Square-Root And Higher-Root Singularities

Why a Blow-Up Can Still Have a Finite Integral

A function may become unbounded near an endpoint and still have a finite integral over a finite interval. The key is how quickly the function grows as the endpoint is approached. Reciprocal square-root and higher-root powers provide simple, exact examples: their values grow without bound, but their accumulated area can remain finite.

For instance, $1/\sqrt{x}$ is unbounded as $x$ approaches zero from the right. The expression cannot be evaluated at zero, so the integral from zero must be defined using a limit. That limit exists and has a finite value. By contrast, $1/x$ also grows without bound there, but its corresponding limit diverges.

Definition: If $f$ is continuous on $(0,b]$ but unbounded near $0$, its endpoint integral is defined by $$ \int_0^b f(x)\,dx = \lim_{\varepsilon\to 0^+}\int_\varepsilon^b f(x)\,dx, $$ provided the limit exists as a finite real number. If the limit is infinite or does not exist, the improper integral diverges.

The lower endpoint is not substituted into an antiderivative directly. Instead, first integrate over the ordinary interval $[\varepsilon,b]$, where $\varepsilon>0$, and then take the one-sided limit. This is the method used throughout the examples below.

The Antiderivative Behind Root Singularities

For $p\ne 1$, the power rule gives an antiderivative of $x^{-p}$ on $x>0$:

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

When $0\le p<1$, the exponent $1-p$ is positive, so $x^{1-p}$ tends to zero as $x\to0^+$. This is why the endpoint limit can be finite. In particular, a reciprocal $n$-th root has the form $x^{-1/n}$. For an integer $n>1$, its exponent $1/n$ is less than one, so it falls on the integrable side of the power threshold. The same reasoning applies to $x^{-m/n}$ when $m<n$.

The result can be calculated directly, and its justification is exactly the endpoint limit in the definition.

Theorem: If $b>0$ and $p<1$, then $$ \int_0^b x^{-p}\,dx = \frac{b^{1-p}}{1-p}. $$ In particular, this value is finite for reciprocal-root powers $p=1/n$ with $n>1$, and for powers such as $p=2/3$.

Why the Root-Power Formula Works

To establish the formula, take $\varepsilon>0$. The integrand is continuous on $[\varepsilon,b]$, so the power rule and evaluation at the two ordinary endpoints give

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

Since $p<1$, we have $1-p>0$, and therefore $\varepsilon^{1-p}\to0$ as $\varepsilon\to0^+$. Taking the limit yields

\lim_{\varepsilon\to0^+} \frac{b^{1-p}-\varepsilon^{1-p}}{1-p} = \frac{b^{1-p}}{1-p}.

This proves the stated value from the definition of the improper integral. The proof also shows why the condition $p<1$ matters: it is what makes the lower-endpoint term tend to zero.

Worked Example: A Reciprocal Square-Root Singularity

Evaluate $\displaystyle \int_0^9 \frac{1}{\sqrt{x}}\,dx$.

Write the integrand as $x^{-1/2}$, and replace the singular endpoint by a positive cutoff:

\int_0^9 \frac{1}{\sqrt{x}}\,dx = \lim_{\varepsilon\to0^+}\int_\varepsilon^9 x^{-1/2}\,dx.

An antiderivative is $2x^{1/2}=2\sqrt{x}$, because $\frac{d}{dx}(2\sqrt{x})=x^{-1/2}$ for $x>0$. Thus

\begin{aligned} \int_0^9 \frac{1}{\sqrt{x}}\,dx &= \lim_{\varepsilon\to0^+} \left[2\sqrt{x}\right]_\varepsilon^9\\ &= \lim_{\varepsilon\to0^+}(2\sqrt{9}-2\sqrt{\varepsilon})\\ &= 6-0\\ &=6. \end{aligned}

Check: differentiating $2\sqrt{x}$ gives $1/\sqrt{x}$, and its upper-endpoint value is $2\sqrt{9}=6$; its lower-endpoint limit is zero. The computed improper integral is therefore $6$.

Higher-Root Powers

A reciprocal higher root can also be written as a fractional power. For example, the reciprocal cube root is $x^{-1/3}$, while $1/x^{2/3}$ is the reciprocal of the square of the cube root. The antiderivative rule still applies, as long as the exponent $p$ is not one. For $p=2/3$, an antiderivative is $3x^{1/3}$, since differentiating gives $x^{-2/3}$.

Thus, the word “root” does not by itself determine whether the integral is finite. What matters is the exponent of the resulting power. The examples illustrate both reciprocal roots and a fractional power of a root.

Worked Example: A Reciprocal Cube-Root Power

Evaluate $\displaystyle \int_0^8 \frac{2}{x^{2/3}}\,dx$.

Use the cutoff definition. For $x>0$, $2/x^{2/3}=2x^{-2/3}$, whose antiderivative is $6x^{1/3}$:

\begin{aligned} \int_0^8 \frac{2}{x^{2/3}}\,dx &= \lim_{\varepsilon\to0^+}\int_\varepsilon^8 2x^{-2/3}\,dx\\ &= \lim_{\varepsilon\to0^+} \left[6x^{1/3}\right]_\varepsilon^8\\ &= \lim_{\varepsilon\to0^+} \left(6\sqrt[3]{8}-6\sqrt[3]{\varepsilon}\right)\\ &= 6(2)-0\\ &=12. \end{aligned}

Check: $\frac{d}{dx}(6x^{1/3})=6\cdot\frac13x^{-2/3}=2x^{-2/3}$. Also, $6\sqrt[3]{8}=12$ and $6\sqrt[3]{\varepsilon}\to0$, confirming the value $12$.

The same endpoint-limit procedure works when the singular endpoint is a positive number rather than zero. For a shifted power $(x-a)^{-p}$ on $(a,b]$, where $a<b$, the lower limit is taken as $x\to a^+$. An antiderivative can be evaluated for $x>a$, and the resulting lower-endpoint term is a power of $x-a$.

Worked Example: A Shifted Higher-Root Singularity

Evaluate $\displaystyle \int_3^7 \frac{1}{(x-3)^{3/4}}\,dx$.

The integrand is unbounded as $x\to3^+$. Its antiderivative for $x>3$ is $4(x-3)^{1/4}$, so define the integral by a cutoff $\varepsilon>0$ measured from the singular endpoint:

\begin{aligned} \int_3^7 \frac{1}{(x-3)^{3/4}}\,dx &= \lim_{\varepsilon\to0^+} \int_{3+\varepsilon}^7 (x-3)^{-3/4}\,dx\\ &= \lim_{\varepsilon\to0^+} \left[4(x-3)^{1/4}\right]_{3+\varepsilon}^7\\ &= \lim_{\varepsilon\to0^+} \left(4\cdot4^{1/4}-4\varepsilon^{1/4}\right)\\ &= 4\sqrt{2}-0\\ &=4\sqrt{2}. \end{aligned}

Here $4^{1/4}=\sqrt{2}$, because $(\sqrt{2})^4=4$. Check: differentiating $4(x-3)^{1/4}$ gives $4\cdot\frac14(x-3)^{-3/4}=(x-3)^{-3/4}$. Its upper value is $4\sqrt{2}$, and its lower-endpoint limit is zero, as required.

Worked Example: A Reciprocal Fourth Root

Evaluate $\displaystyle \int_0^{16} \frac{1}{x^{1/4}}\,dx$.

For $x>0$, the integrand is $x^{-1/4}$, with antiderivative $\frac43x^{3/4}$. Apply the endpoint definition:

\begin{aligned} \int_0^{16} \frac{1}{x^{1/4}}\,dx &= \lim_{\varepsilon\to0^+}\int_\varepsilon^{16}x^{-1/4}\,dx\\ &= \lim_{\varepsilon\to0^+} \left[\frac43x^{3/4}\right]_\varepsilon^{16}\\ &= \frac43(16^{3/4})-\lim_{\varepsilon\to0^+}\frac43\varepsilon^{3/4}\\ &= \frac43(8)-0\\ &=\frac{32}{3}. \end{aligned}

To verify the exact upper value, $16^{1/4}=2$, so $16^{3/4}=2^3=8$. Differentiating $\frac43x^{3/4}$ gives $\frac43\cdot\frac34x^{-1/4}=x^{-1/4}$. The lower-endpoint term tends to zero, confirming $\frac{32}{3}$.

Why the Reciprocal First Power Is Different

The power-rule antiderivative used above requires $p\ne1$. At the boundary value $p=1$, it cannot be used: the denominator $1-p$ would be zero. The antiderivative of $1/x$ for $x>0$ is $\ln x$. Applying the cutoff definition on $(0,1]$ gives

\begin{aligned} \int_0^1 \frac{1}{x}\,dx &= \lim_{\varepsilon\to0^+}\int_\varepsilon^1\frac{1}{x}\,dx\\ &= \lim_{\varepsilon\to0^+}(\ln 1-\ln\varepsilon)\\ &=+\infty. \end{aligned}

Thus, this improper integral diverges. The contrast is not that $1/x$ is unbounded while the reciprocal-root powers are bounded; all of these integrands are unbounded near zero. The difference is that the power $x^{-1}$ does not produce a lower-endpoint term tending to zero.

Takeaway: A reciprocal root $x^{-1/n}$, for $n>1$, has a finite integral at zero because its power is less than one. The reciprocal first power $x^{-1}$ is exactly at the threshold and has a divergent integral there.

A Common Mistake: Using the Power Rule at the Threshold

A tempting but invalid calculation is to treat $1/x$ as $x^{-1}$, apply the formula $\int x^{-p}\,dx=x^{1-p}/(1-p)$ at $p=1$, and then declare the result zero because the numerator involves $x^0$. This formula explicitly requires $p\ne1$; at $p=1$, its denominator is zero, so it gives no antiderivative at all.

The correction is to use the correct antiderivative and keep the cutoff until taking the limit:

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

There is a related error even for an integrable root: writing $\left[2\sqrt{x}\right]_0^9$ as if the original integrand were defined at zero. The numerical result happens to agree, but the reasoning skips the definition of the improper integral. The rigorous calculation first evaluates from $\varepsilon$ to $9$, then takes $\varepsilon\to0^+$.

How to Decide and Evaluate

For a power singularity at a finite endpoint, identify the exponent before calculating. If it is below one, the endpoint term in the power antiderivative tends to zero and the integral is finite. If it equals one, use the logarithmic antiderivative and test its limit; for $1/x$, that limit diverges. For an integrable root, the calculation then follows a consistent sequence:

Replace the singular endpoint with a cutoff.
Write the improper integral as a one-sided limit of integrals over intervals where the integrand is continuous.

Find and check an antiderivative.
Differentiate it on the open interval where the integrand is defined.

Evaluate at ordinary endpoints.
Keep the cutoff in the expression rather than substituting the singular endpoint.

Take the one-sided limit.
Decide whether the lower-endpoint term tends to zero, a finite nonzero value, or infinity.

State convergence and the value.
A finite limit means the improper integral converges; an infinite limit means it diverges.

These integrals matter because they show that unboundedness alone does not decide whether an improper integral is finite. The endpoint limit, not merely the appearance of the integrand, determines convergence. Reciprocal-root powers are especially useful examples: the antiderivative is explicit, the limiting step is visible, and the threshold case $1/x$ makes clear why the exponent condition cannot be ignored.

Check Your Understanding

For each question, use an endpoint limit when the integrand is unbounded.

Evaluate $\displaystyle \int_0^4 x^{-1/2}\,dx$, showing the cutoff limit.
Does $\displaystyle \int_0^1 x^{-3/4}\,dx$ converge? If so, find its value.
Find an antiderivative of $x^{-2/3}$ and use it to evaluate $\displaystyle \int_0^1 x^{-2/3}\,dx$.
Why is the power-rule formula for $\int x^{-p}\,dx$ invalid when $p=1$?
Evaluate $\displaystyle \int_0^1 \frac{1}{x}\,dx$ as an improper integral and state whether it converges.