Tutorials › Real Analysis › Improper Riemann Integrals

Home  /  Tutorials  /  Riemann Integration

Riemann Integration · Tutorial 497 of 1000

Improper Riemann Integrals

Learn to define and evaluate improper Riemann integrals, distinguish convergence from divergence, and apply a Cauchy criterion to tails and endpoint behavior.

Advanced 10 min read

What You'll Learn

  • Define improper integrals at infinite endpoints and at finite singularities using limits of proper integrals.
  • Determine why each side of an interior singularity must converge separately.
  • Apply the Cauchy criterion to tails and truncated endpoint integrals.
  • Evaluate power integrals and identify their convergence thresholds.
  • Verify conditional convergence for an oscillatory improper integral.
  • Distinguish an ordinary improper integral from a symmetric principal value.

From Proper Integrals to Improper Integrals

The integration by parts formula in the previous tutorial applies on a closed, bounded interval where the functions have the required regularity. Some useful integrals do not fit that setting directly: the interval may be unbounded, or the integrand may become unbounded at an endpoint. The standard approach is to integrate only over closed, bounded intervals where the integrand is Riemann integrable, and then take a limit.

The limit is part of the definition, not merely a convenient way to compute. An improper integral exists only when the relevant limit is finite. When an integrand is singular at an interior point, both sides must have finite limits independently; cancellation between divergent pieces does not establish convergence.

Definitions by Truncation

Suppose \(f\) is Riemann integrable on every closed, bounded subinterval of \([a,\infty)\). The integrals \(\int_a^R f(x)\,dx\) are then proper integrals for finite \(R>a\). Their limit defines the integral over the unbounded interval if that limit exists as a finite real number.

Definition (Improper Integral on an Infinite Interval): If \(f\) is Riemann integrable on every closed, bounded subinterval of \([a,\infty)\), define
$$ \int_a^\infty f(x)\,dx = \lim_{R\to\infty}\int_a^R f(x)\,dx $$
when the limit exists and is finite. In that case, the improper integral converges. If there is no finite limit, it diverges. Similarly,
$$ \int_{-\infty}^b f(x)\,dx = \lim_{L\to-\infty}\int_L^b f(x)\,dx, \qquad \int_{-\infty}^{\infty}f(x)\,dx = \int_{-\infty}^c f(x)\,dx+\int_c^\infty f(x)\,dx, $$
where the integral over the whole real line is defined only when both integrals on the right converge, for some finite \(c\).

For a finite interval, the same limiting idea handles an endpoint singularity. The function need only be Riemann integrable on each truncated interval. For instance, if the possible singularity is at \(a\), the integral is defined by moving the lower limit away from \(a\).

Definition (Improper Integral at a Finite Endpoint): Suppose \(f\) is Riemann integrable on every interval \([a+\varepsilon,b]\) with \(0<\varepsilon<b-a\). The integral at \(a\) is
$$ \int_a^b f(x)\,dx = \lim_{\varepsilon\downarrow0}\int_{a+\varepsilon}^b f(x)\,dx $$
if this is a finite limit. For a possible singularity at \(b\), use
$$ \int_a^b f(x)\,dx = \lim_{\varepsilon\downarrow0}\int_a^{b-\varepsilon} f(x)\,dx. $$
If \(f\) has an interior singularity at \(c\), define the integral only if both one-sided improper integrals converge, and then set
$$ \int_a^b f(x)\,dx = \int_a^c f(x)\,dx+\int_c^b f(x)\,dx. $$

A function can have more than one improper feature. If it has a singularity at an endpoint and is also integrated over an infinite interval, the relevant truncations must be handled so that each limit required by the definition exists. The displayed definitions give the basic one-sided cases from which these situations are formed.

The Cauchy Criterion for Improper Integrals

A limit at infinity can be difficult to calculate directly. The Cauchy criterion instead asks whether the integral over every sufficiently distant interval is small. At a finite singular endpoint, it asks whether the integral over every sufficiently short truncated segment next to that endpoint is small.

Theorem (Cauchy Criterion for Improper Riemann Integrals): Suppose \(f\) is Riemann integrable on every closed, bounded subinterval of \([a,\infty)\). The improper integral \(\int_a^\infty f(x)\,dx\) converges if and only if, for every \(\varepsilon>0\), there is \(A>a\) such that
$$ \left|\int_R^S f(x)\,dx\right|<\varepsilon \quad\text{whenever } S\geq R\geq A. $$
For an endpoint singularity at \(a\), convergence is equivalent to the following condition: for every \(\varepsilon>0\), there is \(\delta_0>0\) such that
$$ \left|\int_{a+\delta}^{a+\eta} f(x)\,dx\right|<\varepsilon \quad\text{whenever }0<\delta\leq\eta<\delta_0. $$

Proof. First consider the infinite interval. If the improper integral converges to \(I\), then for a given \(\varepsilon>0\), the definition of a limit gives \(A>a\) such that

$$ \left|\int_a^T f(x)\,dx-I\right|<\frac{\varepsilon}{2} \quad\text{for every }T\geq A. $$

For \(S\geq R\geq A\), additivity of the proper Riemann integral gives

$$ \int_R^S f(x)\,dx = \int_a^S f(x)\,dx-\int_a^R f(x)\,dx. $$

Subtract and add \(I\), then apply the triangle inequality:

$$ \left|\int_R^S f(x)\,dx\right| \leq \left|\int_a^S f(x)\,dx-I\right| + \left|\int_a^R f(x)\,dx-I\right| <\varepsilon. $$

Conversely, suppose the stated tail condition holds. Define \(F(R)=\int_a^R f(x)\,dx\) for \(R>a\). Given any \(\varepsilon>0\), choose \(A\) from the condition. For \(S\geq R\geq A\), additivity gives

$$ |F(S)-F(R)| = \left|\int_R^S f(x)\,dx\right| <\varepsilon. $$

Thus the values \(F(R)\) satisfy the Cauchy condition as \(R\to\infty\). Completeness of the real numbers implies that \(F(R)\) has a finite limit, which is precisely the improper integral.

For the endpoint statement, put \(G(\delta)=\int_{a+\delta}^b f(x)\,dx\) for \(0<\delta<b-a\). If \(G(\delta)\) has a finite limit as \(\delta\downarrow0\), then for sufficiently small \(\delta,\eta>0\), both \(G(\delta)\) and \(G(\eta)\) are close to that limit. When \(\delta\leq\eta\), additivity gives

$$ \int_{a+\delta}^{a+\eta} f(x)\,dx = G(\delta)-G(\eta). $$

The difference is therefore arbitrarily small. Conversely, the stated condition implies that \(G(\delta)\) satisfies the Cauchy condition as \(\delta\downarrow0\): when \(\delta\leq\eta\), the same identity bounds \(|G(\delta)-G(\eta)|\). Completeness of the real numbers gives a finite limit for \(G\). This proves both criteria. \(\square\)

The criterion is especially useful when no antiderivative is available. It also emphasizes that convergence is about the net contribution from distant or short intervals, not about whether \(f(x)\) itself tends to zero.

Worked Applications

Worked Example: A Power on an Infinite Interval

For \(p\in\mathbb{R}\), consider \(\int_1^\infty x^{-p}\,dx\). If \(p\ne1\), the power rule on the proper interval \([1,R]\) gives

$$ \int_1^R x^{-p}\,dx = \frac{R^{1-p}-1}{1-p}. $$

If \(p>1\), then \(R^{1-p}\to0\), so the limit is \(1/(p-1)\). If \(p<1\), then \(R^{1-p}\to\infty\), and the displayed expression tends to \(+\infty\). In the remaining case \(p=1\),

$$ \int_1^R \frac{1}{x}\,dx=\ln R, $$

which tends to \(+\infty\). Therefore,

$$ \int_1^\infty x^{-p}\,dx \text{ converges exactly when }p>1, \qquad \int_1^\infty x^{-p}\,dx=\frac{1}{p-1}\text{ when }p>1. $$

For example, at \(p=2\), the truncated integral is \(1-1/R\), which tends to \(1\). At \(p=1\), it is \(\ln R\), so the integral diverges. These cases illustrate why an unbounded domain requires an actual limit rather than direct substitution of an infinite endpoint into an antiderivative.

Worked Example: A Power Singularity at an Endpoint

Consider \(\int_0^1 x^{-p}\,dx\), with \(p\in\mathbb{R}\). For \(0<\varepsilon<1\) and \(p\ne1\),

$$ \int_\varepsilon^1 x^{-p}\,dx = \frac{1-\varepsilon^{1-p}}{1-p}. $$

If \(p<1\), then \(\varepsilon^{1-p}\to0\), and the limit is \(1/(1-p)\). If \(p>1\), then \(\varepsilon^{1-p}\to\infty\); since \(1-p<0\), the expression tends to \(+\infty\). When \(p=1\),

$$ \int_\varepsilon^1\frac{1}{x}\,dx=-\ln\varepsilon, $$

which also tends to \(+\infty\). Thus the endpoint integral converges exactly when \(p<1\). In particular,

$$ \int_0^1 \frac{1}{\sqrt{x}}\,dx = \lim_{\varepsilon\downarrow0}2(1-\sqrt{\varepsilon}) =2. $$

The integrand is unbounded near zero, but boundedness on the entire interval is not required by the definition. Each truncated integral is proper, and the limit is finite.

Worked Example: Conditional Convergence of an Oscillatory Integral

Consider \(\int_1^\infty \sin x/x\,dx\). The integrand has no elementary antiderivative, so use the Cauchy criterion. On a finite interval \([A,B]\), where \(B\geq A\geq1\), integration by parts with \(u(x)=1/x\) and \(v'(x)=\sin x\) gives \(u'(x)=-1/x^2\) and \(v(x)=-\cos x\). Hence

$$ \int_A^B\frac{\sin x}{x}\,dx = \left[-\frac{\cos x}{x}\right]_A^B -\int_A^B\frac{\cos x}{x^2}\,dx. $$

Since \(|\cos x|\leq1\),

$$ \left|\int_A^B\frac{\sin x}{x}\,dx\right| \leq \frac{1}{A}+\frac{1}{B}+\int_A^B\frac{1}{x^2}\,dx \leq \frac{3}{A}. $$

The final bound tends to zero as \(A\to\infty\), uniformly for \(B\geq A\). The Cauchy criterion proves that \(\int_1^\infty \sin x/x\,dx\) converges.

This convergence is not absolute. On each interval \([k\pi,(k+1)\pi]\), where \(k\geq1\) is an integer, \(x\leq(k+1)\pi\) and

$$ \int_{k\pi}^{(k+1)\pi}\frac{|\sin x|}{x}\,dx \geq \frac{1}{(k+1)\pi}\int_{k\pi}^{(k+1)\pi}|\sin x|\,dx = \frac{2}{(k+1)\pi}. $$

The sum of these lower bounds diverges because the harmonic series diverges. Additivity and order preservation for proper integrals therefore show that the truncated integrals of \(|\sin x|/x\) are unbounded. The integral of \(\sin x/x\) converges, but the integral of its absolute value does not.

Worked Example: A Symmetric Cancellation Is Not Convergence

The function \(f(x)=1/x\) is singular at the interior point \(0\) of \([-1,1]\). The ordinary improper integral requires both one-sided limits. For \(0<\varepsilon<1\),

$$ \int_{-1}^{-\varepsilon}\frac{1}{x}\,dx = \ln\varepsilon, \qquad \int_{\varepsilon}^{1}\frac{1}{x}\,dx = -\ln\varepsilon. $$

As \(\varepsilon\downarrow0\), the left integral tends to \(-\infty\) and the right integral tends to \(+\infty\). Neither one-sided improper integral converges, so \(\int_{-1}^1 1/x\,dx\) does not exist as an ordinary improper Riemann integral. The symmetric sum of the two truncated integrals happens to be zero for every \(\varepsilon\), but that cancellation is not the definition of the integral. A symmetric limiting procedure is called a principal value; it is a different notion.

Splitting Intervals and Using Limits Carefully

For an improper integral, splitting at an ordinary interior point is legitimate, provided the relevant limits exist. For example, if \(a<c\), then for \(R>c\), additivity on the proper interval gives

$$ \int_a^R f(x)\,dx = \int_a^c f(x)\,dx+\int_c^R f(x)\,dx. $$

Consequently, if the integral from \(a\) to infinity converges, then the integral from \(c\) to infinity also converges and their values differ by the proper integral from \(a\) to \(c\). Conversely, convergence of the tail from \(c\) implies convergence from \(a\), since the added piece is finite. The same reasoning applies when splitting a finite interval away from its singular endpoint.

Theorem (Linearity for Convergent Improper Integrals): Suppose \(\int_a^\infty f(x)\,dx\) and \(\int_a^\infty g(x)\,dx\) converge, and let \(\alpha,\beta\in\mathbb{R}\). Then \(\int_a^\infty(\alpha f(x)+\beta g(x))\,dx\) converges and
$$ \int_a^\infty(\alpha f(x)+\beta g(x))\,dx = \alpha\int_a^\infty f(x)\,dx + \beta\int_a^\infty g(x)\,dx. $$

Proof. For every finite \(R>a\), linearity of the proper Riemann integral gives

$$ \int_a^R(\alpha f(x)+\beta g(x))\,dx = \alpha\int_a^R f(x)\,dx+\beta\int_a^R g(x)\,dx. $$

By hypothesis, the two truncated integrals on the right have finite limits as \(R\to\infty\). The limit laws for sums and scalar multiples show that the right-hand side converges to the stated finite value. The left-hand side therefore has that finite limit, which proves both convergence and the identity. \(\square\)

This result requires convergence of both component integrals. It does not say that divergent integrals may be combined and their divergences cancelled. The example involving \(1/x\) shows why that distinction matters. More generally, every singular point divides the interval into separate one-sided questions, and every infinite endpoint requires control of its own tail.

Check Your Understanding

Use the definitions and results above to check your understanding of improper Riemann integrals.

  1. What limit defines \(\int_a^\infty f(x)\,dx\), and what must be true of the limit for the integral to converge?
  2. State the Cauchy criterion for an improper integral on \([a,\infty)\) in terms of integrals over \([R,S]\).
  3. For which values of \(p\) does \(\int_0^1 x^{-p}\,dx\) converge? For which values does \(\int_1^\infty x^{-p}\,dx\) converge?
  4. Why must both one-sided integrals at an interior singularity converge separately?
  5. What does the estimate for \(\int_A^B \sin x/x\,dx\) establish, and how does it differ from absolute convergence?

Previous

Next

Back to Tutorials