Why the Order of Steps Matters
Integration by parts is an identity for a finite interval. An improper integral, by contrast, is defined through a limit of integrals over finite intervals. To combine these ideas safely, first use integration by parts where its hypotheses apply, and only then take the limit that defines the improper integral.
This order matters because an endpoint at infinity is not an ordinary endpoint, and an integrand that becomes unbounded at a finite endpoint may not be defined there. In either situation, a boundary term must be evaluated as a limit. It cannot simply be omitted or assigned a value by inspection.
Integration by Parts on a Finite Interval
If \(u\) and \(v\) are differentiable on a finite interval \([a,b]\), with the derivatives and products involved continuous there, the product rule gives \((uv)'=u'v+uv'\). Integrating both sides over \([a,b]\) and using the Fundamental Theorem of Calculus yields the definite integration-by-parts formula.
The notation \(\bigl[u(x)v(x)\bigr]_a^b\) means \(u(b)v(b)-u(a)v(a)\). For an improper integral, the formula is first applied only where the interval is finite and the functions meet the required conditions.
For example, if \(f\) is unbounded at \(a\) but continuous on \((a,b]\), then \(\int_a^b f(x)\,dx\) means \(\lim_{c\to a^+}\int_c^b f(x)\,dx\), when that limit exists and is finite. Integration by parts must be performed on \([c,b]\) before taking \(c\to a^+\).
A General Limiting Procedure
Suppose the upper endpoint is infinite and the integrand is a product that suggests \(u\) and \(dv\). For finite \(b>a\), apply integration by parts to \(\int_a^b u\,dv\). The improper integral exists precisely when the resulting complete expression has a finite limit as \(b\to\infty\); the calculation does not license dropping a term merely because it looks small.
Replace the infinite or singular endpoint with a finite endpoint, such as \(b\) or \(c\).
Keep the boundary expression with both endpoint values.
Evaluate the limit of the boundary term and the remaining integral as required.
A finite limit establishes convergence and gives the value. If no finite limit exists, the improper integral diverges.
The finite-interval identity is justified by the product rule and the Fundamental Theorem of Calculus. Taking a limit afterward preserves that identity whenever the limit of the complete right-hand side exists. This is why the boundary term is part of the argument, not an optional correction.
Worked Example: A Logarithm over an Infinite Interval
Worked Example: Evaluate \(\int_1^\infty \frac{\ln x}{x^2}\,dx\)
By definition, evaluate the limit of \(\int_1^b \frac{\ln x}{x^2}\,dx\) as \(b\to\infty\). Choose \(u=\ln x\) and \(dv=x^{-2}\,dx\). Then \(du=\frac{1}{x}\,dx\), and \(v=-\frac{1}{x}\).
Apply integration by parts on \([1,b]\):
Since \(\ln 1=0\), the boundary term is \(-\frac{\ln b}{b}\). The remaining integral is \(\int_1^b x^{-2}\,dx=[-1/x]_1^b=1-\frac{1}{b}\). Therefore,
As \(b\to\infty\), \(\frac{1}{b}\to0\) and \(\frac{\ln b}{b}\to0\). Thus the improper integral converges, and
Check: An antiderivative is \(-\frac{\ln x+1}{x}\), since
Evaluating this antiderivative from \(1\) to \(b\) gives \(-\frac{\ln b+1}{b}-(-1)=1-\frac{\ln b+1}{b}\), the same finite-interval expression.
Worked Example: A Logarithmic Singularity at a Finite Endpoint
Worked Example: Evaluate \(\int_0^1 \ln x\,dx\)
The function \(\ln x\) is unbounded as \(x\to0^+\), so write the integral as \(\lim_{a\to0^+}\int_a^1\ln x\,dx\). On \([a,1]\), take \(u=\ln x\) and \(dv=dx\). Then \(du=\frac{1}{x}\,dx\) and \(v=x\).
To evaluate the endpoint term, write \(a\ln a=\frac{\ln a}{1/a}\). By L’Hôpital’s Rule, \(\lim_{a\to0^+}\frac{\ln a}{1/a}=\lim_{a\to0^+}\frac{1/a}{-1/a^2}=\lim_{a\to0^+}(-a)=0\). Hence \(-a\ln a\to0\), and \(1-a\to1\). Therefore,
Check: An antiderivative is \(x\ln x-x\), because \(\frac{d}{dx}(x\ln x-x)=\ln x+1-1=\ln x\). Its truncated value is \([x\ln x-x]_a^1=-1-a\ln a+a\), which tends to \(-1\), as obtained above.
Worked Example: An Exponential Product
Worked Example: Evaluate \(\int_1^\infty xe^{-x}\,dx\)
Use the definition \(\int_1^\infty xe^{-x}\,dx=\lim_{b\to\infty}\int_1^b xe^{-x}\,dx\). On the finite interval, choose \(u=x\) and \(dv=e^{-x}\,dx\). Then \(du=dx\) and \(v=-e^{-x}\).
Since \((b+1)e^{-b}\to0\) as \(b\to\infty\), the integral converges and its value is \(\frac{2}{e}\).
Check: An antiderivative is \(-(x+1)e^{-x}\). Differentiating gives \(-e^{-x}+(x+1)e^{-x}=xe^{-x}\). Its truncated value is \(-(b+1)e^{-b}-(-2e^{-1})=\frac{2}{e}-(b+1)e^{-b}\), matching the calculation.
Boundary Terms Cannot Be Guessed Away
A common error is to keep the remaining integral after integration by parts but forget to evaluate the boundary expression at both endpoints. For the preceding example, the mistaken calculation might be:
This drops \([ -xe^{-x}]_1^\infty\). The upper-end contribution is \(0\), but the lower-end contribution is \(-(-1/e)=1/e\). The full calculation is therefore \(\frac{1}{e}+\frac{1}{e}=\frac{2}{e}\). A boundary term that vanishes at infinity can still have a nonzero value at the finite endpoint.
It is also unsafe to conclude that an improper integral converges merely because one part of an integration-by-parts expression has a finite limit. The limit must be taken for the full identity obtained on the truncated interval. For instance, if a boundary term and a remaining integral both fail to have finite limits separately, their combination cannot be assessed by discarding either one; the original truncated integral is governed by their complete sum.
When This Method Is Useful
Integration by parts is especially useful when a product contains a factor that becomes simpler after differentiation and another factor that can be integrated directly. In improper integrals, it can expose a boundary term whose limit is easy to evaluate and leave a simpler integral. The method still requires careful attention to the type of improper endpoint: an infinite limit and a finite singular endpoint each require their own one-sided limiting process.
For an interval with an interior singularity, split the integral at that point and treat the two one-sided integrals separately. Do not use integration by parts across a point where the functions fail to satisfy the finite-interval conditions. Likewise, if a limit is not finite, the corresponding improper integral does not converge simply because the formal integration-by-parts identity can be written on truncated intervals.
Check Your Understanding
For each question, make the finite-interval step explicit before taking any improper limit.
- For \(\int_1^\infty \frac{\ln x}{x^3}\,dx\), state the truncated integral and the integration-by-parts choices \(u\) and \(dv\).
- When evaluating \(\int_0^1\ln x\,dx\), why must the lower endpoint be replaced by a positive number before applying integration by parts?
- In \(\int_1^\infty xe^{-x}\,dx\), what is the value of the boundary contribution at \(x=1\), and why does it not vanish?
- What does it mean for an improper integral to converge?
- Explain why an integration-by-parts formula on an infinite interval must be obtained by first applying the formula on finite intervals.
