Why Comparison Arguments Need Care
Integral comparison is powerful because it can replace a difficult integrand with a simpler one whose convergence is known. But a comparison conclusion is valid only when its hypotheses hold on the region that matters. In particular, the functions must have the right signs, the comparison must be made near the relevant endpoint or tail, and every improper part of the integral must be checked.
These details are easy to overlook. A pointwise inequality does not automatically compare signed integrals in the way it compares nonnegative ones. A limit of zero in a limit-comparison test is not the same as a positive finite limit. And information about an integrand near infinity says nothing by itself about a singularity inside the interval.
The Comparison Rules and Their Sign Conditions
Suppose \(f\) and \(g\) are nonnegative and integrable on every closed subinterval of \([a,\infty)\). If \(0\le f(x)\le g(x)\) for all sufficiently large \(x\), then convergence of \(\int_a^\infty g(x)\,dx\) implies convergence of \(\int_a^\infty f(x)\,dx\). In the other direction, if \(0\le g(x)\le f(x)\) for all sufficiently large \(x\), divergence of \(\int_a^\infty g(x)\,dx\) implies divergence of \(\int_a^\infty f(x)\,dx\).
The same rules apply near a finite singular endpoint. For example, on \((a,b]\), if \(0\le f(x)\le g(x)\) sufficiently near \(a\), convergence of \(\int_a^b g(x)\,dx\) implies convergence of \(\int_a^b f(x)\,dx\). A compact portion away from the singular endpoint does not affect convergence, provided the integrand is integrable there.
Here is why the first rule works. For \(R>a\), nonnegativity and the pointwise inequality give \(0\le\int_a^R f(x)\,dx\le\int_a^R g(x)\,dx\), once the inequality holds throughout the relevant range (or after separating off an initial finite interval). If \(\int_a^\infty g(x)\,dx\) converges, its partial integrals are bounded above. The partial integrals of \(f\) are nondecreasing and bounded above as well, so they have a finite limit. The argument for the divergence rule is similar: if \(f\ge g\ge0\), then the partial integrals of \(f\) are at least those of \(g\), which grow without bound.
For signed functions, these conclusions cannot be used as stated. A standard safe alternative for absolute convergence is to compare the absolute value: if \(|f(x)|\le g(x)\), \(g(x)\ge0\), and \(\int g\) converges, then \(\int |f|\) converges.
Worked Example: A Signed Inequality Does Not Prove Divergence
Worked Example: A Signed Inequality Does Not Prove Divergence
Consider \(f(x)=-1/x^2\) and \(g(x)=1/x\) for \(x\ge1\). It is true that \(f(x)\le g(x)\), and \(\int_1^\infty g(x)\,dx\) diverges. But that does not imply that \(\int_1^\infty f(x)\,dx\) diverges.
Compute the integral of \(f\) using a finite upper limit \(R\):
Taking \(R\to\infty\) gives \(\int_1^\infty f(x)\,dx=-1\), a finite value. In contrast, \(\int_1^R 1/x\,dx=\ln R\to\infty\). The inequality compares functions of different signs, so the nonnegative divergence rule does not apply. The correct check is that \(f\) is not nonnegative; for absolute convergence, \(\int_1^\infty |f(x)|\,dx=1\).
Limit Comparison: Read the Limit Before Drawing a Conclusion
For positive functions near the improper endpoint, a common method is to compare their ratio. If \(f(x)>0\), \(g(x)>0\), and \(f(x)/g(x)\to c\) as \(x\) approaches the relevant endpoint, where \(0<c<\infty\), then \(f\) and \(g\) have the same convergence behavior there.
The reason is that, sufficiently close to the endpoint, the ratio lies between two positive constants. For example, it can be made to satisfy \(c/2<f(x)/g(x)<3c/2\). Multiplying by \(g(x)>0\) gives \((c/2)g(x)<f(x)<(3c/2)g(x)\). The direct comparison rules then transfer convergence or divergence in both directions.
Worked Example: A Zero Ratio Limit Is Not an Equivalence
Worked Example: A Zero Ratio Limit Is Not an Equivalence
On \([1,\infty)\), let \(f(x)=1/x^2\) and \(g(x)=1/x\). Both are positive, and
Now calculate the two improper integrals. For \(R>1\),
whereas
Thus the smaller function has a convergent integral while the larger one diverges. The zero ratio limit does not say that the integrals behave the same. It is consistent with the valid comparison rule, but it is not the positive finite limit required for two-way limit comparison.
Split Every Improper Integral at Its Improper Parts
An improper integral is defined by one-sided limits. For example, \(\int_a^\infty f(x)\,dx\) is the limit of \(\int_a^R f(x)\,dx\) as \(R\to\infty\), if that limit is finite. If an integral has both an infinite endpoint and an interior singularity, or has singularities at multiple points, it must be split into separate improper integrals.
Every piece must converge for the ordinary improper integral to converge. One part cannot repair another by cancellation. For an interior singularity at \(c\), the expression \(\int_a^b f(x)\,dx\) requires both one-sided limits at \(c\) to exist finitely, in addition to any other improper endpoints. A symmetric principal value, if considered, is a different limiting procedure and does not establish ordinary improper-integral convergence.
Worked Example: A Convergent Tail Does Not Repair a Divergent Endpoint
Worked Example: A Convergent Tail Does Not Repair a Divergent Endpoint
Define \(f(x)=1/x\) for \(0<x\le1\), and \(f(x)=e^{-x}\) for \(x>1\). The integral over \((0,\infty)\) has an improper endpoint at zero and an infinite tail, so split at \(1\).
For the part near zero, let \(\varepsilon>0\). Then
For the tail, let \(R>1\). Then
The tail converges, but the part near zero diverges. Therefore the full improper integral diverges. This conclusion follows from checking both pieces, not from the tail alone.
Asymptotic Equivalence Is Local to the Limit Being Taken
For positive functions, \(f(x)\sim g(x)\) as \(x\to\infty\) means that \(f(x)/g(x)\to1\) as \(x\to\infty\). This relation supports a convergence comparison on a sufficiently far tail. It does not assert that the functions are close everywhere on their domain, nor does it control a singularity at a finite interior point.
Likewise, when the relevant singularity is at a finite endpoint \(a\), the ratio must be examined as \(x\to a^+\) (or from the appropriate side). A limit at another endpoint or at infinity cannot settle convergence near \(a\). If there are several improper locations, make a local comparison for each one.
Worked Example: Equivalence at Infinity Misses an Interior Singularity
Worked Example: Equivalence at Infinity Misses an Interior Singularity
For \(x>0\), define \(g(x)=1\) when \(0<x<2\) and \(g(x)=e^{-x}\) when \(x\ge2\). Define \(f(x)=1/|x-1|\) when \(0<x<2\), except at \(x=1\), and \(f(x)=e^{-x}\) when \(x\ge2\). The functions are identical for \(x\ge2\), so
Nevertheless, \(\int_0^\infty g(x)\,dx\) converges: its integral on \((0,2)\) is \(2\), and its tail is \(\int_2^\infty e^{-x}\,dx=e^{-2}\). The function \(f\) has an interior singularity at \(x=1\). On either side, the singular contribution has the form \(\int dx/|x-1|\). In particular, for \(0<\varepsilon<1\),
Thus \(\int_0^\infty f(x)\,dx\) diverges despite equivalence at infinity. The limit comparison correctly describes the tail, which is the same for both functions; it makes no claim about the separate singularity at \(x=1\).
A Practical Check Before Comparing
Before using a comparison, identify which limit defines the improper part. Then check the sign and the direction of the inequality on that part. For a ratio test, evaluate the ratio at that same endpoint and distinguish a positive finite limit from zero, infinity, or a nonexistent limit. Finally, split the integral wherever the definition requires separate limits and verify every piece.
Mark every infinite endpoint and every point where the integrand is singular.
Use direct comparison for nonnegative functions; use absolute values when proving absolute convergence of a signed integrand.
Take the ratio at the endpoint or tail under consideration, not at a different point in the domain.
A positive finite ratio limit gives matching behavior; a zero or infinite limit has more limited implications.
Convergence of one part does not compensate for divergence of another.
Check Your Understanding
For each question, identify the comparison hypothesis that matters before deciding what conclusion is justified.
- If \(0\le f(x)\le g(x)\) on a tail and \(\int g\) converges there, what can be concluded about \(\int f\)?
- Why does \(f(x)\le g(x)\) alone not allow the nonnegative comparison theorem to be applied?
- If \(f(x)/g(x)\to0\) and \(\int g\) diverges, does limit comparison determine whether \(\int f\) converges?
- What must be checked when an integrand has an interior singularity as well as an infinite endpoint?
- If \(f(x)\sim g(x)\) as \(x\to\infty\), what part of the integral does this information control, and what does it not control?