Why an Infinite-Interval Integral Needs a Limit
An ordinary definite integral is taken over a finite interval. An integral whose interval extends without bound cannot be evaluated as an ordinary definite integral all at once: there is no finite endpoint at which to stop the accumulation.
Instead, we first stop at a finite point, calculate the proper integral on that truncated interval, and then let the stopping point move farther and farther out. The limit tells us whether the accumulated signed area approaches a single finite value.
Definition: An Infinite Upper Bound
Suppose \(f\) is continuous on every finite interval \([a,b]\) with \(b\ge a\). For each finite \(b\), the integral \(\int_a^b f(x)\,dx\) is a proper integral. We define the integral with infinite upper bound by taking the limit as that finite endpoint increases without bound.
Here, \(b\to\infty\) means that \(b\) increases without bound. It does not mean that infinity is a number that can be substituted for \(b\). The expression on the right is a limit of values of proper integrals.
There is a corresponding definition for an infinite lower bound. If \(f\) is continuous on every finite interval \([b,a]\) with \(b\le a\), then
provided this limit exists as a finite real number. The finite endpoint \(b\) moves to the left without bound.
What Convergence and Divergence Mean
For a right-hand improper integral, define the accumulated value up to \(b\) by \(A(b)=\int_a^b f(x)\,dx\). The improper integral converges precisely when \(A(b)\) approaches a finite number as \(b\to\infty\). It diverges if the accumulated values grow without bound, decrease without bound, or fail to approach any single value.
Convergence concerns a limit, not whether the integrand itself approaches zero. An integrand may become small and still accumulate an unbounded total. The definition asks what happens to the integrals over longer and longer finite intervals.
| Behavior of the truncation limit | Improper integral |
|---|---|
| The limit is a finite real number | Converges to that number |
| The limit increases without bound | Diverges to \(+\infty\) |
| The limit decreases without bound | Diverges to \(-\infty\) |
| The limit does not exist | Diverges |
The value, when it exists, is a signed accumulation: regions where \(f(x)\) is negative contribute negatively. Convergence does not mean that every part of the accumulated area is positive.
Worked Example: A Finite Limit
Worked Example: A Finite Limit
Determine whether \(\int_0^\infty \frac{1}{(1+x)^2}\,dx\) converges, and find its value if it does.
Step 1: Truncate the interval. For a finite \(b>0\), form the proper integral:
Step 2: Evaluate the proper integral. An antiderivative is \(-\frac{1}{1+x}\), since
Therefore, using the finite endpoints \(0\) and \(b\),
Step 3: Take the limit.
The limit is finite, so the improper integral converges to \(1\). The antiderivative check above verifies the integration used in the calculation.
Worked Example: Divergence to Infinity
Worked Example: Divergence to Infinity
Determine whether \(\int_0^\infty 1\,dx\) converges.
Step 1: Use a finite upper bound. For \(b>0\),
Step 2: Take the limit.
The limit is not a finite real number. Thus \(\int_0^\infty 1\,dx\) diverges to \(+\infty\); it does not have a finite improper-integral value. The finite-interval calculation is consistent because the derivative of \(x\) is \(1\).
Worked Example: Divergence Because the Limit Does Not Exist
Worked Example: Divergence Because the Limit Does Not Exist
Determine whether \(\int_0^\infty \cos x\,dx\) converges.
Step 1: Evaluate the proper integral to a finite endpoint. Since the derivative of \(\sin x\) is \(\cos x\),
Step 2: Test whether the truncation values approach a limit. Along the sequence \(b_n=\frac{\pi}{2}+2\pi n\), where \(n\) is a nonnegative integer, \(\sin b_n=1\). Along \(c_n=\frac{3\pi}{2}+2\pi n\), \(\sin c_n=-1\). Both sequences of endpoints increase without bound, but the corresponding integral values stay at \(1\) and \(-1\), respectively.
Because the truncation values do not approach a single limit, the improper integral diverges. Checking both sequences verifies that no finite limit exists.
Why the Limit Definition Works
For each finite \(b\), \(\int_a^b f(x)\,dx\) measures the signed accumulation from the fixed starting point \(a\) to the current stopping point \(b\). Moving \(b\) farther out includes more of the interval. If these accumulated values settle toward a finite number, that number is the well-defined total accumulation over the unbounded interval.
This definition also specifies exactly what must be checked: the limit of the finite-interval integrals. Merely finding an antiderivative is not enough. A finite limit of the evaluated truncations establishes convergence; an infinite limit or a nonexistent limit establishes divergence.
Replace the infinite bound by a variable endpoint such as \(b\).
Work on the finite interval and retain the variable endpoint in the result.
Determine whether the result approaches one finite real number as the endpoint moves without bound.
Give the finite value if the limit exists; otherwise identify the type of divergence when possible.
Integrals with Two Infinite Bounds
An integral over the entire real line has two unbounded ends. It is defined by treating those ends separately. Choose any finite number \(c\), split the interval there, and require both one-sided improper integrals to converge.
The split point \(c\) can be any finite real number. If both one-sided integrals converge, the sum is independent of the choice of \(c\): for two choices \(c<d\), additivity of proper integrals gives \(\int_c^d f(x)\,dx\) as the piece included in one side and removed from the other, so the total is unchanged.
Both sides must converge separately. Taking a single symmetric limit \(\lim_{R\to\infty}\int_{-R}^{R}f(x)\,dx\) is not, by itself, the definition of the improper integral over the whole real line.
Common Pitfall: Treating Infinity as an Endpoint
A tempting but invalid shortcut is to substitute infinity directly into an antiderivative or to let two divergent sides cancel. For example, symmetric truncations of \(\int_{-\infty}^{\infty}x\,dx\) give
But the two required one-sided limits do not converge:
Thus the improper integral over the whole real line diverges under its definition. The symmetric value \(0\) results from pairing truncations and cancelling their contributions; it does not establish that the two one-sided integrals have finite values.
Check Your Understanding
For each question, focus on the finite-interval truncation and the limit that defines the improper integral.
- Write the limit definition of \(\int_2^\infty f(x)\,dx\).
- What does it mean for an infinite-interval integral to diverge to \(+\infty\)?
- If the truncation values alternate between \(1\) and \(-1\) along two sequences of endpoints tending to infinity, can the improper integral converge? Explain.
- How must \(\int_{-\infty}^{\infty} f(x)\,dx\) be defined, and what must be true of both one-sided integrals?
- Why does a finite symmetric limit alone not establish convergence of an integral over the whole real line?
