From a Finite Interval to an Infinite One
An integral whose upper endpoint is infinity cannot be evaluated by simply substituting infinity into an antiderivative. Instead, first integrate over a finite interval and then ask what happens as its upper endpoint increases without bound. The limit both determines the value, when there is one, and tells us whether the improper integral converges.
For example, \(\int_0^\infty e^{-x}\,dx\) is not an ordinary definite integral with a real-number endpoint. Its meaning comes from the finite integrals \(\int_0^b e^{-x}\,dx\), considered as \(b\) grows. Their limit is \(1\), so the improper integral converges to \(1\).
The definition uses a limit of finite-interval integrals. The symbol \(\infty\) is not a number that can be substituted for \(b\); it describes how \(b\) changes.
The Antiderivative-and-Limit Method
Suppose \(F\) is an antiderivative of \(f\) on \([a,\infty)\), meaning \(F'(x)=f(x)\) there. For each finite \(b>a\), the Fundamental Theorem of Calculus gives \(\int_a^b f(x)\,dx=F(b)-F(a)\). Therefore, evaluating the improper integral reduces to taking the limit of this difference.
This follows directly from applying the finite-interval theorem first and then using the definition of the improper integral:
For any real \(b>a\), continuity of \(f\) on \([a,b]\) allows the Fundamental Theorem of Calculus to be applied.
Since \(F'(x)=f(x)\), we have \(\int_a^b f(x)\,dx=F(b)-F(a)\).
The definition says to take the limit of the finite integrals as \(b\to\infty\). Substituting their exact values gives \(\lim_{b\to\infty}(F(b)-F(a))\).
The endpoint value \(F(a)\) is a fixed finite number. Thus, convergence depends on whether \(F(b)\) approaches a finite value as \(b\to\infty\). The antiderivative itself does not need to approach zero; it needs to have a finite limiting value so that the difference has a finite limit.
Worked Examples
Worked Example: An Exponential Integral That Converges
Evaluate \(\int_0^\infty e^{-3x}\,dx\) and determine whether it converges.
An antiderivative is \(F(x)=-\frac{1}{3}e^{-3x}\), because
For a finite upper endpoint \(b\), the Fundamental Theorem of Calculus gives
As \(b\to\infty\), \(e^{-3b}\to0\). Therefore,
The limit is finite, so the integral converges. Check: differentiating \(F(x)=-\frac13e^{-3x}\) gives the original integrand, and the finite-interval values approach \(\frac13\) as required.
Worked Example: A Rational Function with a Finite Limit
Evaluate \(\int_2^\infty \frac{5}{x^3}\,dx\) and determine whether it converges. The integrand is continuous for every \(x\ge2\). Rewrite it as \(5x^{-3}\) to find an antiderivative:
For finite \(b>2\),
Since \(1/b^2\to0\) as \(b\to\infty\),
The limit is finite, so the integral converges to \(\frac58\). Check: \(F'(x)=5/x^3\), and the finite-interval result tends to \(\frac58\) because its term \(-5/(2b^2)\) tends to zero.
Worked Example: A Logarithmic Antiderivative That Diverges
Determine whether \(\int_1^\infty \frac{1}{x}\,dx\) converges. An antiderivative on \([1,\infty)\) is \(F(x)=\ln x\), since
For finite \(b>1\),
As \(b\to\infty\), \(\ln b\to+\infty\). Thus the limit is not finite, and
Check: the derivative of \(\ln x\) is the given integrand, and the finite-interval integrals grow without bound rather than approaching a finite value. The integral has no finite value under the definition of convergence.
How to State Convergence Correctly
After calculating the limit, state both the outcome and its interpretation. If it is finite, report the integral's value and say that it converges. If it tends to \(+\infty\) or \(-\infty\), say that it diverges; an infinite limit is not a finite value of the improper integral. If the limit fails to exist, the integral also diverges.
For an integrand with a known antiderivative, the steps are always the same:
- Check that the integrand is defined and continuous on every finite interval from the stated lower endpoint onward.
- Find an antiderivative \(F\) on that interval.
- Evaluate \(F(b)-F(a)\) with a finite variable endpoint \(b\).
- Take the limit as \(b\to\infty\).
- Classify the result: a finite real limit means convergence; otherwise the integral diverges.
It is important not to omit the lower-endpoint term. Even when \(F(b)\) has a simple limit, the value of the convergent integral is that limit minus \(F(a)\), not just the limit of \(F(b)\).
Common Mistake: Substituting Infinity Directly
A tempting but incorrect calculation is to write \(\int_1^\infty \frac1x\,dx=[\ln x]_1^\infty=\ln(\infty)-\ln(1)\), and then treat \(\ln(\infty)\) as an ordinary number. Infinity is not an endpoint at which the antiderivative can be evaluated. This notation also hides the question of whether a finite limit exists.
The correct calculation keeps the upper endpoint finite until after evaluating:
Therefore the integral diverges. The expression \(+\infty\) describes the behavior of the finite-interval integrals; it is not the value of a convergent improper integral.
Why the Limit Is Essential
On each finite interval, antiderivatives evaluate definite integrals through the Fundamental Theorem of Calculus. But an infinite interval is not one of those finite intervals. Taking the limit is what extends the finite-interval calculation while preserving the definition of the improper integral.
This distinction matters even when the antiderivative is easy to find. A formula can produce finite values for every finite \(b\) while those values grow without bound as \(b\) increases, as in \(\int_1^b 1/x\,dx=\ln b\). Conversely, the finite values may approach a definite real number, as in the exponential and rational examples. It is the limiting behavior—not simply the existence of an antiderivative—that determines convergence.
Check Your Understanding
For each integral, use a finite upper endpoint before taking the limit. State whether it converges and give its value if it does.
- Evaluate \(\int_0^\infty e^{-2x}\,dx\).
- Determine whether \(\int_3^\infty \frac{2}{x^2}\,dx\) converges.
- Determine whether \(\int_1^\infty \frac{4}{x}\,dx\) converges.
- What condition on \(\lim_{b\to\infty}(F(b)-F(a))\) determines convergence when \(F'=f\)?
- Explain why \(\infty\) cannot be substituted directly into an antiderivative as if it were a real endpoint.
