Why Compare the Absolute Value?
For a positive integrand, an upper bound by a known integrable function can prove that an improper integral converges. A sign-changing integrand is different: positive and negative contributions may cancel. The signed integral can therefore converge even when the total size of those contributions is unbounded.
To establish convergence without relying on cancellation, we compare the absolute value of the integrand with a positive function whose improper integral is known to be finite. This proves the stronger conclusion of absolute convergence.
For example, on an infinite interval, convergence means that the limit of \(\int_a^R f(x)\,dx\) as \(R\) tends to infinity exists and is finite. Absolute convergence means that the corresponding limit for \(\int_a^R |f(x)|\,dx\) is finite. At a finite singular endpoint \(b\), the limits are instead taken as \(x\) approaches \(b\) from within the interval.
An absolutely convergent integral is convergent. The reverse implication is not part of the definition: convergence of the signed integral alone does not establish convergence of the integral of the absolute value.
The Absolute-Value Comparison Test
The comparison only needs to hold on the part of the interval where the improper behavior occurs. Any remaining closed, bounded part on which \(f\) is continuous contributes a finite integral. The comparison function \(g\) must be nonnegative, and it is the inequality involving \(|f|\)—not merely \(f\)—that supplies the required bound.
Why the Test Works
Here is the argument for an infinite interval \([a,\infty)\). Assume that \(0\leq |f(x)|\leq g(x)\) for \(x\geq a\), and that \(\int_a^\infty g(x)\,dx\) is finite. For finite \(R>a\), comparison gives
As \(R\) increases, the integrals of \(|f|\) do not decrease, because the integrand is nonnegative. They are also bounded above by the finite integral of \(g\). Thus they have a finite limit, which is precisely the improper integral of \(|f|\). To see that the signed integral converges as well, take \(S>R\). Then
The last quantity tends to zero as \(R\) tends to infinity, uniformly for \(S>R\), because the tail of the convergent positive integral of \(g\) tends to zero. Hence the signed integrals satisfy the Cauchy criterion and converge to a finite value. The same reasoning applies at a finite improper endpoint, with intervals approaching that endpoint in place of tails.
Find a nonnegative comparison function \(g\) such that \(|f(x)|\leq g(x)\) near the improper endpoint.
Verify that the improper integral of \(g\) is finite on the relevant interval.
The integral of \(|f|\) is finite, so the integral of \(f\) is absolutely convergent and therefore convergent.
Worked Example: A Sign Change Near a Singular Endpoint
Worked Example: A Sign Change Near a Singular Endpoint
Determine whether the integral of \(f(x)=\frac{x-\frac12}{\sqrt{x}}\) from \(0\) to \(1\) converges absolutely. The integrand is continuous for \(0<x\leq 1\), changes sign at \(x=\frac12\), and has an improper endpoint at \(0\).
For \(0<x\leq 1\), \(\left|x-\frac12\right|\leq\frac12\). Therefore
The comparison integral is finite:
By the absolute-value comparison test, the integral of \(|f|\) is finite, so the original integral converges absolutely. Its signed value can also be checked directly. An antiderivative on \(x>0\) is \(\frac23x^{3/2}-\sqrt{x}\), since
Thus the signed improper integral is \(\lim_{\varepsilon\to0^+}\left[\frac23x^{3/2}-\sqrt{x}\right]_{\varepsilon}^{1}=-\frac13\). The antiderivative has the stated derivative, and substituting the endpoints gives a finite value, consistent with the comparison test.
Worked Example: An Infinite Tail With One Sign Change
Worked Example: An Infinite Tail With One Sign Change
Consider \(f(x)=\frac{x-1}{(x+1)^3}\) on \([0,\infty)\). It changes sign at \(x=1\). On \(0\leq x\leq1\), \(|x-1|\leq1\) and \((x+1)^3\geq1\), so \(|f(x)|\leq1\). On \(x\geq1\), \(|x-1|\leq x+1\), giving
Both comparison integrals are finite:
Hence \(\int_0^\infty |f(x)|\,dx\leq\frac32\), so the integral of \(f\) is absolutely convergent. To check its signed value, rewrite
An antiderivative is \(-\frac{1}{x+1}+\frac{1}{(x+1)^2}\), whose derivative is \(\frac{1}{(x+1)^2}-\frac{2}{(x+1)^3}=f(x)\). Its value at \(0\) is \(0\), and its limit at infinity is \(0\). Therefore \(\int_0^\infty f(x)\,dx=0\). This finite signed value does not rely on cancellation to establish convergence; the absolute-value bound has already proved the stronger result.
Worked Example: A Singularity at the Right Endpoint
Worked Example: A Singularity at the Right Endpoint
Consider \(f(x)=\frac{x-\frac12}{\sqrt{1-x}}\) on \(0\leq x<1\). It changes sign at \(x=\frac12\) and is improper at \(x=1\). Since \(\left|x-\frac12\right|\leq\frac12\) throughout this interval,
The positive comparison integral is finite:
Thus the integral of \(f\) is absolutely convergent. For a direct signed-value check, an antiderivative is \(-\sqrt{1-x}+\frac23(1-x)^{3/2}\). Its derivative is \(\frac{1}{2\sqrt{1-x}}-\sqrt{1-x}=\frac{x-\frac12}{\sqrt{1-x}}\). The value of this antiderivative tends to \(0\) at \(x=1\), while its value at \(x=0\) is \(-1+\frac23=-\frac13\). Therefore the signed integral equals \(\frac13\), a finite value consistent with absolute convergence.
Absolute Convergence and Conditional Convergence
The examples establish absolute convergence by proving that the integral of \(|f|\) is finite. Conditional convergence is a different conclusion: the signed improper integral must first be shown to have a finite limit, and the integral of \(|f|\) must then be shown to diverge. Failure to find an integrable comparison function does not, by itself, prove conditional convergence. It may mean only that the chosen bound was not strong enough, or that the integral diverges.
| What is established? | Conclusion |
|---|---|
| The improper integral of \(|f|\) is finite. | The integral of \(f\) is absolutely convergent, hence convergent. |
| The improper integral of \(f\) is finite, but the integral of \(|f|\) diverges. | The integral of \(f\) is conditionally convergent. |
| Only an upper bound on \(f\), rather than on \(|f|\), is known. | Absolute convergence has not been established. |
Common Pitfall: Comparing the Signed Function
A tempting but invalid argument is to compare \(f\) directly with a positive integrable function and infer that the absolute integral is finite. For a concrete example, let \(f(x)=\frac{2}{x}-1\) for \(x\geq1\). It changes sign at \(x=2\). For \(x\geq2\), \(f(x)\leq0\leq\frac{1}{x^2}\), and \(\int_2^\infty \frac{1}{x^2}\,dx=\frac12\). But this signed inequality does not bound \(|f|\).
Indeed, for \(x\geq4\), \(|f(x)|=1-\frac2x\geq\frac12\), so \(\int_2^\infty |f(x)|\,dx\) diverges. The signed integral from \(2\) to \(R\) also fails to converge:
The correction is to test the inequality \(|f(x)|\leq g(x)\) and to verify that the comparison integral of \(g\) is finite. The inequality \(f(x)\leq g(x)\) alone does not provide the needed control of the magnitude of a sign-changing function.
Summary
For an improper integral with a sign-changing integrand, absolute convergence is established by bounding \(|f|\) above by a nonnegative function with a finite improper integral. This comparison controls the total size of the contributions, not merely their signed balance. It proves both that the absolute integral is finite and that the original signed integral converges.
Check Your Understanding
Use the definitions and comparison principle to answer these questions.
- What additional integral must be finite for an improper integral to be absolutely convergent?
- State the comparison inequality needed to prove absolute convergence using a nonnegative function \(g\).
- Why does a finite signed integral not by itself prove absolute convergence?
- If the signed integral converges but the integral of the absolute value diverges, what is the convergence called?
- Why is \(f(x)\leq g(x)\) insufficient for proving absolute convergence of a sign-changing integrand?
