The Power Singularity Threshold

Why the Exponent Determines Convergence

The function $x^{-p}$ is defined for $x>0$, but its behavior near zero depends on the real exponent $p$. For some exponents it becomes unbounded as $x$ approaches zero, yet its integral can still have a finite value. For others, the accumulated area grows without bound.

The distinction is determined by evaluating the integral with its singular endpoint temporarily cut off. As the cutoff approaches zero, the resulting expression either approaches a finite number or fails to do so. The critical exponent is $p=1$, and it must be treated separately because the antiderivative changes form there.

Definition and the Truncated Integral

For a real number $p$, the integral from zero to one is defined as an endpoint limit. For each $\varepsilon$ with $0<\varepsilon<1$, the truncated integral is an ordinary integral on the interval $[\varepsilon,1]$. The improper integral converges if the limit of these truncated integrals as $\varepsilon$ approaches zero exists and is finite.

Definition: For $p\in\mathbb{R}$, define $$ \int_0^1 x^{-p}\,dx = \lim_{\varepsilon\to 0^+}\int_\varepsilon^1 x^{-p}\,dx, $$ provided this limit exists as a finite real number. If the truncated integrals grow without bound, the improper integral diverges.

This definition focuses on the endpoint $x=0$. For any fixed $\varepsilon>0$, the function $x^{-p}$ is continuous on $[\varepsilon,1]$, so the ordinary antiderivative evaluation applies there. The question is what happens to that exact value as $\varepsilon$ tends to zero.

The Antiderivative and the Exceptional Exponent

When $p\ne 1$, the power rule gives an antiderivative of $x^{-p}$:

\int x^{-p}\,dx = \frac{x^{1-p}}{1-p}+C, \qquad p\ne 1.

The condition $p\ne 1$ matters: the denominator $1-p$ would be zero at $p=1$. In that case, $x^{-p}=x^{-1}$, whose antiderivative is $\ln x$ for $x>0$. Thus we evaluate the truncated integral in two cases.

\int_\varepsilon^1 x^{-p}\,dx = \begin{cases} \dfrac{1-\varepsilon^{1-p}}{1-p}, & p\ne 1,\\[6pt] -\ln\varepsilon, & p=1. \end{cases}

The displayed expressions come from evaluating at both endpoints, not from substituting zero into an antiderivative. The lower endpoint remains $\varepsilon>0$ until after the evaluation, so every expression is defined.

Deriving the Threshold

For $p\ne 1$, the only term in the truncated value whose behavior remains to be determined is $\varepsilon^{1-p}$. Its limit depends on the sign of $1-p$. If $p<1$, that exponent is positive, so the power tends to zero. If $p>1$, the exponent is negative, so the power grows without bound as $\varepsilon$ approaches zero.

When $p<1$.
Because $1-p>0$, $\lim_{\varepsilon\to0^+}\varepsilon^{1-p}=0$. The truncated integral therefore has a finite limit.

When $p>1$.
Because $1-p<0$, write $\varepsilon^{1-p}=1/\varepsilon^{p-1}$. This tends to positive infinity, so the truncated integral tends to positive infinity.

When $p=1$.
The truncated integral is $-\ln\varepsilon$, which tends to positive infinity as $\varepsilon\to0^+$. This case diverges as well.

Theorem: For real $p$, the improper integral $\int_0^1 x^{-p}\,dx$ converges if and only if $p<1$. When it converges, its value is $$ \int_0^1 x^{-p}\,dx=\frac{1}{1-p}. $$ For $p\ge 1$, the truncated integrals tend to positive infinity.

To verify the finite value in the convergent case, take the limit of the exact truncated expression. If $p<1$, then

\lim_{\varepsilon\to0^+}\int_\varepsilon^1 x^{-p}\,dx = \lim_{\varepsilon\to0^+}\frac{1-\varepsilon^{1-p}}{1-p} = \frac{1-0}{1-p} = \frac{1}{1-p}.

For $p>1$, the same expression can be rewritten to make its divergence clear:

\frac{1-\varepsilon^{1-p}}{1-p} = \frac{\varepsilon^{1-p}-1}{p-1} = \frac{1/\varepsilon^{p-1}-1}{p-1} \longrightarrow+\infty.

Here $p-1>0$, and $\varepsilon^{p-1}$ tends to zero, so its reciprocal tends to positive infinity. At $p=1$, the separate expression $-\ln\varepsilon$ also tends to positive infinity. These cases establish both directions of the criterion: every $p<1$ gives a finite limit, and every $p\ge1$ does not.

Worked Examples

Worked Example: A Convergent Square-Root Singularity

Evaluate $\int_0^1 x^{-1/2}\,dx$ as an improper integral. Here $p=\frac12<1$, so the theorem predicts convergence. Begin with a cutoff:

\int_\varepsilon^1 x^{-1/2}\,dx = \left[2x^{1/2}\right]_\varepsilon^1 = 2(1)^{1/2}-2\varepsilon^{1/2} = 2-2\sqrt{\varepsilon}.

As $\varepsilon\to0^+$, $\sqrt{\varepsilon}\to0$, and therefore

\int_0^1 x^{-1/2}\,dx = \lim_{\varepsilon\to0^+}(2-2\sqrt{\varepsilon}) = 2.

Check the antiderivative: $\frac{d}{dx}(2\sqrt{x})=2\cdot\frac12x^{-1/2}=x^{-1/2}$ for $x>0$. Evaluating it at $1$ and $\varepsilon$ gives exactly the truncated value above. The endpoint limit then gives the claimed answer.

Worked Example: The Borderline Case $p=1$

Evaluate $\int_0^1 x^{-1}\,dx$. Since $x^{-1}=1/x$, use the logarithmic antiderivative and retain the cutoff:

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

As $\varepsilon\to0^+$, $\ln\varepsilon\to-\infty$, so $-\ln\varepsilon\to+\infty$. Thus

\int_0^1 \frac{1}{x}\,dx \quad\text{diverges to }+\infty.

Check the antiderivative by differentiation: $\frac{d}{dx}\ln x=1/x=x^{-1}$ for $x>0$. The evaluation at $1$ and $\varepsilon$ is therefore correct, and its limit confirms divergence rather than a finite integral value.

Worked Example: A Power Beyond the Threshold

Determine whether $\int_0^1 x^{-3/2}\,dx$ converges. Here $p=\frac32>1$. For $\varepsilon>0$, an antiderivative is

\frac{x^{1-3/2}}{1-3/2} = \frac{x^{-1/2}}{-1/2} = -2x^{-1/2}.

Evaluate the truncated integral:

\int_\varepsilon^1 x^{-3/2}\,dx = \left[-2x^{-1/2}\right]_\varepsilon^1 = -2(1)^{-1/2}-\bigl(-2\varepsilon^{-1/2}\bigr) = -2+\frac{2}{\sqrt{\varepsilon}}.

As $\varepsilon\to0^+$, $1/\sqrt{\varepsilon}\to+\infty$, so the truncated integral tends to $+\infty$ and the improper integral diverges. Check by differentiation:

\frac{d}{dx}\bigl(-2x^{-1/2}\bigr) = -2\left(-\frac12\right)x^{-3/2} = x^{-3/2}.

The derivative recovers the integrand, and the endpoint evaluation gives the divergent expression used in the limit.

A Common Mistake: Substituting Zero Too Soon

The antiderivative formula for $p\ne1$ is valid on $x>0$, but it does not justify direct substitution of zero when the antiderivative is unbounded there. For example, applying the endpoint notation directly to $x^{-2}$ might lead to the invalid calculation

\left[-\frac1x\right]_0^1 = -1-\left(-\frac10\right).

The expression $-1/0$ is undefined. It is not a legitimate endpoint value, so this calculation does not evaluate the improper integral. The correction is to keep the cutoff positive, evaluate, and only then take a limit:

\int_\varepsilon^1 x^{-2}\,dx = \left[-\frac1x\right]_\varepsilon^1 = -1+\frac1\varepsilon \longrightarrow+\infty.

This gives the correct conclusion: the integral diverges. The same discipline applies at $p=1$, where the power antiderivative formula cannot be used at all; the correct antiderivative is $\ln x$, and its cutoff limit diverges.

How to Use the Criterion

For an integral of this exact form, the exponent alone decides convergence. The size of the singularity at individual points does not by itself settle the question: the truncated area must be evaluated or its limit analyzed. When $p<1$, the finite value is not merely an estimate; it follows from the exact antiderivative limit. When $p\ge1$, the same cutoff calculation identifies unbounded growth.

Exponent	Behavior of the truncated integral	Conclusion
$p<1$	Tends to $1/(1-p)$	Converges
$p=1$	Equals $-\ln\varepsilon$, tending to $+\infty$	Diverges
$p>1$	Grows like a positive multiple of $\varepsilon^{1-p}$	Diverges

The threshold is strict: convergence requires $p<1$, not $p\le1$. The borderline case is not obtained by substituting $p=1$ into $1/(1-p)$; that expression is undefined there. Instead, $p=1$ has its own antiderivative and its own limit calculation. Keeping these cases distinct prevents the most common error in using the power singularity criterion.

Check Your Understanding

Use a positive cutoff $\varepsilon$ before taking any endpoint limit.

For which real values of $p$ does $\int_0^1 x^{-p}\,dx$ converge?
Evaluate the integral when $p=0$, showing the truncated expression and its limit.
What is the antiderivative needed when $p=1$, and what happens to the truncated integral?
For $p>1$, why does $\varepsilon^{1-p}$ grow without bound as $\varepsilon\to0^+$?
Why is direct substitution of $x=0$ into an antiderivative not a valid way to evaluate a singular endpoint integral?

The Power Singularity Threshold

What You'll Learn

Why the Exponent Determines Convergence

Definition and the Truncated Integral

The Antiderivative and the Exceptional Exponent

Deriving the Threshold

Worked Examples

Worked Example: A Convergent Square-Root Singularity

Worked Example: The Borderline Case \(p=1\)

Worked Example: A Power Beyond the Threshold

A Common Mistake: Substituting Zero Too Soon

How to Use the Criterion

Check Your Understanding

Exponent	Behavior of the truncated integral	Conclusion
\(p<1\)	Tends to \(1/(1-p)\)	Converges
\(p=1\)	Equals \(-\ln\varepsilon\), tending to \(+\infty\)	Diverges
\(p>1\)	Grows like a positive multiple of \(\varepsilon^{1-p}\)	Diverges

The Power Singularity Threshold

What You'll Learn

Why the Exponent Determines Convergence

Definition and the Truncated Integral

The Antiderivative and the Exceptional Exponent

Deriving the Threshold

Worked Examples

Worked Example: A Convergent Square-Root Singularity

Worked Example: The Borderline Case \(p=1\)

Worked Example: A Power Beyond the Threshold

A Common Mistake: Substituting Zero Too Soon

How to Use the Criterion

Check Your Understanding

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