Why an Infinite Interval Needs a Definition
An ordinary definite integral is taken over a finite interval. For example, \(\int_a^R f(x)\,dx\) has two finite endpoints when \(a\) and \(R\) are real numbers with \(a<R\). The notation \(\int_a^\infty f(x)\,dx\) looks similar, but infinity is not a real number and cannot be used as an ordinary endpoint.
To make sense of the notation, we first integrate only as far as a finite cutoff \(R\). We then ask what happens to those finite-interval integrals as the cutoff moves farther to the right without bound. The limit, if it exists as a finite real number, is the value assigned to the right-infinite integral.
Continuity of \(f\) on every finite interval \([a,R]\) is a familiar sufficient condition for the finite-cutoff integrals to be defined. The definition above concerns what happens as the interval grows; it does not require the integral over the infinite interval to be treated as an ordinary definite integral.
Convergence and Divergence
For each finite cutoff \(R>a\), define the cutoff integral \(A(R)=\int_a^R f(x)\,dx\). The improper integral converges precisely when \(A(R)\) approaches a finite real number as \(R\to\infty\). That number is the value of the integral.
There are two important ways the required finite limit can fail. The cutoff integrals may grow without bound, or they may fail to settle toward any limit. In either case the improper integral diverges. A limit of \(+\infty\) or \(-\infty\) is not a finite real limit, so it does not count as convergence under this definition.
| Behavior of the cutoff integrals | Classification |
|---|---|
| \(A(R)\) approaches a finite real number \(L\) | The integral converges, and its value is \(L\). |
| \(A(R)\) grows without bound, for example \(A(R)\to+\infty\) | The integral diverges. |
| \(A(R)\) has no limit as \(R\to\infty\) | The integral diverges. |
The integrand itself need not have a finite integral just because its values become small far to the right. Convergence is defined by the accumulated areas represented by \(A(R)\), not by the values of \(f(R)\) alone. The examples below focus on the cutoff integrals, which are the quantities the definition asks us to examine.
Worked Examples
Worked Example: A Convergent Right-Infinite Integral
Determine whether \(\int_0^\infty \frac{1}{(x+1)^2}\,dx\) converges, and find its value if it does.
First take a finite cutoff \(R>0\). An antiderivative is \(-1/(x+1)\), since
Thus the finite-cutoff integral is
Now take the cutoff limit. As \(R\to\infty\), \(1/(R+1)\to0\), so
The limit is finite, so the improper integral converges and equals \(1\). The derivative check above confirms the antiderivative used in the finite-cutoff calculation.
Worked Example: Cutoff Integrals That Grow Without Bound
Determine whether \(\int_0^\infty \frac{1}{x+1}\,dx\) converges.
For any finite \(R>0\), an antiderivative of \(1/(x+1)\) is \(\ln(x+1)\), because
Therefore,
As \(R\to\infty\), \(\ln(R+1)\to+\infty\). The cutoff integrals do not approach a finite real number, so
The finite-cutoff antiderivative is valid; divergence follows from the limit of its evaluated values. In particular, a finite answer cannot be assigned merely because the integrand is defined at every finite \(x\geq0\).
Worked Example: Cutoff Integrals with No Limit
Show that \(\int_0^\infty 2x\cos(x^2)\,dx\) diverges because its cutoff integrals have no limit.
For a finite cutoff \(R>0\), the chain rule gives
Hence
To test for a limit, consider two sequences of cutoffs, each tending to infinity:
Both sequences tend to infinity, but their cutoff integrals are
If the cutoff integrals had a limit as \(R\to\infty\), they would approach the same value along every sequence of cutoffs tending to infinity. These two sequences give different values, \(1\) and \(-1\), so no limit exists. The right-infinite integral therefore diverges.
A Useful Criterion for Nonnegative Functions
When \(f(x)\geq0\) for all \(x\geq a\), the cutoff integrals cannot decrease as the cutoff moves to the right. This gives a useful way to recognize convergence: for a nonnegative continuous function, the right-infinite integral converges if and only if its cutoff integrals are bounded above.
Here is the justification. If \(R_2>R_1>a\), additivity of finite-interval integrals gives
because \(f\) is nonnegative. Thus \(A\) is nondecreasing. If \(A\) is bounded above, let \(L\) be the least upper bound of its values. Given any \(\varepsilon>0\), the definition of least upper bound gives a cutoff \(R_0\) such that \(A(R_0)>L-\varepsilon\). For every \(R\geq R_0\), monotonicity and the fact that \(L\) is an upper bound imply
These inequalities show that \(A(R)\to L\), a finite limit. Conversely, if \(A\) is unbounded above, then for any real number \(M\) there is a cutoff \(R_0\) with \(A(R_0)>M\). For every \(R\geq R_0\), monotonicity gives \(A(R)\geq A(R_0)>M\). This is exactly the statement that \(A(R)\to+\infty\). Therefore the integral diverges in this case.
The nonnegativity condition matters: it is what ensures that the cutoff integrals are nondecreasing. For a function that takes both positive and negative values, the cutoff integrals can move in either direction, and this particular boundedness criterion does not apply as stated.
Common Pitfall: Substituting Infinity as an Endpoint
A common error is to treat \(\infty\) as if it were a real number that can be inserted into an antiderivative. For example, after finding \(\ln(x+1)\) as an antiderivative of \(1/(x+1)\), one might try to “substitute infinity” and assign a value without evaluating a limit. This skips the essential question: what is the limit of the finite-cutoff integrals?
The correct calculation keeps \(R\) finite until after evaluating the definite integral:
Since this limit is not finite, the integral diverges. The notation \(\int_0^\infty\) is shorthand for the stated limiting process; it does not make infinity an ordinary endpoint.
Why the Definition Matters
The limit definition gives a precise meaning to accumulated area over an unbounded interval. It also prevents two different questions from being confused: whether each finite-cutoff integral can be calculated, and whether those answers settle toward a finite value as the cutoff grows. The examples show that finite-cutoff integrals can have a finite limit, grow without bound, or fail to have any limit.
When deciding whether a right-infinite integral converges, keep the order of reasoning clear:
Work with \(\int_a^R f(x)\,dx\), where \(R\) is a real number greater than \(a\).
Use an antiderivative or another valid method on the bounded interval.
Determine whether the evaluated expression approaches a finite real number as \(R\to\infty\).
A finite limit means convergence; an infinite limit or no limit means divergence.
This procedure is the definition, not merely a computational trick. In later applications, the main work may be finding or estimating the finite-cutoff integrals. Whatever method is used, convergence still depends on the limit specified here.
Check Your Understanding
Use the cutoff-limit definition to answer these questions.
- What limit defines \(\int_a^\infty f(x)\,dx\), and what must be true of that limit for the integral to converge?
- If the cutoff integrals tend to \(+\infty\), how is the improper integral classified?
- For a nonnegative continuous function, why are its cutoff integrals nondecreasing?
- Why does boundedness above imply convergence in the nonnegative-function criterion?
- For \(f(x)=2x\cos(x^2)\) on \([0,\infty)\), what two sequences of cutoffs show that the cutoff integrals have no limit?
