Power Singularities and Lp Integrability

Two Places a Power Can Fail to Be Integrable

A power function can cause an \(L^p\) integral to diverge for two different reasons. Near zero, a negative power may become too large on every sufficiently small interval. At infinity, a power may decay too slowly for its integral over an unbounded interval to be finite. The thresholds are different: a singularity at zero requires \(\alpha p<1\), while a tail at infinity requires \(\beta p>1\). Equality in either condition produces a logarithmically divergent integral.

Throughout, \(1\le p<\infty\), and the measure is Lebesgue measure \(m\) on the indicated subset of \(\mathbb R\). The \(L^p\) functions considered are measurable real- or complex-valued functions, and membership means \(\int |f|^p\,dm<\infty\). The power functions below are specified away from zero; assigning any finite value at the single point zero does not change their integrals, since \(\{0\}\) has Lebesgue measure zero.

Definition: A measurable function \(f\) belongs to \(L^p(E,m)\), for \(1\le p<\infty\), if \(\int_E |f|^p\,dm<\infty\). Its \(L^p\) norm is \(\|f\|_p=(\int_E|f|^p\,dm)^{1/p}\). Membership and the norm depend only on the almost-everywhere equivalence class of \(f\).

The Power Integral Test

The core calculation is the integral of \(x^{-q}\). For \(q\ne1\), its antiderivative on \((0,\infty)\) is \(x^{1-q}/(1-q)\); for \(q=1\), it is \(\ln x\). To handle an endpoint at zero or infinity, first integrate over a truncated interval and then take the endpoint limit. Equivalently, the truncated intervals increase to the full interval, so the Monotone Convergence Theorem applies to these nonnegative functions.

Proposition (Power integral test): For \(r>0\), \(\int_0^r x^{-q}\,dx\) is finite exactly when \(q<1\), and then it equals \(r^{1-q}/(1-q)\). For \(R>0\), \(\int_R^\infty x^{-q}\,dx\) is finite exactly when \(q>1\), and then it equals \(R^{1-q}/(q-1)\).

Proof. For the integral near zero, truncate at \(\varepsilon\in(0,r)\). If \(q\ne1\), then \(\int_\varepsilon^r x^{-q}\,dx=(r^{1-q}-\varepsilon^{1-q})/(1-q)\). When \(q<1\), \(1-q>0\), so \(\varepsilon^{1-q}\to0\) as \(\varepsilon\downarrow0\), giving the finite value \(r^{1-q}/(1-q)\). When \(q>1\), \(1-q<0\), so \(\varepsilon^{1-q}\to+\infty\); the truncated integrals tend to \(+\infty\). For \(q=1\), they equal \(\ln r-\ln\varepsilon\), which also tends to \(+\infty\). The truncations increase to \((0,r)\), so Monotone Convergence identifies their limit with the Lebesgue integral.

For the integral at infinity, truncate at \(S>R\). If \(q\ne1\), then \(\int_R^S x^{-q}\,dx=(S^{1-q}-R^{1-q})/(1-q)\). When \(q>1\), \(S^{1-q}\to0\), and the limit is \(R^{1-q}/(q-1)\). When \(q<1\), \(S^{1-q}\to+\infty\), so the integrals diverge. If \(q=1\), they equal \(\ln S-\ln R\), which tends to \(+\infty\). The intervals \((R,S)\) increase to \((R,\infty)\), and Monotone Convergence again gives the asserted Lebesgue integral. \(\square\)

Near Zero: The Singularity Threshold

Theorem: Let \(r>0\), \(1\le p<\infty\), and \(\alpha\in\mathbb R\). On \((-r,r)\) with Lebesgue measure, the function \(f(x)=|x|^{-\alpha}\) for \(x\ne0\) belongs to \(L^p((-r,r),m)\) if and only if \(\alpha p<1\). If this condition holds, then \(\int_{-r}^{r}|f|^p\,dm=2r^{1-\alpha p}/(1-\alpha p)\).

Proof. For \(x\ne0\), \(|f(x)|^p=|x|^{-\alpha p}\). By symmetry, its integral over \((-r,r)\) is twice its integral over \((0,r)\); the point zero has measure zero and does not affect this equality. Apply the power integral test with \(q=\alpha p\). The integral is finite exactly when \(\alpha p<1\), and in that case it is \(2r^{1-\alpha p}/(1-\alpha p)\). By the definition of \(L^p\), this is precisely the membership condition. \(\square\)

Worked Example: A Locally Integrable Singularity

Determine whether \(f(x)=|x|^{-1/4}\) belongs to \(L^2((-1,1),m)\), and compute its norm if it does.

Here \(p=2\) and \(\alpha=1/4\), so \(\alpha p=(1/4)\cdot2=1/2<1\). The theorem applies and gives

\int_{-1}^{1}|f(x)|^2\,dx =2\int_0^1 x^{-1/2}\,dx =2\left[\frac{x^{1/2}}{1/2}\right]_0^1 =2(2-0)=4.

Thus \(f\in L^2((-1,1),m)\) and \(\|f\|_2=\sqrt4=2\). Check: integrating on the positive and negative halves separately gives \(2\) on each half, so the total is \(2+2=4\), as above.

Worked Example: Testing the Critical Exponent

On \((0,1)\) with Lebesgue measure, let \(g(x)=x^{-1/3}\). Decide membership in \(L^2\) and \(L^3\), including the integral calculations at the boundary.

For \(p=2\), the exponent in \(|g|^p\) is \((1/3)\cdot2=2/3<1\). Therefore

\int_0^1 |g(x)|^2\,dx =\int_0^1 x^{-2/3}\,dx =\left[3x^{1/3}\right]_0^1 =3, \qquad \|g\|_2=\sqrt3.

For \(p=3\), the exponent is \((1/3)\cdot3=1\). Truncating the integral at \(\varepsilon>0\) gives \(\int_\varepsilon^1 |g(x)|^3\,dx=\int_\varepsilon^1x^{-1}\,dx=-\ln\varepsilon\). As \(\varepsilon\downarrow0\), this tends to \(+\infty\). Hence \(g\notin L^3((0,1),m)\). Check: the criterion gives \(1/3<1/2\) for \(p=2\), but \(1/3\not<1/3\) for \(p=3\), matching both computations.

At Infinity: The Tail Threshold

Theorem: Let \(R>0\), \(1\le p<\infty\), and \(\beta\in\mathbb R\). On \((R,\infty)\) with Lebesgue measure, the function \(h(x)=x^{-\beta}\) belongs to \(L^p((R,\infty),m)\) if and only if \(\beta p>1\). If this condition holds, then \(\int_R^\infty|h|^p\,dm=R^{1-\beta p}/(\beta p-1)\).

Proof. Since \(x>0\), \(|h(x)|^p=x^{-\beta p}\). Use the power integral test at infinity with \(q=\beta p\). Its integral is finite exactly when \(\beta p>1\), and then equals \(R^{1-\beta p}/(\beta p-1)\). This is the defining condition for \(h\in L^p((R,\infty),m)\). \(\square\)

Worked Example: A Decaying Tail in L²

For \(h(x)=x^{-3/4}\) on \((1,\infty)\), compute its \(L^2\) norm or show that it does not exist as a finite norm.

Here \(p=2\) and \(\beta=3/4\), so \(\beta p=(3/4)\cdot2=3/2>1\). The tail theorem applies. Directly,

\int_1^\infty |h(x)|^2\,dx =\int_1^\infty x^{-3/2}\,dx =\lim_{S\to\infty}\left[-2x^{-1/2}\right]_1^S =\lim_{S\to\infty}\left(2-\frac{2}{\sqrt S}\right) =2.

Therefore \(h\in L^2((1,\infty),m)\) and \(\|h\|_2=\sqrt2\). Check: the formula \(R^{1-\beta p}/(\beta p-1)\), with \(R=1\) and \(\beta p=3/2\), gives \(1^{-1/2}/(1/2)=2\), the same integral.

Combining the Local and Tail Tests

For a function on all of \(\mathbb R\), passing one test does not guarantee \(L^p\) membership: the integral must be finite both near zero and in the tails. More precisely, suppose a measurable function has the form \(f(x)=|x|^{-\alpha}\) for \(0<|x|<1\), and \(f(x)=|x|^{-\beta}\) for \(|x|\ge1\), with any finite assigned value at zero. Then for \(1\le p<\infty\), it belongs to \(L^p(\mathbb R,m)\) exactly when \(\alpha p<1\) and \(\beta p>1\).

To justify that these two conditions suffice, the local theorem gives a finite integral on \((-1,1)\), while the tail theorem gives finite integrals on \((1,\infty)\) and \((-\infty,-1)\); reflection \(x\mapsto -x\) makes the two tail integrals equal. Their sum is finite. Conversely, if either condition fails, the integral on the corresponding subset already diverges, so the integral on \(\mathbb R\), which is at least that subset's integral by monotonicity, diverges as well.

Worked Example: One Profile with Both Behaviors

Define \(f(0)=0\), let \(f(x)=|x|^{-1/4}\) for \(0<|x|<1\), and let \(f(x)=|x|^{-3/4}\) for \(|x|\ge1\). Determine whether \(f\in L^2(\mathbb R,m)\), and compute its squared norm.

Near zero, \(\alpha p=(1/4)\cdot2=1/2<1\). The integral over \((-1,1)\) is \(2/(1-1/2)=4\). On the positive tail, \(\beta p=(3/4)\cdot2=3/2>1\), and the integral from \(1\) to infinity is \(1^{1-3/2}/(3/2-1)=2\). The negative tail contributes another \(2\), because the profile depends only on \(|x|\). Therefore

\int_{\mathbb R}|f|^2\,dm=4+2+2=8, \qquad \|f\|_2=\sqrt8=2\sqrt2.

Thus \(f\in L^2(\mathbb R,m)\). Check: the local integral is \(2\int_0^1x^{-1/2}\,dx=2\cdot2=4\), and each tail integral is \(\int_1^\infty x^{-3/2}\,dx=2\); these independent component calculations sum to \(8\).

The Boundary Case and a Common Mistake

A frequent error is to replace the strict threshold \(\alpha p<1\) by \(\alpha p\le1\), perhaps because the power \(x^{-1}\) has an elementary antiderivative. But the antiderivative itself shows why equality fails: for every \(\varepsilon\in(0,1)\),

\int_\varepsilon^1x^{-1}\,dx =\left[\ln x\right]_\varepsilon^1 =0-\ln\varepsilon =-\ln\varepsilon \longrightarrow+\infty \quad\text{as }\varepsilon\downarrow0.

The correction is to test the exponent of \(|f|^p\), not just the exponent of \(f\), and keep the inequality strict. At infinity the boundary is strict in the other direction. For example, if \(h(x)=x^{-1/2}\) and \(p=2\), then \(\beta p=(1/2)\cdot2=1\), and for \(S>1\),

\int_1^S|h(x)|^2\,dx =\int_1^Sx^{-1}\,dx =\ln S \longrightarrow+\infty \quad\text{as }S\to\infty.

Thus a power with exponent exactly at either threshold is not rescued by being finite at each point of the interval: \(L^p\) membership concerns the integral over the whole set. The two tests are useful precisely because they isolate the obstruction. A compact region bounded away from zero contributes a finite integral for these power functions; only the singular endpoint and the unbounded tail require separate tests.

Check Your Understanding

For each question, compute the exponent of the power after taking the \(p\)-th power, and use the appropriate endpoint test.

For which \(\alpha\in\mathbb R\) does \(x\mapsto |x|^{-\alpha}\) belong to \(L^4((-2,2),m)\)?
Determine whether \(x\mapsto x^{-2/5}\) on \((1,\infty)\) belongs to \(L^2\), and evaluate the integral if it is finite.
For \(g(x)=x^{-1/5}\) on \((0,1)\), compute \(\int_0^1|g|^3\,dm\) and decide whether \(g\in L^3\).
State the two necessary and sufficient exponent conditions for the piecewise profile with power \(|x|^{-\alpha}\) near zero and \(|x|^{-\beta}\) on both tails to belong to \(L^p(\mathbb R,m)\), for finite \(p\ge1\).
Show directly using truncated integrals why \(x^{-1}\) is not integrable on \((1,\infty)\).