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Infinite-Interval Integrals · Tutorial 293 of 998

Substitution with an Infinite Endpoint

Track the transformed cutoff as well as the differential to evaluate half-line integrals correctly by substitution.

Beginner 10 min read

What You'll Learn

  • Define a half-line integral using a finite cutoff and a limit.
  • Transform both the variable of integration and the cutoff under substitution.
  • Apply decreasing exponential substitutions with correctly reversed limits.
  • Apply power substitutions and verify the resulting integral.
  • Use exact cutoff identities to distinguish convergence from divergence.
  • Identify and correct a sign error caused by reversing limits incorrectly.

Why the Cutoff Must Move

Substitution is familiar in ordinary definite integrals: change variables, transform the differential, and update the limits. For an integral over an unbounded interval, there is one extra point to keep track of. The infinite endpoint is defined through a limit, so its finite cutoff must also be transformed.

This matters even when the substitution itself is simple. An increasing substitution sends a large cutoff to a large value of the new variable. A decreasing substitution sends it toward the other end of the new interval, and the order of the limits changes. In both cases, the transformed integral must retain the same cutoff information as the original.

Key principle: First replace the infinite endpoint by a finite cutoff. Then apply ordinary substitution, transforming the differential and both limits. Only after that should the cutoff tend to infinity, or to its corresponding endpoint in the new variable.

Half-Line Integrals as Limits

If \(f\) is continuous on \([a,\infty)\), its improper integral over the half-line is defined by truncating at a finite number \(R\) and taking a limit:

$$ \int_a^\infty f(x)\,dx = \lim_{R\to\infty}\int_a^R f(x)\,dx. $$

The integral converges if this limit exists and is finite. If the limit does not exist or is infinite, the improper integral diverges. The finite-cutoff integrals are ordinary definite integrals, so substitution can be applied to each one before taking the limit.

Suppose an increasing substitution \(x=\phi(u)\) maps \([\alpha,\beta)\) onto \([a,\infty)\), with \(\phi(\alpha)=a\) and \(\phi(u)\to\infty\) as \(u\to\beta^-\). If \(\phi\) is continuously differentiable and \(f\) is continuous on the relevant finite intervals, then the cutoff \(R\) corresponds to the unique \(u_R\) satisfying \(\phi(u_R)=R\). Thus \(u_R\to\beta^-\), and the substitution is made with upper limit \(u_R\), not immediately with \(\beta\).

Definition: For an increasing substitution \(x=\phi(u)\) with \(\phi(\alpha)=a\) and \(\phi(u)\to\infty\) as \(u\to\beta^-\), the transformed half-line integral is interpreted as $$ \lim_{u\to\beta^-}\int_\alpha^u f(\phi(s))\phi'(s)\,ds, $$ provided this limit exists. The substitution formula follows by applying ordinary substitution up to each finite cutoff.

A decreasing substitution has the same finite-cutoff principle, but reverses the direction of travel. For instance, if \(x=\phi(u)\) maps \((0,\beta]\) onto \([a,\infty)\), with \(\phi(\beta)=a\) and \(\phi(u)\to\infty\) as \(u\to0^+\), then the cutoff \(R\) corresponds to a value \(u_R\to0^+\). The transformed finite integral runs from \(\beta\) down to \(u_R\). Since \(\phi'(u)<0\) for a strictly decreasing differentiable substitution, the negative differential and the reversed limits work together.

The Substitution Rule and Its Justification

Theorem: Let \(f\) be continuous on \([a,\infty)\), and let \(x=\phi(u)\) be a continuously differentiable, strictly increasing map from \([\alpha,\beta)\) onto \([a,\infty)\), where \(\phi(\alpha)=a\) and \(\phi(u)\to\infty\) as \(u\to\beta^-\). Then, when the limits exist, $$ \int_a^\infty f(x)\,dx = \lim_{u\to\beta^-}\int_\alpha^u f(\phi(s))\phi'(s)\,ds. $$ For a strictly decreasing substitution onto the same half-line, the same identity holds with the transformed limits ordered from the starting value down toward the value corresponding to the cutoff.

The rule is a consequence of ordinary substitution on every finite interval. The infinite endpoint introduces a limit, but does not change the finite-interval substitution formula.

1
Choose a finite cutoff.
Write the original integral up to \(R\), where \(R>a\) is finite.
2
Find its new-variable value.
For an increasing substitution, let \(u_R\) satisfy \(\phi(u_R)=R\). Then \(u_R\to\beta^-\) as \(R\to\infty\).
3
Substitute on the finite interval.
The ordinary substitution formula gives \(\int_a^R f(x)\,dx=\int_\alpha^{u_R} f(\phi(s))\phi'(s)\,ds\).
4
Take the matching limit.
Let \(R\to\infty\), equivalently \(u_R\to\beta^-\). The two cutoff integrals are equal for every finite \(R\), so their limits agree whenever they exist.

This also explains why a valid substitution preserves convergence or divergence: the original and transformed cutoff integrals are equal at every finite cutoff. Their limiting behavior must therefore agree. The transformed integral is not a separate approximation; it is an exact rewriting of the same family of truncated integrals.

Worked Example: An Exponential Substitution

Worked Example: Transforming an Exponential Tail

Evaluate \(\int_0^\infty e^{-2x}\,dx\) using the decreasing substitution \(u=e^{-x}\). Begin with a finite cutoff \(R\):

$$ I_R=\int_0^R e^{-2x}\,dx. $$

When \(x=0\), \(u=1\); when \(x=R\), \(u=e^{-R}\). Also, \(du=-e^{-x}\,dx=-u\,dx\), so \(dx=-du/u\), and \(e^{-2x}=u^2\). Substituting all of these gives

$$ I_R = \int_1^{e^{-R}} u^2\left(-\frac{du}{u}\right) = \int_1^{e^{-R}}-u\,du = \int_{e^{-R}}^1u\,du. $$

Evaluate the finite integral:

$$ I_R = \left[\frac{u^2}{2}\right]_{e^{-R}}^1 = \frac12-\frac{e^{-2R}}{2} = \frac12\left(1-e^{-2R}\right). $$

As \(R\to\infty\), \(e^{-2R}\to0\), so the improper integral converges to \(1/2\). The transformed cutoff is \(e^{-R}\), which tends to \(0\); it is not replaced by zero until the limit is taken.

Check: Directly, an antiderivative of \(e^{-2x}\) is \(-\frac12e^{-2x}\), since its derivative is \(e^{-2x}\). Therefore \(\int_0^R e^{-2x}\,dx=\frac12(1-e^{-2R})\), matching the transformed calculation and giving the same limit \(1/2\).

Worked Example: A Power Substitution

Worked Example: A Power Tail with \(x=t^2\)

Evaluate \(\int_1^\infty x^{-3/2}\,dx\) using \(x=t^2\). For a finite cutoff \(R>1\), set \(T=\sqrt R\), so \(x=1\) corresponds to \(t=1\) and \(x=R\) corresponds to \(t=T\). Since \(dx=2t\,dt\) and \(x^{-3/2}=(t^2)^{-3/2}=t^{-3}\) for \(t\ge1\),

$$ \int_1^R x^{-3/2}\,dx = \int_1^{\sqrt R}t^{-3}(2t)\,dt = \int_1^{\sqrt R}2t^{-2}\,dt. $$

An antiderivative of \(2t^{-2}\) is \(-2t^{-1}\), because its derivative is \(2t^{-2}\). Therefore

$$ \int_1^{\sqrt R}2t^{-2}\,dt = \left[-\frac{2}{t}\right]_1^{\sqrt R} = -\frac{2}{\sqrt R}-(-2) = 2-\frac{2}{\sqrt R}. $$

As \(R\to\infty\), \(\sqrt R\to\infty\) and \(2/\sqrt R\to0\), so the integral converges to \(2\).

Check: An antiderivative of \(x^{-3/2}\) is \(-2x^{-1/2}\), because \(\frac{d}{dx}(-2x^{-1/2})=x^{-3/2}\). Its finite-cutoff value is \(\left[-2x^{-1/2}\right]_1^R=2-2/\sqrt R\), exactly the value obtained by substitution.

Worked Example: A Substitution Does Not Create Convergence

Worked Example: The Logarithmic Tail Under \(t=1/x\)

Consider \(\int_1^\infty \frac{1}{x}\,dx\). Use the decreasing substitution \(t=1/x\) on a finite cutoff \(R>1\). The limits are \(x=1\) giving \(t=1\), and \(x=R\) giving \(t=1/R\). Since \(x=1/t\) and \(dx=-t^{-2}\,dt\), the integrand and differential combine as

$$ \frac{1}{x}\,dx = t\left(-\frac{1}{t^2}\,dt\right) = -\frac{1}{t}\,dt. $$

Thus the finite-cutoff integral is

$$ \int_1^R\frac{1}{x}\,dx = \int_1^{1/R}-\frac{1}{t}\,dt = \int_{1/R}^1\frac{1}{t}\,dt = [\ln t]_{1/R}^1 = \ln R. $$

As \(R\to\infty\), \(\ln R\to\infty\), so the improper integral diverges. In the new variable, the moving cutoff \(1/R\) approaches \(0\), and the transformed integral grows without bound.

Check: The original integrand has antiderivative \(\ln x\) for \(x>0\), so \(\int_1^R dx/x=\ln R-\ln1=\ln R\). This agrees exactly with the transformed calculation.

A Common Sign and Limit Error

For the exponential example, \(u=e^{-x}\) gives \(dx=-du/u\), and the cutoff limits are \(1\) and \(e^{-R}\). A common incorrect calculation drops the negative sign but keeps the original order of limits:

$$ \int_1^{e^{-R}}u\,du = \frac12(e^{-2R}-1). $$

This expression is negative, even though the original integrand \(e^{-2x}\) is positive throughout the interval. The mistake is that the negative sign in \(dx=-du/u\) was lost. The correct calculation is

$$ \int_1^{e^{-R}}-u\,du = \int_{e^{-R}}^1u\,du = \frac12(1-e^{-2R}). $$

There are two equivalent ways to handle a decreasing substitution: keep the negative differential and the original order of limits, or reverse the limits and remove the negative sign. Doing both, or doing neither, gives the wrong sign.

Practical check: At every finite cutoff, compare the sign and value of the transformed integral with the original finite integral. The transformed limits must correspond to the original endpoints, and the differential must include its sign.

Why the Cutoff Viewpoint Matters

The finite-cutoff method separates two tasks that are easy to confuse. First, substitution rewrites an ordinary definite integral exactly. Second, the improper integral is obtained by taking a limit. If the transformed cutoff is not tracked, it is easy to reverse limits incorrectly or to assume a finite transformed integral before establishing that its endpoint limit exists.

The examples show both possibilities. In the exponential and power examples, the transformed finite integrals approach finite limits. In the reciprocal example, the transformed cutoff approaches zero while the integral grows without bound. Substitution changes the representation of an integral; it does not change whether the original cutoff integrals converge.

For a new substitution on a half-line, make the following checks before evaluating: identify the finite cutoff, compute its new-variable value, transform the differential including its sign, and write the transformed limits in the correct order. Then evaluate the finite integral and take the corresponding endpoint limit.

Check Your Understanding

For each question, keep the finite cutoff visible until the final limit.

  1. Under \(u=e^{-x}\), what are the transformed limits for \(\int_2^R f(x)\,dx\)?
  2. For \(x=t^3\) on \([1,R]\), what is the new upper cutoff in the \(t\)-variable?
  3. Why must the differential sign be included when using a decreasing substitution?
  4. Using \(t=1/x\), write the transformed finite-cutoff integral for \(\int_1^R x^{-2}\,dx\).
  5. What does equality of the original and transformed finite-cutoff integrals imply about their limits, if those limits exist?

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