Evaluating Infinite-Interval Integrals by Cutoff Limits

From an Infinite Interval to a Limit

An integral over an interval that extends without bound is not evaluated by treating infinity as an ordinary endpoint. Instead, choose a finite cutoff, evaluate the resulting proper integral, and then let the cutoff increase without bound. The limit records whether the accumulated integral approaches a finite value.

This approach is useful because it keeps the two distinct tasks visible: first perform an ordinary definite integral over a finite interval; then determine the behavior of its value as the endpoint moves farther away. A finite limit gives the value of the improper integral. If there is no finite limit, the integral diverges.

Definition: Let $f$ be continuous on $[a,\infty)$, where $a$ is finite. The right-infinite integral is defined by $$ \int_a^\infty f(x)\,dx = \lim_{R\to\infty}\int_a^R f(x)\,dx, $$ provided this limit exists as a finite real number. If the limit is not finite or does not exist, the improper integral diverges.

The variable $R$ is a finite cutoff, and the limit $R\to\infty$ means that $R$ increases without bound. For each fixed $R>a$, the integral from $a$ to $R$ is an ordinary definite integral. The improper integral is the limit of those ordinary integral values, not an integral evaluated directly at an infinite endpoint.

The continuity assumption ensures that each finite-cutoff integral is well-defined by the usual definite-integral rules. This tutorial considers integrands that are continuous at the finite endpoint as well as throughout the half-line. An integral with a singularity at a finite endpoint requires separate care.

The Cutoff Procedure

Suppose $F$ is an antiderivative of $f$ on $[a,\infty)$, so $F'(x)=f(x)$ there. For any finite $R>a$, the Fundamental Theorem of Calculus gives

\int_a^R f(x)\,dx=F(R)-F(a).

Only after obtaining this finite-cutoff expression do we take its limit as $R\to\infty$. Since $F(a)$ is a fixed finite number, the improper integral converges precisely when $F(R)$ has a finite limit, and its value is that limit minus $F(a)$.

Introduce a cutoff.
Replace the infinite endpoint by a finite number $R$, with $R>a$.

Integrate on the finite interval.
Find an antiderivative and evaluate the definite integral from $a$ to $R$.

Take the cutoff limit.
Compute the limit as $R\to\infty$. A finite limit is the integral's value; otherwise, it diverges.

Why the Procedure Works

The method follows directly from the definition and the Fundamental Theorem of Calculus. It is not a special rule for replacing infinity with a number. Each finite cutoff gives an ordinary integral, and the definition of the improper integral asks whether these ordinary values approach a finite limit.

Derivation: Let $f$ be continuous on $[a,\infty)$, and let $F'(x)=f(x)$. For every finite $R>a$, the Fundamental Theorem of Calculus yields $$ \int_a^R f(x)\,dx=F(R)-F(a). $$ Taking $R\to\infty$, if $\lim_{R\to\infty}F(R)=L$ for a finite real number $L$, then $$ \lim_{R\to\infty}\int_a^R f(x)\,dx = \lim_{R\to\infty}\bigl(F(R)-F(a)\bigr) = L-F(a). $$ Thus the improper integral converges and equals $L-F(a)$. If $F(R)$ has no finite limit, subtracting the fixed number $F(a)$ cannot create one, so the improper integral does not converge to a finite value.

The finite-cutoff step matters even when the antiderivative has a simple expression. The symbol $\infty$ is not a real number that can be substituted into $F$; the limit describes what happens as real cutoffs $R$ grow.

Worked Examples

Worked Example: A Convergent Power Integral

Evaluate $\int_1^\infty \frac{3}{x^4}\,dx$ using a cutoff.

For finite $R>1$, first compute the ordinary integral. Since an antiderivative of $3x^{-4}$ is $-x^{-3}$, we obtain

\int_1^R \frac{3}{x^4}\,dx = \left[-\frac{1}{x^3}\right]_1^R = -\frac{1}{R^3}-\left(-1\right) = 1-\frac{1}{R^3}.

As $R\to\infty$, $1/R^3\to 0$. Therefore,

\int_1^\infty \frac{3}{x^4}\,dx = \lim_{R\to\infty}\left(1-\frac{1}{R^3}\right) = 1.

Check the antiderivative: differentiating $-x^{-3}$ gives $3x^{-4}$. The cutoff expression also has the expected endpoint behavior: at $R=1$ its value is $0$, and as $R$ increases it approaches $1$.

Worked Example: A Decaying Exponential

Evaluate $\int_0^\infty 5e^{-5x}\,dx$.

Use a finite cutoff $R>0$. Because the derivative of $-e^{-5x}$ is $5e^{-5x}$, the finite integral is

\int_0^R 5e^{-5x}\,dx = \left[-e^{-5x}\right]_0^R = -e^{-5R}-(-1) = 1-e^{-5R}.

Since $e^{-5R}\to 0$ as $R\to\infty$, the cutoff limit is

\int_0^\infty 5e^{-5x}\,dx = \lim_{R\to\infty}(1-e^{-5R}) = 1.

Check by differentiation: $\frac{d}{dx}(-e^{-5x})=5e^{-5x}$, so the antiderivative is correct. The finite-cutoff value starts at $0$ when $R=0$ and approaches $1$, in agreement with the computed improper integral.

Worked Example: A Shifted Power Integral

Evaluate $\int_2^\infty \frac{6}{(x+1)^4}\,dx$.

For a finite cutoff $R>2$, an antiderivative is $-2(x+1)^{-3}$, because

\frac{d}{dx}\left(-\frac{2}{(x+1)^3}\right) = -2\left(-3\right)(x+1)^{-4} = \frac{6}{(x+1)^4}.

Now evaluate at both finite endpoints:

\int_2^R \frac{6}{(x+1)^4}\,dx = \left[-\frac{2}{(x+1)^3}\right]_2^R = -\frac{2}{(R+1)^3}+\frac{2}{27}.

The term $2/(R+1)^3$ tends to zero as $R\to\infty$. Hence

\int_2^\infty \frac{6}{(x+1)^4}\,dx = \lim_{R\to\infty}\left(\frac{2}{27}-\frac{2}{(R+1)^3}\right) = \frac{2}{27}.

The derivative check above confirms the antiderivative, and substituting the lower endpoint into it gives $-2/27$. Subtracting that lower-endpoint value produces the positive cutoff value shown, which tends to $2/27$.

Worked Example: A Cutoff Limit That Is Not Finite

Determine whether $\int_0^\infty e^x\,dx$ converges.

For finite $R>0$, the antiderivative of $e^x$ is $e^x$, so

\int_0^R e^x\,dx = \left[e^x\right]_0^R = e^R-1.

As $R\to\infty$, $e^R-1\to\infty$, so the cutoff values do not approach a finite real number. Therefore, the improper integral diverges; it does not have a finite value. The check is direct: $\frac{d}{dx}e^x=e^x$, and the finite integral's value grows without bound as the cutoff increases.

A Common Mistake: Substituting Infinity Directly

A frequent error is to write an antiderivative at an infinite endpoint as though infinity were a number. For example, after finding an antiderivative of $x^{-2}$, one might write

\int_1^\infty \frac{1}{x^2}\,dx = \left[-\frac{1}{x}\right]_1^\infty = -\frac{1}{1} = -1.

This calculation is wrong: it treats the symbol $\infty$ as if it were a number and never forms the difference of endpoint values. The displayed $-1$ is only the antiderivative evaluated at the lower endpoint, $-1/1$. The corrected calculation keeps the cutoff finite until after evaluating both endpoints:

\int_1^\infty \frac{1}{x^2}\,dx = \lim_{R\to\infty}\left(\left[-\frac{1}{x}\right]_1^R\right) = \lim_{R\to\infty}\left(-\frac{1}{R}+1\right) = 1.

Differentiating $-1/x$ gives $1/x^2$, confirming the antiderivative. The error was not in the integration; it was in failing to evaluate the finite-cutoff expression correctly before taking its limit.

What the Limit Tells You

A cutoff calculation can lead to three broad outcomes. The values may approach a finite number, as in the convergent examples; they may grow without bound, as for $\int_0^\infty e^x\,dx$; or they may fail to approach any limit. In the latter two situations, the improper integral diverges under the definition above. Finding an antiderivative alone does not decide which outcome occurs: the behavior of the cutoff expression must still be examined.

For a convergent integral, the answer comes from the complete endpoint difference $F(R)-F(a)$, followed by its limit. Keeping this order prevents sign errors and makes the role of the finite lower endpoint explicit. It also gives a reliable way to check a result: differentiate the proposed antiderivative, evaluate both endpoints for a finite cutoff, and only then take the limit.

Check Your Understanding

For each question, begin with a finite cutoff and state what happens as that cutoff tends to infinity.

Evaluate $\int_1^\infty \frac{2}{x^3}\,dx$. Show the finite-cutoff expression before taking its limit.
Evaluate $\int_0^\infty 3e^{-3x}\,dx$, and verify the antiderivative by differentiation.
For $\int_0^\infty 1\,dx$, write the cutoff integral and explain why the improper integral does not converge to a finite value.
State the condition on the cutoff limit that makes a right-infinite integral converge.
Explain why an antiderivative evaluated at infinity is not a substitute for taking a limit.

Evaluating Infinite-Interval Integrals by Cutoff Limits

What You'll Learn

From an Infinite Interval to a Limit

The Cutoff Procedure

Why the Procedure Works

Worked Examples