Why compare integrands near an endpoint?
When an integrand becomes unbounded near a finite endpoint, its values at individual points do not determine whether its improper integral is finite. What matters is how quickly the integrand grows as the endpoint is approached. A useful way to assess that growth is to compare the integrand with a simpler function whose integral is already known.
Limit comparison makes this idea precise: if two nonnegative integrands have a ratio approaching a positive finite number, then near the endpoint they differ by at most constant factors. Those bounds are enough to show that either both local integrals converge or both diverge.
Local integrability and the power test
Let \(a<c\), and suppose \(f\) is nonnegative and integrable on every closed subinterval contained in \((a,c]\). The integral at \(a\) is defined by the one-sided limit
provided this limit is finite. If the limit is infinite, the improper integral diverges. We say that \(f\) is locally integrable at \(a\) when this integral is finite. This definition focuses on the behavior near \(a\); changing the integrand away from the endpoint does not affect local integrability there.
The basic reference functions are powers of the distance \(x-a\). For \(p\in\mathbb R\) and \(\delta>0\), their endpoint integrals satisfy
For \(p<1\), this follows by integrating from \(a+\varepsilon\) and letting \(\varepsilon\to0^+\). When \(p\ne1\), the antiderivative is \((x-a)^{1-p}/(1-p)\). At \(p=1\), the antiderivative is \(\ln(x-a)\), and the endpoint limit diverges. Thus the critical exponent for a finite endpoint is \(p=1\).
The limit comparison theorem at a finite endpoint
Assume \(f\) and \(g\) are nonnegative and integrable away from \(a\), and that \(g(x)>0\) sufficiently close to \(a\). The ratio test applies when the ratio has a finite, positive limit.
The theorem follows from the fact that a ratio close to \(L\) gives two-sided bounds. Choose a positive number \(\eta\) smaller than \(L\), for example \(\eta=L/2\). By the definition of the limit, sufficiently close to \(a\),
Rearranging yields \(L/2<f(x)/g(x)<3L/2\). Since \(g(x)>0\), multiplication by \(g(x)\) preserves the inequalities.
These bounds hold on an interval \((a,a+\delta)\) for some \(\delta>0\). The direct comparison theorem now gives the conclusion: if the integral of \(g\) is finite near \(a\), the upper bound makes the integral of \(f\) finite there; if the integral of \(g\) diverges, the lower bound forces the integral of \(f\) to diverge. The same bounds also work in the reverse direction, so the two integrals have the same convergence behavior.
Use a function \(g\) whose local integrability is known, often a power of the distance to the endpoint.
Evaluate \(\lim_{x\to a^+}f(x)/g(x)\), checking that the ratio is defined near \(a\).
A finite positive limit gives the same convergence behavior; then use the reference integral to classify \(f\).
Worked comparisons with power singularities
Worked Example: A convergent power singularity near zero
Determine whether \(f(x)=(1+3x)x^{-2/3}\) is locally integrable at \(0\). Compare it with \(g(x)=x^{-2/3}\), which is positive for \(x>0\).
First compute the ratio:
The limit is finite and positive, so limit comparison applies. The reference integral has exponent \(p=2/3<1\). Directly, for \(0<\varepsilon<1\),
Thus the reference integral converges, and the integral of \(f\) also converges near zero. As a check on the ratio calculation, multiplying \(g(x)\) by \(1+3x\) recovers \(f(x)\) exactly.
Worked Example: A divergent power singularity at a right endpoint
Determine whether \(f(x)=(5+(4-x))(4-x)^{-3/2}\) is locally integrable as \(x\to4^-\). Use \(g(x)=(4-x)^{-3/2}\), which is positive for \(x<4\).
The ratio and its endpoint limit are
Because the limit is \(5\), limit comparison says that \(f\) and \(g\) have the same local integrability. To evaluate the reference integral, set \(u=4-x\). The portion from \(x=3\) to \(x=4-\varepsilon\) becomes
This tends to \(+\infty\) as \(\varepsilon\to0^+\). Hence the reference integral diverges, and so does the integral of \(f\) at \(4\). The substitution also confirms that the distance to the endpoint is \(u=4-x\), so the relevant power is \(u^{-3/2}\), with exponent greater than \(1\).
When the ratio limit is zero or infinite
A ratio limit of zero or infinity does not give the two-sided bounds used in the theorem. The direction of a one-sided comparison matters. If \(f(x)/g(x)\to0\), then \(f(x)\le Cg(x)\) near the endpoint for some constant \(C>0\). Consequently, convergence of the integral of \(g\) implies convergence of the integral of \(f\). But divergence of the integral of \(g\) does not by itself classify \(f\). The corresponding useful implication for an infinite ratio is reversed: if \(f(x)/g(x)\to\infty\) and the integral of \(g\) diverges, then the integral of \(f\) diverges. Other combinations need further information.
Worked Example: A zero ratio with an integrable reference
Use \(f(x)=x^{-1/4}\) and \(g(x)=x^{-1/2}\) to determine whether \(f\) is locally integrable at zero. The ratio is
The reference integral is convergent because its power exponent is \(1/2<1\). In detail,
Since \(f/g\to0\), for all sufficiently small \(x>0\) we have \(0\le f(x)\le g(x)\). The comparison therefore proves that \(f\) is locally integrable. Checking directly gives \(\int_0^1x^{-1/4}\,dx=4/3\), a finite value, consistent with the conclusion.
A common mistake: drawing too much from a zero limit
Suppose \(f(x)/g(x)\to0\). It is incorrect to conclude that \(f\) is integrable merely because this ratio tends to zero. The implication requires the reference function \(g\) to have a convergent integral. Here is a concrete case where the reference diverges and the smaller integrand still diverges.
The reference integral diverges, since its power exponent is \(2>1\). But that fact alone does not decide the integral of \(f\); we must check it. For \(0<\varepsilon<1\),
Thus \(f\) also diverges, despite the zero ratio. The ratio limit establishes a one-sided bound, not equivalence of convergence behavior. By contrast, a finite positive ratio limit gives bounds in both directions and is precisely what makes the limit comparison theorem decisive.
Using distance from either endpoint
At a left endpoint \(a\), the natural power reference is \((x-a)^{-p}\); at a right endpoint \(b\), it is \((b-x)^{-p}\). In either case, the quantity inside the power is the positive distance from \(x\) to the singular endpoint. When a ratio has a finite positive limit against the appropriate power, the power threshold immediately classifies the integral: it converges for \(p<1\) and diverges for \(p\ge1\).
Limit comparison is most useful when the original formula is more complicated than a power but has a clear leading size near the endpoint. The ratio removes the shared singular factor and tests whether only a positive constant remains in the limit. Always verify the endpoint direction, the sign and positivity conditions, and the convergence of the reference integral before stating a conclusion.
Check Your Understanding
For each question, identify the endpoint and the comparison needed before deciding what the ratio tells you.
- What condition on \(\lim_{x\to a^+} f(x)/g(x)\) guarantees that \(f\) and \(g\) have the same local integrability?
- Does \((x-a)^{-4/5}\) have a convergent integral near \(a\)? What about \((x-a)^{-1}\)?
- If \(f(x)/g(x)\to0\) and \(g\) is locally integrable, what can you conclude about \(f\)?
- If \(f(x)/g(x)\to0\) but \(g\) is not locally integrable, can the ratio alone classify \(f\)?
- For a singularity at a right endpoint \(b\), what distance should appear in a power reference function?
