Why an Unbounded Endpoint Needs Special Treatment
A definite integral is usually evaluated on an interval where the integrand is defined and continuous, including at the endpoints. But sometimes a function becomes unbounded as \(x\) approaches a finite endpoint. For example, \(1/\sqrt{x}\) grows without bound as \(x\) approaches \(0\) from the right.
The unbounded value at the endpoint does not, by itself, tell us whether the integral is finite. The right way to investigate is to keep the endpoint just inside the interval, evaluate an ordinary definite integral, and then take a one-sided limit.
Definition at a Finite Endpoint
Suppose \(a<b\), and \(f\) is continuous on \((a,b]\) but may be unbounded or undefined at \(a\). For each \(t\) with \(a<t\le b\), the truncated integral \(\int_t^b f(x)\,dx\) avoids the problematic endpoint. The improper integral at the lower endpoint is defined by taking \(t\) toward \(a\) from the right.
For an unbounded integrand at the upper endpoint, use a variable upper bound instead. Suppose \(f\) is continuous on \([a,b)\), so that \(\int_a^t f(x)\,dx\) is an ordinary integral whenever \(a\le t<b\).
The direction of each limit matters: approach a lower endpoint from the right, and an upper endpoint from the left. The integrals with variable bounds are ordinary definite integrals; the possible difficulty arises when the variable bound approaches the singular endpoint.
Why the Variable-Bound Limit Works
The integral on a truncated interval measures accumulation without including the point where the function is unbounded. As the variable bound approaches the endpoint, these truncated integrals include more and more of the interval. The improper integral is defined to be their limiting value, if there is a finite one.
This definition also explains how to evaluate the integral using an antiderivative. For a lower-endpoint singularity, take an antiderivative \(F\) of \(f\) on \((a,b]\). The Fundamental Theorem of Calculus applies on each nonsingular interval \([t,b]\), giving
The improper integral is then the limit of this expression as \(t\to a^+\). The antiderivative is evaluated at ordinary points first; only afterward is the one-sided limit taken. For an upper-endpoint singularity, the corresponding calculation is \(\int_a^t f(x)\,dx=F(t)-F(a)\), followed by the limit as \(t\to b^-\).
Decide whether the integrand is unbounded or undefined near the lower endpoint, the upper endpoint, or both.
Use \(t\to a^+\) for a lower endpoint \(a\), or \(t\to b^-\) for an upper endpoint \(b\).
Use an antiderivative on the interval where the integrand is defined.
A finite real limit gives the value of the convergent improper integral; failure to have a finite limit means it diverges.
Worked Examples
Worked Example: A Lower-Endpoint Singularity with a Finite Integral
Evaluate \(\int_0^4 \frac{1}{\sqrt{x}}\,dx\) as an improper integral.
The integrand is unbounded as \(x\to0^+\), so replace the lower endpoint by \(t>0\):
An antiderivative of \(x^{-1/2}\) is \(2\sqrt{x}\), since
Evaluate the truncated integral and then take the limit:
The limit is finite, so the improper integral converges and equals \(4\). The derivative check confirms that \(2\sqrt{x}\) is the required antiderivative on \(x>0\).
Worked Example: A Singular Upper Endpoint
Evaluate \(\int_0^2 \frac{1}{\sqrt{2-x}}\,dx\). Here the integrand is unbounded as \(x\to2^-\), so move the upper endpoint inward:
An antiderivative is \(-2\sqrt{2-x}\). Checking by differentiation,
Now evaluate at the ordinary bounds \(0\) and \(t<2\):
The one-sided limit is finite, so the integral converges to \(2\sqrt{2}\). The derivative check verifies the antiderivative throughout \(0\le x<2\), where it is used.
Worked Example: A Lower-Endpoint Singularity That Diverges
Determine whether \(\int_0^1 \frac{1}{x}\,dx\) converges.
Since \(1/x\) is unbounded as \(x\to0^+\), use the lower-endpoint definition:
An antiderivative on \(x>0\) is \(\ln x\), because \(\frac{d}{dx}\ln x=1/x\). Thus
The limit is not a finite real number, so the improper integral diverges. Differentiating \(\ln x\) verifies the antiderivative, but it does not make the limiting value finite.
Convergence Is About the Limit, Not the Endpoint Value
At a singular endpoint, \(f(a)\) or \(f(b)\) may not exist as a finite number. That does not prevent us from defining the integral by a limit. What determines convergence is the behavior of the truncated integrals, not whether the integrand has a finite value at the endpoint.
The first worked example illustrates this distinction: \(1/\sqrt{x}\) becomes unbounded near \(0\), yet the accumulated integral approaches \(4\). In the third example, \(1/x\) is also unbounded near \(0\), but its truncated integrals grow without bound. Unboundedness alone therefore does not decide whether an improper integral converges.
The limit must be finite. A limit of \(+\infty\) or \(-\infty\) is not a convergent improper integral. Nor does a limit that fails to exist as a real number give a convergent integral. In each case, the definition—not a guess based only on the appearance of the integrand—settles the question.
Common Mistake: Substituting the Singular Endpoint Directly
A tempting shortcut is to find an antiderivative and immediately substitute the singular endpoint into it. For \(\int_0^4 x^{-1/2}\,dx\), this might be written incorrectly as
The numerical result happens to be correct, but this calculation skips the defining limit. Direct endpoint substitution is not justified when the original definite integral is improper; the ordinary endpoint formula applies first to a truncated interval. The correct calculation is
This distinction matters especially when an antiderivative has no finite value at the endpoint. For the integral of \(1/x\), \(\ln 0\) is not a finite number, and the limiting calculation shows that the integral diverges. Do not use endpoint substitution to assume convergence or divergence; evaluate the one-sided limit.
A Reliable Checklist
When evaluating an integral with a finite-endpoint singularity, make each part of the definition visible in your work:
- State which endpoint is singular.
- Replace it by a variable bound that stays inside the interval.
- Evaluate the resulting ordinary definite integral.
- Take the required one-sided limit.
- State whether that limit is finite, and therefore whether the integral converges.
- Check the antiderivative by differentiating it.
Writing the limit explicitly is more than a formatting choice: it shows which value is being tested and prevents a singular endpoint from being treated as an ordinary bound. These improper integrals extend definite integration to cases where the integrand is not bounded at an endpoint, while preserving a precise criterion for when the accumulated value is finite.
Check Your Understanding
For each problem, identify the singular endpoint, write the correct one-sided limit, and decide whether the integral converges.
- Evaluate \(\int_0^9 \frac{1}{\sqrt{x}}\,dx\).
- Write the limit definition needed to investigate \(\int_1^3 \frac{1}{\sqrt{3-x}}\,dx\).
- Determine whether \(\int_0^1 x^{-1/3}\,dx\) converges, and find its value if it does.
- Explain why \(\int_0^1 \frac{1}{x}\,dx\) cannot be evaluated by simply substituting \(0\) and \(1\) into \(\ln x\).
- What must be true about the one-sided limit for an improper integral at a finite endpoint to converge?
