Evaluating Endpoint Integrals By Antiderivative Limits

Why an Antiderivative Must Be Paired With a Limit

An integrand may become unbounded at an endpoint and still have a finite integral. For instance, $1/\sqrt{x}$ is unbounded as $x$ approaches zero from the right. The usual Fundamental Theorem of Calculus cannot be applied by simply substituting zero into an antiderivative, because the integrand is not defined at that endpoint.

The remedy is to stay inside the interval where the integrand is defined. First integrate from a nearby endpoint, where the integral is proper. Then take the one-sided limit as that nearby endpoint approaches the singular endpoint. This gives both a way to evaluate the integral and a test for whether it converges.

Key idea: For an endpoint-singular integral, find an antiderivative on the open interval and evaluate the integral using the limit of that antiderivative at the endpoint. Do not substitute into the antiderivative at the singular endpoint unless that substitution is justified by a limit.

The Definition and Evaluation Rule

Suppose $f$ is continuous on $(a,b]$, but may be unbounded or undefined at $a$. The improper integral at the left endpoint is defined by the one-sided limit

\int_a^b f(x)\,dx = \lim_{t\to a^+}\int_t^b f(x)\,dx,

provided this limit exists as a finite real number. If it does, the improper integral converges to that number. If the limit is infinite or does not exist, the integral diverges. This is the one-sided-limit definition of an endpoint integral.

Now suppose $F$ is an antiderivative of $f$ throughout $(a,b]$, meaning $F'(x)=f(x)$ there. For each $t$ strictly between $a$ and $b$, the ordinary Fundamental Theorem of Calculus applies on the closed interval $[t,b]$. Thus

\int_t^b f(x)\,dx=F(b)-F(t).

Taking the required limit gives the evaluation rule.

Theorem: Let $f$ be continuous on $(a,b]$, and let $F$ be an antiderivative of $f$ on $(a,b]$. If the finite limit $L=\lim_{t\to a^+}F(t)$ exists, then $$ \int_a^b f(x)\,dx=F(b)-L. $$ If $F(t)$ has no finite limit as $t\to a^+$, then the improper integral does not converge to a finite value.

The corresponding rule at a right endpoint is analogous. If $f$ is continuous on $[a,b)$, $F'=f$ on that interval, and $L=\lim_{t\to b^-}F(t)$ is finite, then

\int_a^b f(x)\,dx=L-F(a).

Why the Rule Works

The argument is short, but it explains why an antiderivative can be used even though the original endpoint is singular. The Fundamental Theorem is applied only to intervals that do not include the singular endpoint. The improper integral is then obtained by taking a limit of those ordinary integrals.

Truncate the interval.
Choose $t$ with $a<t<b$, so $f$ is continuous on $[t,b]$.

Apply the Fundamental Theorem.
On this proper interval, $\int_t^b f(x)\,dx=F(b)-F(t)$.

Take the one-sided limit.
Let $t\to a^+$. If $F(t)\to L$ finitely, the integrals tend to $F(b)-L$.

In particular, this derivation shows that the finite-limit condition is not merely a convenient shortcut. The truncated integrals equal $F(b)-F(t)$, so they can approach a finite number only if $F(t)$ approaches a finite limit. A divergent primitive limit means that the defining limit for the improper integral cannot be a finite real number.

Worked Example: A Square-Root Singularity at Zero

Worked Example: Integrating One Over the Square Root of x

Evaluate $\int_0^4 \frac{1}{\sqrt{x}}\,dx$ as an improper integral.

The integrand is unbounded at zero, so begin at a positive number $\varepsilon$, where $0<\varepsilon<4$:

\int_0^4 \frac{1}{\sqrt{x}}\,dx = \lim_{\varepsilon\to0^+}\int_\varepsilon^4 x^{-1/2}\,dx.

An antiderivative on $x>0$ is $F(x)=2\sqrt{x}$, since

F'(x)=2\cdot\frac{1}{2\sqrt{x}}=\frac{1}{\sqrt{x}}.

Apply the Fundamental Theorem on $[\varepsilon,4]$:

\int_\varepsilon^4 x^{-1/2}\,dx = 2\sqrt{4}-2\sqrt{\varepsilon} = 4-2\sqrt{\varepsilon}.

As $\varepsilon\to0^+$, $\sqrt{\varepsilon}\to0$, so the integral converges and equals $4$.

Check: Differentiating $2\sqrt{x}$ gives $1/\sqrt{x}$ on $x>0$, and the truncated integral tends to $4$, as claimed.

Worked Example: The Natural Logarithm Near Zero

Worked Example: Integrating the Natural Logarithm from Zero to One

Evaluate $\int_0^1 \ln(x)\,dx$. The function $\ln(x)$ is defined for $x>0$ and decreases without bound as $x\to0^+$, so write the integral as

\int_0^1 \ln(x)\,dx = \lim_{\varepsilon\to0^+}\int_\varepsilon^1 \ln(x)\,dx.

An antiderivative on $x>0$ is $F(x)=x\ln(x)-x$. Indeed, using the product rule,

F'(x)=\ln(x)+x\cdot\frac{1}{x}-1=\ln(x).

Evaluate the truncated integral:

\int_\varepsilon^1 \ln(x)\,dx = \bigl(1\cdot\ln(1)-1\bigr) - \bigl(\varepsilon\ln(\varepsilon)-\varepsilon\bigr) = -1-\varepsilon\ln(\varepsilon)+\varepsilon.

We need the limit of $\varepsilon\ln(\varepsilon)$. Set $u=1/\varepsilon$. Then $\varepsilon\ln(\varepsilon)=-\ln(u)/u$, and $u\to\infty$. For $u\ge1$,

0\le \ln(u)=\int_1^u \frac{1}{v}\,dv \le \int_1^u \frac{1}{\sqrt{v}}\,dv =2(\sqrt{u}-1).

Dividing by $u$ shows $0\le\ln(u)/u\le2/\sqrt{u}$, whose upper bound tends to zero. Therefore $\varepsilon\ln(\varepsilon)\to0$. Also $\varepsilon\to0$, so the integral converges to $-1$.

Check: The derivative of $x\ln(x)-x$ is $\ln(x)$, and the truncated values approach $-1$. The negative answer is consistent with $\ln(x)<0$ for $0<x<1$.

Worked Example: A Singular Right Endpoint

Worked Example: Integrating Up to a Square-Root Singularity

Evaluate $\int_1^2 \frac{1}{\sqrt{2-x}}\,dx$. This time the integrand is unbounded as $x\to2^-$, so truncate the interval at $r<2$:

\int_1^2 \frac{1}{\sqrt{2-x}}\,dx = \lim_{r\to2^-}\int_1^r \frac{1}{\sqrt{2-x}}\,dx.

An antiderivative on $x<2$ is $F(x)=-2\sqrt{2-x}$. By the chain rule,

F'(x) = -2\cdot\frac{1}{2\sqrt{2-x}}\cdot(-1) = \frac{1}{\sqrt{2-x}}.

Thus

\int_1^r \frac{1}{\sqrt{2-x}}\,dx = -2\sqrt{2-r}-\bigl(-2\sqrt{2-1}\bigr) = 2-2\sqrt{2-r}.

As $r\to2^-$, $2-r\to0^+$, so the final term tends to zero. The integral converges to $2$.

Check: The derivative of the proposed antiderivative is the given integrand, and the truncated integral tends to $2$.

When the Antiderivative Limit Does Not Converge

The same procedure can identify divergence. For example, on $x>0$, an antiderivative of $1/x$ is $\ln(x)$. Truncating at $\varepsilon>0$ gives

\int_\varepsilon^1 \frac{1}{x}\,dx = \ln(1)-\ln(\varepsilon) = -\ln(\varepsilon).

As $\varepsilon\to0^+$, $-\ln(\varepsilon)\to+\infty$. The defining limit is not finite, so $\int_0^1 \frac{1}{x}\,dx$ diverges. This example illustrates the procedure; it does not replace the required limit test with a guess based only on how the integrand behaves.

A Common Mistake: Substituting at the Singular Endpoint

A tempting but invalid step in the logarithm example is to write $F(0)=0$ for $F(x)=x\ln(x)-x$, on the grounds that $x$ is zero. But $\ln(0)$ is undefined, so the expression $0\cdot\ln(0)$ is not a valid substitution. The correction is to evaluate at a positive $\varepsilon$ and take a limit:

\lim_{\varepsilon\to0^+} \bigl(\varepsilon\ln(\varepsilon)-\varepsilon\bigr) = 0.

The limit happens to be finite, but it must be established; it cannot be inferred by directly substituting zero into an expression that is undefined there. More generally, an antiderivative may fail to have a finite endpoint limit, as the $1/x$ example demonstrates. Always check the limit that the definition of the improper integral requires.

A Reliable Evaluation Routine

For a singularity at the left endpoint, use the following routine. For a singularity at the right endpoint, reverse the direction of the one-sided limit and evaluate the corresponding truncated integral.

Identify which endpoint is singular and write the integral as a one-sided limit of proper integrals.
Find an antiderivative on the interval where the integrand is defined, and verify it by differentiation.
Apply the Fundamental Theorem only to the truncated, proper interval.
Take the required one-sided limit of the resulting endpoint expression.
State whether the integral converges and give its value, or state that it diverges.

The important separation is between finding an antiderivative and evaluating the improper integral. An antiderivative supplies the truncated values; the endpoint limit determines whether those values approach a finite answer.

Check Your Understanding

For each question, first identify the one-sided limit that defines the integral.

For $\int_0^9 \frac{1}{\sqrt{x}}\,dx$, what antiderivative can be used, and what is the value of the improper integral?
Why is it necessary to truncate the interval before applying the Fundamental Theorem to $\int_0^1\ln(x)\,dx$?
What limit of an antiderivative determines convergence at a left endpoint $a$?
What does the limit of $\int_\varepsilon^1 \frac{1}{x}\,dx$ show as $\varepsilon\to0^+$?
For a singularity at the right endpoint $b$, from which direction must the truncation point approach $b$?