Why the Truncation Must Change
Substitution changes more than the expression for the integrand. It also changes the point at which an improper integral is cut off. If the original variable is \(x\) and we set \(x=\phi(t)\), then a truncation \(x=a\) becomes the transformed bound \(t=\phi^{-1}(a)\), when the inverse exists. Keeping the old numerical cutoff after changing variables generally gives a different truncated integral.
This matters because an integral with a singular endpoint is defined by a limit of proper integrals. Substitution is applied to each proper, truncated integral; the resulting transformed integrals are then considered in the corresponding limit. The substitution need not remove the singularity, and it does not make the original improper integral automatically convergent.
Endpoint Integrals and Their Truncations
Suppose \(f\) is continuous on \((c,d]\), but may be unbounded near \(c\). The integral from \(c\) to \(d\) is defined by cutting off a small interval next to \(c\) and taking a one-sided limit.
For each \(a>c\), the truncated integral \(\int_a^d f(x)\,dx\) is an ordinary proper integral, provided \(f\) is continuous on \([a,d]\). The question of convergence concerns the limit as \(a\) approaches \(c\), not an integral that is evaluated directly at the singular endpoint.
A Substitution Rule with the Endpoint Limit Included
Here is a useful setting for substitutions at a lower singular endpoint. Let \(\phi\) be continuous and strictly increasing on \([\alpha,\beta]\), with \(\phi(\alpha)=c\) and \(\phi(\beta)=d\). Assume that for every \(u\) with \(\alpha<u<\beta\), \(\phi\) is continuously differentiable on \([u,\beta]\). In particular, this condition requires continuous differentiability up to the nonsingular endpoint \(\beta\) on every such truncated interval. Suppose also that \(f\) is continuous on \((c,d]\).
The hypotheses ensure that the displayed equality is an application of the substitution rule for proper integrals: on \([u,\beta]\), the map \(\phi\) is continuously differentiable, and the composite integrand is continuous. No differentiability at the singular endpoint \(\alpha\) is required. In particular, one must not assume that the substitution rule applies to the full interval \([\alpha,\beta]\) as a proper-integral substitution.
Choose \(u>\alpha\). The corresponding original cutoff is \(a=\phi(u)>c\).
Use the proper-integral substitution rule on \([u,\beta]\), where the required differentiability holds.
As \(u\to\alpha^+\), continuity and strict increase give \(\phi(u)\to c^+\), so the two sides approach the corresponding improper integrals.
Why the Rule Works
For a fixed \(u>\alpha\), define an antiderivative on the truncated range by \(F(x)=\int_{\phi(u)}^x f(s)\,ds\). The fundamental theorem of calculus gives \(F'(x)=f(x)\) there. The chain rule then gives
Integrating this identity from \(u\) to \(\beta\) yields
This proves the equality for each truncated interval. Since \(a=\phi(u)\) tends to \(c\) as \(u\) tends to \(\alpha\), the equality holds between the corresponding limits whenever those limits exist. This argument does not assert that either limit exists; convergence must still be determined.
Worked Example: A Square-Root Singularity
Worked Example: A Square-Root Singularity
Evaluate \(\int_0^1 x^{-1/2}\,dx\) by setting \(x=t^2\). The integrand is unbounded at \(x=0\), so first truncate at \(x=\varepsilon\), where \(0<\varepsilon<1\). Since \(x=t^2\) and \(t\geq 0\), the matching transformed cutoff is \(t=\sqrt{\varepsilon}\), not \(t=\varepsilon\).
On the proper truncated interval, \(dx=2t\,dt\) and \(x^{-1/2}=(t^2)^{-1/2}=1/t\), since \(t>0\). Therefore
Taking the limit gives
As a check, an antiderivative for \(x>0\) is \(2\sqrt{x}\), and its derivative is \(1/\sqrt{x}\). Its endpoint difference is \(2\sqrt{1}-\lim_{x\to0^+}2\sqrt{x}=2\), agreeing with the truncated calculation.
Worked Example: A Logarithm at the Endpoint
Worked Example: A Logarithm at the Endpoint
Evaluate \(\int_0^1 \frac{\ln x}{\sqrt{x}}\,dx\). Truncate at \(x=\varepsilon\), then use \(x=t^2\). The transformed lower bound is \(t=\sqrt{\varepsilon}\). For \(t>0\), \(\ln(t^2)=2\ln t\), so the truncated integral becomes
An antiderivative of \(4\ln t\) is \(4(t\ln t-t)\), as differentiation gives \(4(\ln t+1-1)=4\ln t\). Thus
As \(t\to0^+\), \(t\ln t\to0\). Therefore both terms involving \(\sqrt{\varepsilon}\) tend to zero, and the improper integral is \(-4\).
To check the result directly, write the corresponding antiderivative in the original variable as \(G(x)=2\sqrt{x}\ln x-4\sqrt{x}\). Differentiating for \(x>0\) gives
Also, \(G(1)=-4\) and \(\lim_{x\to0^+}G(x)=0\), consistent with the limit calculation.
Worked Example: A Higher-Root Substitution
Worked Example: A Higher-Root Substitution
Evaluate \(\int_0^1 \frac{x^{-2/3}}{1+x^{1/3}}\,dx\). The factors involving cube roots suggest setting \(x=t^3\). Truncate at \(x=\varepsilon\), so the corresponding transformed cutoff is \(t=\varepsilon^{1/3}\). On this truncated interval \(t>0\), and \(dx=3t^2\,dt\). The integrand and differential combine as follows:
Hence
As \(\varepsilon\to0^+\), the second logarithm tends to \(\ln 1=0\). The improper integral therefore equals \(3\ln 2\).
For a check, differentiate \(H(x)=3\ln(1+x^{1/3})\) for \(x>0\):
Since \(H(1)=3\ln 2\) and \(\lim_{x\to0^+}H(x)=0\), this confirms the value.
A Common Error: Reusing the Old Cutoff
Under \(x=t^2\), the equality \(x=\varepsilon\) implies \(t=\sqrt{\varepsilon}\) on the positive interval. It does not imply \(t=\varepsilon\). For example, if \(\varepsilon=0.01\), the original truncated integral in the square-root example starts at \(x=0.01\), which maps to \(t=0.1\). The exact transformed truncated integral is
whereas using the old numerical cutoff as the transformed bound gives \(\int_{0.01}^1 2\,dt=1.98\). That is not the same truncated integral. In this particular example both choices of lower bound tend to zero as \(\varepsilon\to0^+\), so the incorrect cutoff happens to give the same final limit. That coincidence does not make the truncated equality valid. The reliable procedure is to solve the substitution equation for the new bound before integrating.
Endpoint Regularity Is Part of the Argument
A substitution may have an unbounded derivative near one endpoint even when it is differentiable at every interior point. The proper-integral substitution rule cannot then be applied to a closed interval containing that endpoint on the strength of interior differentiability alone. Instead, first truncate so that the entire interval used for substitution lies where the map is continuously differentiable, including at its nonsingular endpoint. This is why the theorem above requires \(\phi\) to be continuously differentiable on each \([u,\beta]\), not merely on the open interval \((\alpha,\beta)\).
The power substitutions used in the examples meet that requirement on every truncated interval: \(t\mapsto t^2\) and \(t\mapsto t^3\) are continuously differentiable on the relevant closed intervals. Their derivatives need not be used to justify a proper-integral substitution at the singular endpoint itself. The endpoint is reached only through the limit of truncated integrals.
Choosing and Checking a Substitution
A useful substitution often makes a root or logarithm simpler, but its usefulness does not replace the limit argument. Before evaluating, check the following points:
- Endpoint mapping: Does the substitution send the transformed endpoint to the original singular endpoint?
- Monotonicity: Is the substitution one-to-one on the interval being used, so that the transformed bounds have an unambiguous order?
- Truncated regularity: Is it continuously differentiable on each closed interval after the singular endpoint has been cut off?
- Matching cutoff: What transformed point corresponds exactly to the chosen original cutoff?
- Limit: Does the transformed truncated integral approach a finite value?
These checks keep the algebra and the improper-integral definition connected. A substitution can simplify an integrand dramatically, as in the examples, while the convergence question remains a question about the limit of the truncated integrals.
Check Your Understanding
For each question, keep track of the original cutoff as well as the transformed integrand.
- Under \(x=t^2\) on \(0<x\leq1\), what transformed lower bound corresponds to \(x=\varepsilon\)?
- State the condition that makes \(\int_c^d f(x)\,dx\) an endpoint improper integral with finite value \(L\).
- Why does the substitution theorem apply to each truncated interval in the theorem, but not automatically to the entire interval including the singular endpoint?
- Under \(x=t^3\), what lower bound corresponds to \(x=\varepsilon\), and what is \(dx\) in terms of \(t\)?
- Does simplifying the transformed integrand by itself prove that the original improper integral converges? Explain briefly.
