Why Bounded Multipliers Help
An improper integral can be difficult to evaluate exactly, even when its convergence is easier to determine. A useful strategy is to separate the integrand into a familiar nonnegative function and a multiplier whose size is controlled. If that multiplier stays between two positive constants near the problematic endpoint, it cannot change whether the integral is finite.
The lower bound matters as much as the upper bound. An upper bound can show that a product is integrable when a suitable comparison function is integrable. But to conclude that the product diverges whenever the comparison function diverges, we also need a positive lower bound.
The Bounded-Multiplier Comparison Theorem
For an infinite tail, “sufficiently near the endpoint” means for all \(x\ge A\), for some finite \(A\). For an endpoint at \(a\), it means for all \(x\) in some interval \((a,a+\delta)\), where \(\delta>0\). The same result applies at a right-hand finite endpoint by using a left-hand neighborhood. The integrals must be well behaved away from the endpoint; a compact portion of the interval does not affect convergence if its integral is finite.
The constants \(m\) and \(M\) do not need to be the best possible bounds. They only need to be positive and finite, and to work throughout the relevant tail or endpoint neighborhood.
Why the Theorem Works
Consider first an improper integral at infinity. Choose \(A\) so that the bounds on \(g\) hold for \(x\ge A\). For any \(R>A\), multiply those bounds by \(f(x)\ge0\) and integrate. This gives:
These inequalities explain both directions of the result. If the integral of \(f\) converges, its truncated integrals stay bounded as \(R\) increases. The upper inequality then keeps the truncated integrals of \(fg\) bounded as well. Since \(fg\ge0\), those integrals increase with \(R\), so they have a finite limit. Conversely, if the integral of \(fg\) converges, the lower inequality gives \(\int_A^R f(x)\,dx\le \frac{1}{m}\int_A^R f(x)g(x)\,dx\). The truncated integrals of \(f\) are therefore bounded and, being increasing, have a finite limit.
If either integral diverges, the same inequalities show that the other cannot converge: a divergent integral of \(f\) forces the integral of \(fg\) to diverge by the lower bound, and a divergent integral of \(fg\) forces the integral of \(f\) to diverge by the upper bound. A finite initial interval contributes only a finite amount under the stated local-integrability conditions.
Replace the infinite limit, or singular endpoint, with a finite cutoff.
Multiplying \(m\le g(x)\le M\) by \(f(x)\ge0\) preserves the inequality signs.
The resulting bounds show that the two nonnegative improper integrals are finite together, or diverge together.
The proof near a finite endpoint is the same: truncate the integral a positive distance from the endpoint, apply the pointwise bounds, and then let the cutoff approach the endpoint.
Worked Examples
Worked Example: A Trigonometric Multiplier on an Infinite Tail
Determine whether \(\int_1^\infty \frac{2+\sin x}{x^2}\,dx\) converges.
For every real \(x\), \(-1\le\sin x\le1\). Adding 2 gives \(1\le2+\sin x\le3\). Also \(x^{-2}\ge0\) for \(x\ge1\). Thus the integrand is between \(1/x^2\) and \(3/x^2\):
For a finite cutoff \(R>1\), the upper comparison has integral \(\int_1^R 3x^{-2}\,dx=3(1-1/R)\le3\). The truncated integrals of the nonnegative target integrand are therefore bounded above by 3. As \(R\) increases, they increase, so they have a finite limit. The improper integral converges.
Check: The multiplier has the required positive bounds \(m=1\) and \(M=3\), and the comparison integral has the exact finite limit \(\int_1^\infty x^{-2}\,dx=1\). The target integral is consequently finite (in fact, the inequalities also bound it below by 1).
Worked Example: A Trigonometric Multiplier Near a Finite Endpoint
Determine whether \(\int_0^1 \frac{3+\cos x}{\sqrt{x}}\,dx\) converges at \(x=0\).
Since \(-1\le\cos x\le1\), we have \(2\le3+\cos x\le4\). The factor \(x^{-1/2}\) is nonnegative for \(0<x\le1\), so
For \(0<\varepsilon<1\), the upper comparison integral is \(\int_\varepsilon^1 4x^{-1/2}\,dx=8(1-\sqrt{\varepsilon})\le8\). Thus the truncated target integrals are bounded above and increase as \(\varepsilon\) decreases toward zero. They have a finite limit, so the improper integral converges.
Check: The comparison integral has the finite value \(\int_0^1 4x^{-1/2}\,dx=8\), while the multiplier \(3+\cos x\) stays between the positive constants 2 and 4. Both conditions required for the theorem hold near zero.
Worked Example: An Algebraic Multiplier and a Divergent Tail
Determine whether \(\int_1^\infty \frac{\sqrt{x^2+1}}{x^2}\,dx\) converges.
For \(x\ge1\), \(x^2\le x^2+1\le2x^2\), because \(1\le x^2\). Taking square roots and dividing by \(x>0\) gives \(1\le\frac{\sqrt{x^2+1}}{x}\le\sqrt{2}\). Rewrite the integrand as a multiplier times \(1/x\):
For \(R>1\), the lower comparison gives \(\int_1^R \frac{\sqrt{x^2+1}}{x^2}\,dx \ge\int_1^R \frac1x\,dx=\ln R\). As \(R\to\infty\), \(\ln R\to\infty\), so the target integral diverges.
Check: The algebraic multiplier is bounded below by 1 and above by \(\sqrt2\) for every \(x\ge1\). The comparison integral \(\int_1^\infty dx/x\) diverges, and the pointwise lower bound confirms that the target integral also diverges.
When the Conditions Fail
A multiplier being bounded above is not enough to preserve convergence behavior. It might become arbitrarily small, reducing the product enough to make a divergent integral converge. For example, on \(0<x\le1\), take \(f(x)=1/x\) and \(g(x)=x\). The factor \(g\) is positive and bounded above by 1, but it has no positive lower bound near zero. Direct calculation gives:
So the first integral diverges and the second converges. The theorem does not apply because there is no constant \(m>0\) such that \(m\le x\) throughout a neighborhood of zero.
A related mistake is to treat a factor’s occasional size as a uniform bound. To use the theorem, the same constants must bound the multiplier at every point sufficiently near the endpoint in question. A factor that is positive at each individual point may still approach zero, or grow without bound, as the endpoint is approached.
Why This Comparison Is Useful
Trigonometric and algebraic expressions often look complicated while their size is simple to control. Bounds such as \(-1\le\sin x\le1\), or inequalities obtained by comparing the leading powers inside a square root, can turn a difficult-looking integrand into a familiar comparison. Once the multiplier is trapped between positive constants, the convergence question reduces to the underlying nonnegative function.
For an integral with more than one improper endpoint, analyze each endpoint separately. The theorem can be applied near an endpoint only if its multiplier bounds hold there; bounds near one endpoint do not automatically apply near another. The central check is always local: identify the endpoint, verify nonnegativity, and find positive lower and finite upper bounds that hold throughout a full neighborhood or tail.
Check Your Understanding
For each question, identify which bounds or comparison facts are needed before drawing a conclusion.
- State the two-sided bound required of a multiplier near an improper endpoint.
- If \(f(x)\ge0\) and \(2\le g(x)\le5\) for all sufficiently large \(x\), what can be said about the convergence behavior of \(\int f(x)\,dx\) and \(\int f(x)g(x)\,dx\) on that tail?
- Why does an upper bound on a positive multiplier alone fail to establish the theorem’s conclusion?
- For \(x\ge1\), show that \(\sqrt{x^2+4}/x\) has positive lower and finite upper bounds.
- In the example \(f(x)=1/x\), \(g(x)=x\) on \((0,1]\), which condition fails, and what are the two improper integrals’ convergence behaviors?
