Power and Logarithmic Integrability on Lebesgue Measure

Power and Logarithm Singularities

A power singularity can determine whether a function has a finite integral, while a logarithmic factor can decide the outcome precisely at the critical power. These distinctions matter when checking whether a measurable function is integrable: for a real-valued function \(f\), integrability on a measurable set \(E\) means \(\int_E |f|\,dm<\infty\). The examples here are nonnegative, so this condition is simply finiteness of their integrals.

We work with Lebesgue measure \(m\) on \(\mathbb{R}\), whose measurable sets form the Lebesgue sigma-algebra \(\mathcal{L}\). All functions considered are nonnegative and Lebesgue measurable on the stated intervals; integrals may initially take the value \(+\infty\). Values assigned at individual endpoints do not affect these integrals, since singletons have Lebesgue measure zero.

The intervals are chosen so the logarithmic factors have no singularity at the other endpoint. On any compact interval contained in \((0,\infty)\), these functions are continuous and hence bounded and integrable. Thus the question is entirely about the behavior near zero or infinity.

The Exact Integrability Thresholds

Proof. First consider the integral near zero. On a truncated interval \((\varepsilon,e^{-1})\), make the substitution \(t=\log(1/x)\), so \(x=e^{-t}\) and \(dx=-e^{-t}\,dt\). Reversing the endpoints gives the corresponding integral over \(1<t<\log(1/\varepsilon)\), with integrand \(e^{-(1-\alpha)t}t^\gamma\). As \(\varepsilon\) decreases to zero, these intervals increase to \((1,\infty)\). By the Monotone Convergence Theorem applied to the nonnegative integrand, the original integral is finite exactly when \(\int_1^\infty e^{-(1-\alpha)t}t^\gamma\,dt\) is finite.

For the integral near infinity, truncate at \(R>e\) and use \(t=\log x\), so \(x=e^t\) and \(dx=e^t\,dt\). The integral becomes \(\int_1^{\log R}e^{(1-\alpha)t}t^\gamma\,dt\). Again, the Monotone Convergence Theorem on the increasing truncated intervals shows that its finiteness is equivalent to that of \(\int_1^\infty e^{(1-\alpha)t}t^\gamma\,dt\).

It remains to classify these transformed integrals. For \(c>0\) and any real \(\gamma\), the integral \(\int_1^\infty e^{-ct}t^\gamma\,dt\) is finite, whereas \(\int_1^\infty e^{ct}t^\gamma\,dt\) is infinite. To verify the comparisons, note that \(\log t/t\to0\). Therefore, for all sufficiently large \(t\), \(|\gamma|\log t\le ct/2\). This implies \(t^\gamma\le e^{ct/2}\) in the decaying case, so \(e^{-ct}t^\gamma\le e^{-ct/2}\), which has finite integral. In the growing case it implies \(t^\gamma\ge e^{-ct/2}\), so \(e^{ct}t^\gamma\ge e^{ct/2}\), whose integral is infinite. The finite initial interval does not affect either conclusion.

When the exponential factor is \(1\), the remaining integral is \(\int_1^\infty t^\gamma\,dt\). For \(\gamma\ne-1\), its integral up to \(T\) is \((T^{\gamma+1}-1)/(\gamma+1)\), which has a finite limit as \(T\to\infty\) exactly when \(\gamma<-1\). For \(\gamma=-1\), the integral is \(\log T\), which diverges. Applying these facts to the two substitutions proves every case in the statement. \(\square\)

The power decides the result except at \(\alpha=1\). At that boundary, the logarithmic exponent must be strictly less than \(-1\). In particular, a logarithmic factor with exponent \(-1\) still gives divergence, though only logarithmically.

Computing Integrals with Logarithmic Factors

Proof. Make the substitution \(t=-\log x\) first on a truncated interval \((\varepsilon,1)\). Since \(x=e^{-t}\) and \(dx=-e^{-t}\,dt\), the resulting integral, as \(\varepsilon\downarrow0\), is \(\int_0^\infty e^{-(a+1)t}t^n\,dt\). The truncated intervals increase to the full interval, and the integrand is nonnegative, so the Monotone Convergence Theorem justifies taking their limit. Put \(c=a+1>0\), and write \(I_n(c)=\int_0^\infty e^{-ct}t^n\,dt\). Direct integration gives \(I_0(c)=1/c\). For \(n\ge1\), integration by parts on finite intervals and then passage to the endpoints gives \(I_n(c)=\frac{n}{c}I_{n-1}(c)\): the boundary term at zero vanishes, and the term \(t^ne^{-ct}\) tends to zero at infinity by exponential decay. Repeating this recurrence yields \(I_n(c)=n!/c^{n+1}\), as claimed. \(\square\)

Reading the Boundary Cases Correctly

The threshold theorem also gives the pure power tests. Setting \(\gamma=0\), a power \(x^{-\alpha}\) is integrable near zero exactly when \(\alpha<1\), and integrable near infinity exactly when \(\alpha>1\). For a pure logarithm, set \(\alpha=0\): every \((\log(1/x))^\gamma\) is integrable near zero, while \((\log x)^\gamma\) alone is not integrable on \((e,\infty)\) for any real \(\gamma\). At infinity, no fixed logarithmic decay can compensate for the missing power decay.

A common mistake is to treat the two critical cases as if any negative logarithmic exponent guarantees integrability. For example, the incorrect calculation \(\int_0^{e^{-1}}\frac{1}{x\log(1/x)}\,dm=1\) may come from confusing the antiderivative \(-1/t\) of \(t^{-2}\) with the antiderivative of \(t^{-1}\). The correct substitution gives

The correction is that the critical logarithmic exponent must satisfy \(\gamma<-1\), not merely \(\gamma<0\). These are statements about Lebesgue integrability, not Riemann integrability on a closed interval containing a singular endpoint: a finite Lebesgue integral can exist even when the function is unbounded near that endpoint.