Improper integrals and the comparison test

Intuition [Beginner]

Some areas stretch to infinity. Imagine a speed curve that never quite drops to zero but keeps getting smaller. The car keeps moving, and the total distance keeps growing. Does the distance approach a finite number, or does it grow without bound?

When the region under a curve extends infinitely far to the right, you chop off the tail at some large number and compute the area up to that point. Then you push the cutoff farther and farther to the right. If the partial areas settle toward a specific number, the improper integral converges. If they grow without bound, it diverges.

This concept exists because many quantities in physics, probability, and engineering involve infinite ranges. The total energy radiated by a decaying source, the probability accumulated over all outcomes, and the area under $1/x^2$ from $1$ to infinity all require this limiting process.

Visual [Beginner]

The picture shows a curve $y=1/x^2$ above the horizontal axis from $x=1$ extending to the right. A shaded region fills the area under the curve. A vertical dashed line marks a cutoff point at $x=N$, and the shaded area to the left of the cutoff grows but approaches a limit. A second, faster-decaying curve $y=1/x$ is drawn for comparison — its shaded area keeps growing without bound.

The comparison between the two curves shows that a faster-decaying function can have a finite total area while a slower-decaying one does not.

Worked example [Beginner]

Consider the area under $f(x)=1/x^2$ from $1$ to infinity.

Step 1. Chop off at $N=5$. The partial area from $1$ to $5$ is $1-1/5=4/5=0.8$.

Step 2. Chop off at $N=10$. The partial area from $1$ to $10$ is $1-1/10=0.9$.

Step 3. Chop off at $N=100$. The partial area from $1$ to $100$ is $1-1/100=0.99$. As $N$ grows, the partial area approaches $1$. The improper integral converges to $1$.

What this tells us: even though the region stretches infinitely far, the curve decays fast enough that the total area is finite.

Check your understanding [Beginner]

Formal definition [Intermediate+]

Definition (Improper integral, Type I). Let $f:[a,\infty )\to \mathbb{R}$ be integrable on $[a,N]$ for every $N>a$. The improper integral of $f$ over $[a,\infty )$ is

$${\int }_a^{\infty }f(x)dx=\underset{N\to \infty }{\lim }{\int }_a^{N}f(x)dx$$

provided the limit exists and is finite. If the limit exists, the integral converges; otherwise it diverges.

Definition (Improper integral, Type II). Let $f:(a,b]\to \mathbb{R}$ have a singularity at $a$ (that is, $f$ is unbounded near $a$) and be integrable on $[c,b]$ for every $c\in (a,b)$. The improper integral is

$${\int }_a^{b}f(x)dx=\underset{c\to a^{+}}{\lim }{\int }_c^{b}f(x)dx.$$

An analogous definition applies for singularities at the right endpoint.

Counterexamples to common slips [Intermediate+]

• Convergence of ${\int }_a^{\infty }|f|$ does not imply convergence of ${\int }_a^{\infty }f$. In fact the reverse holds: absolute convergence implies convergence, but the converse fails. The integral of $\sin (x)/x$ from $1$ to $\infty$ converges conditionally, but the integral of $|\sin (x)/x|$ diverges.

• Treating $\infty -\infty$ as zero. If ${\int }_a^{\infty }{f}_{+}$ and ${\int }_a^{\infty }{f}_{-}$ both diverge, writing ${\int }_a^{\infty }f=\infty -\infty$ is meaningless. The limit ${\lim }_{N\to \infty }{\int }_a^{N}f$ must exist as a single finite number.

• Assuming the integrand must be non-negative. The comparison test requires non-negative integrands, but the definition of convergence applies to any integrable function on finite subintervals.

Key theorem with proof [Intermediate+]

Theorem (Comparison test for improper integrals). Let $f,g:[a,\infty )\to \mathbb{R}$ with $0\le f(x)\le g(x)$ for all $x\ge a$, and suppose both are integrable on $[a,N]$ for every $N>a$.

(i) If ${\int }_a^{\infty }g(x)dx$ converges, then ${\int }_a^{\infty }f(x)dx$ converges.

(ii) If ${\int }_a^{\infty }f(x)dx$ diverges, then ${\int }_a^{\infty }g(x)dx$ diverges.

Proof of (i). Define $F(N)={\int }_a^{N}f(x)dx$ and $G(N)={\int }_a^{N}g(x)dx$. Since $0\le f\le g$, we have $0\le F(N)\le G(N)$ for all $N\ge a$.

The sequence $(F(N))$ is monotone increasing (because $f\ge 0$), and bounded above by $G(N)\le {\lim }_{M\to \infty }G(M)={\int }_a^{\infty }g(x)dx$, which is finite by hypothesis. By the monotone convergence theorem for sequences of real numbers, $F(N)$ converges to ${\sup }_{N}F(N)$, which is finite. Hence ${\int }_a^{\infty }f(x)dx$ converges.

Proof of (ii). This is the contrapositive of (i). If ${\int }_a^{\infty }g$ converged, then by (i) the integral ${\int }_a^{\infty }f$ would also converge, contradicting the hypothesis. $\square$

Corollary (p-test). The improper integral ${\int }_1^{\infty }\frac{1}{x^p}dx$ converges if $p>1$ and diverges if $p\le 1$.

Proof. For $p>1$: ${\int }_1^N{x}^{-p}dx=\frac{N^{1-p}-1}{1-p}\to \frac{1}{p-1}$ as $N\to \infty$. For $p=1$: ${\int }_1^N{x}^{-1}dx=\ln N\to \infty$. For $p<1$: ${\int }_1^N{x}^{-p}dx=\frac{N^{1-p}-1}{1-p}\to \infty$ since $1-p>0$. $\square$

Bridge. The comparison test is the foundational reason that convergence of an improper integral can be decided by comparing with a known benchmark, and this is exactly the same machinery as the comparison test for infinite series 02.03.03. The central insight is that the monotone convergence of partial areas mirrors the monotone convergence of partial sums. The p-test builds toward the Gamma function 02.09.01 where the integral ${\int }_0^{\infty }{t}^{s-1}{e}^{-t}dt$ is tested for convergence by splitting at $t=1$ and using the p-test near $0$ and exponential decay near $\infty$, and appears again in 02.11.04 where the Lebesgue dominated convergence theorem provides the measure-theoretic generalisation of the comparison test.

Exercises [Intermediate+]

Advanced results [Master]

Theorem 1 (Absolute convergence implies convergence). If ${\int }_a^{\infty }|f(x)|dx$ converges, then ${\int }_a^{\infty }f(x)dx$ converges.

The proof writes $f=f^{+}-f^{-}$ where $f^{+}=\max (f,0)$ and $f^{-}=\max (-f,0)$. Since $0\le f^{\pm }\le |f|$, the comparison test gives convergence of both ${\int }_a^{\infty }{f}^{+}$ and ${\int }_a^{\infty }{f}^{-}$, hence ${\int }_a^{\infty }f={\int }_a^{\infty }{f}^{+}-{\int }_a^{\infty }{f}^{-}$ converges. An integral where $\int |f|$ diverges but $\int f$ converges is called conditionally convergent.

Theorem 2 (Limit comparison test). Let $f,g\ge 0$ on $[a,\infty )$ with ${\lim }_{x\to \infty }f(x)/g(x)=L\in (0,\infty )$. Then ${\int }_a^{\infty }f$ converges if and only if ${\int }_a^{\infty }g$ converges.

For $x$ sufficiently large, $(L/2)g(x)\le f(x)\le (2L)g(x)$, and the comparison test applied in both directions gives equivalence.

Theorem 3 (Dirichlet's test for improper integrals). Let $f:[a,\infty )\to \mathbb{R}$ have a bounded antiderivative ($|F(x)|=|{\int }_a^{x}f(t)dt|\le M$ for all $x$) and let $g:[a,\infty )\to \mathbb{R}$ be monotone decreasing with $g(x)\to 0$ as $x\to \infty$. Then ${\int }_a^{\infty }f(x)g(x)dx$ converges.

The proof uses the second mean value theorem for integrals (Bonnet's theorem): on $[a,b]$, ${\int }_a^{b}f(x)g(x)dx=g(a){\int }_a^{c}f(x)dx$ for some $c\in [a,b]$ when $g$ is non-negative and decreasing. The boundedness of $F$ and $g\to 0$ then give the Cauchy criterion.

Theorem 4 (The Gamma function). For $s>0$, the integral $\Gamma (s)={\int }_0^{\infty }{t}^{s-1}{e}^{-t}dt$ converges and defines a function satisfying $\Gamma (s+1)=s\Gamma (s)$ and $\Gamma (1)=1$, hence $\Gamma (n)=(n-1)!$ for positive integers $n$.

Near $t=0$, $t^{s-1}{e}^{-t}\le t^{s-1}$, and ${\int }_0^1{t}^{s-1}dt$ converges for $s>0$ (p-test at the singularity). Near $\infty$, $t^{s-1}{e}^{-t}$ is dominated by $e^{-t/2}$ for large $t$, and ${\int }_1^{\infty }{e}^{-t/2}dt$ converges.

Theorem 5 (Cauchy criterion for improper integrals). The improper integral ${\int }_a^{\infty }f(x)dx$ converges if and only if for every $ϵ>0$ there exists $A$ such that $\left|{\int }_a^{b}f(x)dx-{\int }_a^{c}f(x)dx\right|<ϵ$ for all $b,c>A$.

This is the direct translation of the Cauchy criterion for the limit ${\lim }_{N\to \infty }{\int }_a^{N}f$.

Theorem 6 (Integration by parts for improper integrals). If $u$ and $v$ are differentiable on $[a,\infty )$ with $u(x)v(x)\to 0$ as $x\to \infty$ and ${\int }_a^{\infty }{u}^{\prime }v$ converges, then ${\int }_a^{\infty }u{v}^{\prime }$ converges and equals $-u(a)v(a)-{\int }_a^{\infty }{u}^{\prime }v$.

Synthesis. The comparison test is the foundational reason that convergence of improper integrals reduces to comparing with known benchmarks, and this is exactly the same principle as the comparison test for infinite series 02.03.03. The central insight is that monotonicity and boundedness force convergence, and the bridge is from partial areas to partial sums. Putting these together with the p-test, the Gamma function convergence proof splits the domain at a transition point and uses different arguments on each side — singularity control near $0$ and exponential decay near $\infty$. This pattern generalises to every improper integral with both a singularity and an infinite endpoint, and appears again in 02.11.04 where the Lebesgue dominated convergence theorem replaces the comparison test with an $L^1$ bound. Dirichlet's test provides the conditional-convergence machinery that the comparison test cannot reach, and this is exactly the tool needed to handle oscillatory integrands like $\sin (x)/x$.

Full proof set [Master]

Proposition (Dirichlet's test for improper integrals). Let $F(x)={\int }_a^{x}f(t)dt$ be bounded on $[a,\infty )$ with $|F(x)|\le M$, and let $g$ be monotone decreasing with $g(x)\to 0$. Then ${\int }_a^{\infty }f(x)g(x)dx$ converges.

Proof. Given $ϵ>0$, choose $A$ so that $g(A)<ϵ/(4M)$. For $b>a>A$, by the second mean value theorem for integrals (Bonnet's form), there exists $\xi \in [a,b]$ with

$${\int }_a^{b}f(x)g(x)dx=g(a){\int }_a^{\xi }f(x)dx.$$

Therefore:

$$\left|{\int }_a^{b}f(x)g(x)dx\right|=g(a)\left|{\int }_a^{\xi }f(x)dx\right|=g(a)|F(\xi )-F(a)|\le g(a)\cdot 2M.$$

Since $a>A$ and $g$ is decreasing, $g(a)\le g(A)<ϵ/(4M)$, so:

$$\left|{\int }_a^{b}f(x)g(x)dx\right|<\frac{ϵ}{4M}\cdot 2M=\frac{ϵ}{2}<ϵ.$$

By the Cauchy criterion, ${\int }_a^{\infty }f(x)g(x)dx$ converges. $\square$

Proposition ($\Gamma (s)$ is well-defined for $s>0$). The integral ${\int }_0^{\infty }{t}^{s-1}{e}^{-t}dt$ converges for all $s>0$.

Proof. Split at $t=1$. On $[0,1]$: $e^{-t}\le 1$, so $t^{s-1}{e}^{-t}\le t^{s-1}$. The integral ${\int }_0^1{t}^{s-1}dt=1/s$ converges since $s>0$. By the comparison test, ${\int }_0^1{t}^{s-1}{e}^{-t}dt$ converges.

On $[1,\infty )$: for $t\ge 1$, $e^{-t/2}\ge e^{-t}$, so $t^{s-1}{e}^{-t}=t^{s-1}{e}^{-t/2}\cdot e^{-t/2}$. For any fixed $s$, the polynomial $t^{s-1}$ grows more slowly than $e^{t/4}$, so there exists $T$ with $t^{s-1}{e}^{-t/2}\le 1$ for $t\ge T$. Hence for $t\ge T$, $t^{s-1}{e}^{-t}\le e^{-t/2}$.

Since ${\int }_T^{\infty }{e}^{-t/2}dt=2e^{-T/2}<\infty$, the comparison test gives convergence of ${\int }_1^{\infty }{t}^{s-1}{e}^{-t}dt$.

Both halves converge, so $\Gamma (s)$ is well-defined. $\square$

Connections [Master]

• Infinite series: convergence and the standard tests 02.03.03. The comparison test, p-test, absolute convergence, and the Cauchy criterion for improper integrals are direct parallels of the same results for infinite series. The series $\sum 1/n^p$ converges for $p>1$ for the same structural reason that ${\int }_1^{\infty }1/x^{p}dx$ converges for $p>1$.

• Fundamental theorems of calculus 02.04.04. Every improper integral is defined as a limit of proper integrals, and each proper integral is evaluated using FTC2. The fundamental theorem provides the antiderivative that makes the limit computable.

• Banach spaces and $L^1$ 02.11.04. The Lebesgue dominated convergence theorem generalises the comparison test to pointwise limits of measurable functions. The space $L^1([a,\infty ))$ is a Banach space whose norm $\Vert f{\Vert }_1={\int }_a^{\infty }|f|$ is defined precisely when the improper integral of $|f|$ converges.

Historical & philosophical context [Master]

Cauchy 1827, in his Exercices de mathematiques ^[Cauchy1827], gave the first systematic treatment of improper integrals, defining ${\int }_a^{\infty }f$ as the limit of ${\int }_a^{N}f$ and establishing basic convergence criteria. Dirichlet 1863, in his Vorlesungen uber Zahlentheorie ^{[Dirichlet1863]}, developed the conditional-convergence test for integrals of oscillatory functions, motivated by number-theoretic applications to Fourier integrals.

Artin 1964, in The Gamma Function ^[Artin1964], presented the Gamma function as the canonical application of improper-integral theory: $\Gamma (s)={\int }_0^{\infty }{t}^{s-1}{e}^{-t}dt$ is the unique log-convex extension of the factorial, and its convergence proof is a paradigmatic application of the comparison test split between a singularity at zero and an infinite endpoint. The Gamma function connects improper integrals to the functional equation $\Gamma (s+1)=s\Gamma (s)$ and, through the reflection formula $\Gamma (s)\Gamma (1-s)=\pi /\sin (\pi s)$, to complex analysis.

Bibliography [Master]

@book{Cauchy1827,
  author    = {Cauchy, Augustin-Louis},
  title     = {Exercices de math\'ematiques},
  publisher = {De Bure fr\`eres},
  year      = {1827}
}

@book{Dirichlet1863,
  author    = {Dirichlet, Peter Gustav Lejeune},
  title     = {Vorlesungen \"{u}ber Zahlentheorie},
  publisher = {Vieweg},
  year      = {1863},
  editor    = {Dedekind, Richard}
}

@book{Artin1964,
  author    = {Artin, Emil},
  title     = {The Gamma Function},
  publisher = {Holt, Rinehart and Winston},
  year      = {1964},
  note      = {Translated by Michael Butler}
}

@book{Apostol1967,
  author    = {Apostol, Tom M.},
  title     = {Calculus, Vol. 1},
  publisher = {Wiley},
  year      = {1967}
}

@book{Rudin1976,
  author    = {Rudin, Walter},
  title     = {Principles of Mathematical Analysis},
  publisher = {McGraw-Hill},
  edition   = {3},
  year      = {1976}
}

@book{Hardy1908,
  author    = {Hardy, G. H.},
  title     = {A Course of Pure Mathematics},
  publisher = {Cambridge University Press},
  year      = {1908}
}