity.
- Replace a singular endpoint by a variable.
- Use the correct one-sided limit.
- Evaluate each piece independently.
- If every required piece converges, add them.
- If any required piece diverges, the entire integral diverges.
Common Error: Treating \(\infty\) as a Number
Do not write \(F(\infty)-F(a)\) as though infinity were an ordinary endpoint.
Write
\[ \boxed{\lim_{b\to\infty}[F(x)]_a^b}. \]Common Error: Substituting a Singular Endpoint Directly
For
\[ \int_0^1\frac1{\sqrt{x}}\,dx, \]the correct setup is
\[ \boxed{\lim_{t\to0^+}\int_t^1\frac1{\sqrt{x}}\,dx}. \]Common Error: Ignoring an Interior Singularity
Inspect the entire interval before integrating. If a denominator becomes zero or another singularity occurs inside the interval, split the integral there.
Common Error: Using One Limit for Two Infinite Endpoints
For \(\int_{-\infty}^{\infty}f(x)\,dx\), split into two improper integrals and test each side separately.
Common Error: Assuming \(f(x)\to0\) Guarantees Convergence
The function \(1/x\) approaches zero, but its improper integral from \(1\) to infinity diverges.
Example: Full Infinite-Interval Workflow
Determine whether
\[ \int_2^\infty\frac1{x^3}\,dx \]converges.
\[ \lim_{b\to\infty}\int_2^b x^{-3}\,dx =\lim_{b\to\infty}\left[-\frac1{2x^2}\right]_2^b. \] \[ =\lim_{b\to\infty}\left(-\frac1{2b^2}+\frac18\right) =\boxed{\frac18}. \]Example: Full Endpoint-Singularity Workflow
\[ \int_0^4\frac1{\sqrt{x}}\,dx =\lim_{t\to0^+}\int_t^4x^{-1/2}\,dx. \] \[ =\lim_{t\to0^+}[2\sqrt{x}]_t^4 =\lim_{t\to0^+}(4-2\sqrt{t}) =\boxed{4}. \]Practice 1: Infinite Interval
Evaluate or state divergence:
\[ \int_1^\infty\frac1{x^3}\,dx. \]Practice 2: Harmonic-Type Integral
\[ \int_2^\infty\frac1x\,dx. \]Practice 3: Exponential Tail
\[ \int_0^\infty e^{-2x}\,dx. \]Practice 4: Endpoint Singularity
\[ \int_0^1x^{-1/3}\,dx. \]Practice 5: Divergent Endpoint Singularity
\[ \int_0^1x^{-3/2}\,dx. \]Practice 6: Interior Singularity
\[ \int_{-2}^{2}\frac1{x^2}\,dx. \]Practice 7: \(p\)-Integral at Infinity
For which values of \(p\) does
\[ \int_1^\infty\frac1{x^p}\,dx \]converge?
Practice 8: \(p\)-Integral Near Zero
For which values of \(p\) does
\[ \int_0^1\frac1{x^p}\,dx \]converge?
Practice 9: Comparison
Determine whether
\[ \int_1^\infty\frac1{x^2+5}\,dx \]converges.
Practice 10: Leading-Term Comparison
\[ \int_1^\infty\frac{5x^2+1}{x^4+3}\,dx. \]Practice 11: Whole Real Line
\[ \int_{-\infty}^{\infty}e^{-|x|}\,dx. \]Practice 12: Identify the Error
A student writes
\[ \int_1^\infty\frac1{x^2}\,dx=\left[-\frac1x\right]_1^\infty=-\frac1\infty+1. \]Explain how this should be written correctly.
Practice 13: Hidden Singularity
Explain why
\[ \int_0^2\frac1{x-1}\,dx \]is improper and determine whether it converges.
Practice 14: Concept Check
Can an improper integral converge if its interval is infinite? Can it converge if its integrand becomes unbounded? Explain.
Practice Answers
Practice 1
\[ \int_1^\infty x^{-3}\,dx=\boxed{\frac12}. \]Practice 2
\[ \boxed{\text{diverges}}. \]Practice 3
\[ \boxed{\frac12}. \]Practice 4
Since \(p=1/3<1\), the near-zero \(p\)-integral converges:
\[ \boxed{\frac32}. \]Practice 5
Here \(p=3/2\ge1\), so
\[ \boxed{\text{diverges}}. \]Practice 6
The integrand has an interior singularity at \(x=0\). Both sides diverge.
\[ \boxed{\text{diverges}}. \]Practice 7
\[ \boxed{p>1}. \]Practice 8
\[ \boxed{p<1}. \]Practice 9
Since
\[ 0<\frac1{x^2+5}\le\frac1{x^2} \]and \(\int_1^\infty1/x^2\,dx\) converges,
\[ \boxed{\text{converges}}. \]Practice 10
For large \(x\),
\[ \frac{5x^2+1}{x^4+3}\sim\frac5{x^2}. \]Thus,
\[ \boxed{\text{converges}}. \]Practice 11
Since \(e^{-|x|}\) is even,
\[ \int_{-\infty}^{\infty}e^{-|x|}\,dx=2\int_0^\infty e^{-x}\,dx=\boxed{2}. \]Practice 12
Infinity is not a number. Write
\[ \int_1^\infty\frac1{x^2}\,dx=\lim_{b\to\infty}\left[-\frac1x\right]_1^b =\lim_{b\to\infty}\left(1-\frac1b\right)=\boxed{1}. \]Practice 13
The denominator vanishes at \(x=1\), so split there. Both sides have logarithmic divergence near the singularity.
\[ \boxed{\text{diverges}}. \]Practice 14
Yes to both. Infinite width can still yield finite accumulation if the integrand decays sufficiently fast, and an unbounded integrand can still yield finite accumulation if its singularity is sufficiently mild.
What to Remember
Improper integrals are defined through limits.
\[ \boxed{\int_a^\infty f(x)\,dx=\lim_{b\to\infty}\int_a^b f(x)\,dx} \]and
\[ \boxed{\int_a^b f(x)\,dx=\lim_{t\to a^+}\int_t^b f(x)\,dx} \]when the left endpoint is singular.
The key \(p\)-integral tests are
\[ \boxed{\int_1^\infty\frac1{x^p}\,dx\text{ converges iff }p>1} \]and
\[ \boxed{\int_0^1\frac1{x^p}\,dx\text{ converges iff }p<1}. \]Never substitute infinity or a singular point directly. Replace the problematic behavior with a limit and test convergence explicitly.
Where We Go Next
Improper integrals extend exact integration to infinite intervals and singular behavior. The later lesson Numerical Integration develops approximation techniques for definite integrals that may be difficult to evaluate symbolically.