Absolute Convergence Near A Finite Singularity

Why Compare an Integral With Its Absolute Value?

An unbounded integrand can change sign as it approaches a singular point. Positive and negative contributions may then affect the ordinary improper integral differently from the integral of the integrand's magnitude. Comparing the two integrals answers an important question: does convergence persist even when all cancellation is removed?

For a function \(f\) with a singularity at the finite endpoint \(0\), the ordinary improper integral is defined by taking a one-sided limit of truncated integrals. The integral of \(|f|\) is defined in the same way. When the latter is finite, the former must also converge. The reverse implication need not hold: signed contributions can cancel, even when their absolute magnitudes have infinite total integral.

Definition: The improper integral \(\int_0^a f(x)\,dx\) converges if the finite limit \(\lim_{\varepsilon\to0^+}\int_\varepsilon^a f(x)\,dx\) exists. It converges absolutely if \(\int_0^a |f(x)|\,dx\) converges. If the ordinary integral converges but the absolute-value integral diverges, the ordinary integral is called conditionally convergent.

Here \(a>0\), and \(f\) is assumed to be integrable on each closed interval \([\varepsilon,a]\) for \(\varepsilon>0\). This allows the truncated integrals to be evaluated in the usual way. The same definitions apply at any finite singular endpoint after shifting the variable.

Absolute Convergence Implies Ordinary Convergence

The implication from absolute to ordinary convergence is a consequence of a simple inequality. It does not require the integrand to have one sign.

Theorem: If \(\int_0^a |f(x)|\,dx\) converges, then \(\int_0^a f(x)\,dx\) converges.

To see why, let \(F(\varepsilon)=\int_\varepsilon^a f(x)\,dx\). For \(0<\delta<\varepsilon\), the difference between two truncated integrals is an integral over the interval between their lower limits. Its magnitude is bounded by the corresponding integral of \(|f|\):

|F(\delta)-F(\varepsilon)| = \left|\int_\delta^\varepsilon f(x)\,dx\right| \leq \int_\delta^\varepsilon |f(x)|\,dx.

If \(\int_0^a |f(x)|\,dx\) is finite, then the right-hand side can be made arbitrarily small by choosing both lower limits sufficiently close to \(0\). Thus the truncated integrals \(F(\varepsilon)\) satisfy the Cauchy criterion for a finite limit. This proves convergence of the ordinary improper integral.

Worked Example: An Absolutely Convergent Signed Power

Consider \(f(x)=-2x^{-2/3}\) on \(0<x\leq1\). Its sign is negative, but absolute convergence is determined by the positive function \(|f(x)|=2x^{-2/3}\). For \(\varepsilon>0\), evaluate the ordinary truncated integral:

\int_\varepsilon^1 -2x^{-2/3}\,dx = \left[-6x^{1/3}\right]_\varepsilon^1 = -6+6\varepsilon^{1/3}.

Taking \(\varepsilon\to0^+\) gives \(\int_0^1 f(x)\,dx=-6\). For the absolute-value integral,

\int_\varepsilon^1 2x^{-2/3}\,dx = \left[6x^{1/3}\right]_\varepsilon^1 = 6-6\varepsilon^{1/3} \longrightarrow 6.

Both integrals converge, so the ordinary integral converges absolutely. As a check, differentiating the antiderivatives gives \(\frac{d}{dx}(-6x^{1/3})=-2x^{-2/3}\) and \(\frac{d}{dx}(6x^{1/3})=2x^{-2/3}=|f(x)|\).

Power Singularities: The Sign Alone Does Not Change the Threshold

For \(p>0\), the power integral is evaluated directly from a truncated interval. If \(p\neq1\), then

\int_\varepsilon^a x^{-p}\,dx = \frac{a^{1-p}-\varepsilon^{1-p}}{1-p}.

When \(p<1\), the term \(\varepsilon^{1-p}\) tends to \(0\), so the limit is finite. When \(p>1\), \(\varepsilon^{1-p}\) grows without bound, and the integral diverges. At \(p=1\), direct integration gives \(\int_\varepsilon^a x^{-1}\,dx=\log(a)-\log(\varepsilon)\), which also diverges as \(\varepsilon\to0^+\). These evaluations give the threshold needed below.

Worked Example: A Negative Power With a Divergent Absolute Integral

Take \(f(x)=-x^{-1}\) on \(0<x\leq1\). Its truncated ordinary integral is

\int_\varepsilon^1 -\frac{1}{x}\,dx = [-\log x]_\varepsilon^1 = \log\varepsilon.

As \(\varepsilon\to0^+\), \(\log\varepsilon\to-\infty\), so the ordinary improper integral does not converge to a finite value. Since \(|f(x)|=x^{-1}\),

\int_\varepsilon^1 |f(x)|\,dx = [\log x]_\varepsilon^1 = -\log\varepsilon \longrightarrow+\infty.

The absolute integral also diverges. This is not conditional convergence: the ordinary integral itself fails to converge. Differentiating \(-\log x\) and \(\log x\) confirms that the antiderivatives used above have derivatives \(-1/x\) and \(1/x\), respectively.

For a constant multiple of a power with a fixed sign, taking the absolute value changes only the sign or the constant's magnitude, not the power. Consequently, ordinary and absolute convergence have the same threshold for these examples. To get different outcomes, sign changes must produce cancellation that persists as the singular endpoint is approached.

A Signed Power That Converges Only Conditionally

Oscillation can create that cancellation. A direct example is \(f(x)=\sin(1/x)/x\) on \(0<x\leq1\). Substituting \(u=1/x\) in a truncated integral converts the finite-endpoint limit into a tail integral:

\int_\varepsilon^1 \frac{\sin(1/x)}{x}\,dx = \int_1^{1/\varepsilon}\frac{\sin u}{u}\,du.

The substitution is exact: \(x=1/u\) gives \(dx=-du/u^2\), while \(1/x=u\), so \((\sin(1/x)/x)\,dx=-\sin(u)\,du/u\). Reversing the limits yields the displayed formula.

Worked Example: A Conditionally Convergent Power-Oscillatory Integral

First check that the ordinary integral has a finite limit. For \(B>A\geq1\), integration by parts gives

\int_A^B \frac{\sin u}{u}\,du = \left[-\frac{\cos u}{u}\right]_A^B - \int_A^B \frac{\cos u}{u^2}\,du.

The endpoint terms have total magnitude at most \(1/A+1/B\), and the remaining integral has magnitude at most \(\int_A^B u^{-2}\,du=1/A-1/B\). Therefore

\left|\int_A^B \frac{\sin u}{u}\,du\right| \leq \frac{1}{A}+\frac{1}{B}+\frac{1}{A}-\frac{1}{B} = \frac{2}{A}.

This bound tends to \(0\) as \(A\to\infty\), so the truncated integrals have a finite limit. By the substitution above, the ordinary integral at \(x=0\) converges.

Now consider the absolute integral. The same substitution gives \(\int_1^{1/\varepsilon}|\sin u|/u\,du\). On each interval \([k\pi,(k+1)\pi]\), where \(k\geq1\), one has \(1/u\geq1/((k+1)\pi)\), and \(\int_{k\pi}^{(k+1)\pi}|\sin u|\,du=2\). Hence

\int_{k\pi}^{(k+1)\pi}\frac{|\sin u|}{u}\,du \geq \frac{2}{(k+1)\pi}.

The sum of these lower bounds diverges because it is a constant multiple of the harmonic series. Thus the absolute integral diverges, while the ordinary integral converges: this is conditional convergence. The substitution also checks the transformation of the absolute integrand, since \(|\sin(1/x)/x|\,dx\) becomes \(|\sin u|\,du/u\) after reversing the limits.

Logarithmic Singularities and Their Absolute Values

A logarithmic factor changes the power threshold. For \(0<x\leq e^{-1}\), the quantity \(\log(1/x)\) is positive. For \(q>0\), use \(t=\log(1/x)\), so \(dt=-dx/x\). This gives the direct evaluation

\int_\varepsilon^{e^{-1}} \frac{dx}{x(\log(1/x))^q} = \int_1^{\log(1/\varepsilon)} \frac{dt}{t^q}.

For \(q>1\), the right-hand side tends to \(1/(q-1)\); for \(q=1\), it equals \(\log(\log(1/\varepsilon))\), which tends to infinity. For \(0<q<1\), its antiderivative \(t^{1-q}/(1-q)\) also grows without bound. Thus the positive logarithmic integrand is integrable at \(0\) exactly when \(q>1\).

Worked Example: An Absolutely Convergent Logarithmic Singularity

Consider \(f(x)=1/[x(\log(1/x))^{3/2}]\) on \(0<x\leq e^{-1}\). It is positive, so \(|f|=f\). Using \(t=\log(1/x)\), the truncated integral is

\int_\varepsilon^{e^{-1}} \frac{dx}{x(\log(1/x))^{3/2}} = \int_1^{\log(1/\varepsilon)}t^{-3/2}\,dt = \left[-2t^{-1/2}\right]_1^{\log(1/\varepsilon)} = 2-\frac{2}{\sqrt{\log(1/\varepsilon)}}.

As \(\varepsilon\to0^+\), this tends to \(2\). Therefore both the ordinary and absolute integrals equal \(2\), and the integral converges absolutely. The substitution is checked by differentiating \(t=\log(1/x)\), which gives \(dt=-dx/x\); the transformed antiderivative differentiates to \(t^{-3/2}\).

Logarithmic damping does not prevent conditional convergence if the integrand continues to oscillate. For example, define \(f(x)=\sin(\log(1/x))/[x\log(1/x)]\) on \(0<x\leq e^{-1}\). With \(t=\log(1/x)\), its truncated ordinary integral becomes \(\int_1^{\log(1/\varepsilon)}\sin(t)/t\,dt\), the convergent integral just examined. Its absolute integral becomes \(\int_1^{\log(1/\varepsilon)}|\sin(t)|/t\,dt\), which diverges by the same interval lower bounds. This logarithmic example is therefore conditionally convergent as well.

Common Mistake: Assuming Cancellation Proves Absolute Convergence

A frequent error is to infer convergence of \(\int_0^a |f(x)|\,dx\) from convergence of \(\int_0^a f(x)\,dx\). For \(f(x)=\sin(1/x)/x\), the ordinary integral converges, but replacing the numerator by its absolute value removes the cancellation. The calculation over each interval \([k\pi,(k+1)\pi]\) then gives a lower bound \(2/((k+1)\pi)\), whose sum diverges. Therefore the ordinary integral converges, but the integral of its absolute value does not.

The valid implication goes in the other direction: a finite integral of \(|f|\) guarantees convergence of the signed integral. When absolute convergence fails, the ordinary integral must be tested separately; it may diverge, as for \(-1/x\), or converge conditionally, as for \(\sin(1/x)/x\).

Check Your Understanding

Use truncated integrals and, where needed, a substitution to answer these questions.

For \(f(x)=3x^{-1/2}\) on \(0<x\leq1\), do the ordinary and absolute integrals converge? Evaluate both.
For \(f(x)=x^{-1}\) on \(0<x\leq1\), what happens to the ordinary and absolute integrals as the lower limit tends to \(0\)?
State the condition on \(p>0\) for \(\int_0^1 x^{-p}\,dx\) to converge.
Why does absolute convergence imply ordinary convergence near a finite singularity?
For \(f(x)=\sin(1/x)/x\), which integral converges: the ordinary integral, the absolute integral, or both?
For \(q>0\), when does \(\int_0^{e^{-1}} dx/[x(\log(1/x))^q]\) converge?