Why the Split Comes Before the Substitution
An interior singularity divides an interval into two sides, and each side has its own limiting behavior. A change of variables can make a singularity easier to analyze, but it does not remove the need to check those two limits separately. The reliable procedure is to split at the singular point, substitute on each truncated interval where the integrand is regular, and only then take the one-sided limits.
This order matters especially when a substitution sends the singular point to a boundary such as \(u=0\) or to an infinite limit. The transformed integral must retain the correct direction and value of each limit. If one side diverges, the ordinary improper integral across the point does not converge, even if expressions from the two sides appear capable of cancelling.
The continuity assumptions at the outer endpoints ensure that the truncated integrals in this definition are proper. If \(f\) is also singular at \(a\) or \(b\), those endpoint singularities require additional improper limits; they are not covered by the displayed definition alone.
The Change-of-Variables Rule on Each Side
For an ordinary proper integral, substitution replaces \(x\) by a differentiable, monotone function of a new variable and includes the derivative factor. At an interior singularity, apply that rule first to intervals whose endpoints stay away from the singular point. Taking a limit comes afterward.
The theorem is an application of the proper change-of-variables rule, not a license to substitute directly through the singularity. On any truncated interval that avoids \(c\), the integrand and substitution are regular, so the proper rule applies. Taking the limit of the resulting equality preserves the relation between the two integrals. This justifies the method without assuming that either improper limit converges.
Write the integral as a left-hand limit and a right-hand limit.
Choose an endpoint \(r<c\) for the left integral or \(s>c\) for the right integral, so that the interval avoids the singularity.
Transform the integrand, differential, and both bounds on that truncated interval.
Check the left and right limits separately before adding them.
Worked Example: A Root Singularity on Both Sides
Worked Example: Substituting at an Integrable Root Singularity
Evaluate the improper integral $$ \int_0^2 \frac{1}{\sqrt{|x-1|}}\,dx. $$ The integrand is unbounded at \(x=1\), so split the integral there. For \(0\le x<1\), \(|x-1|=1-x\); for \(1<x\le2\), \(|x-1|=x-1\). Thus $$ \int_0^2 \frac{1}{\sqrt{|x-1|}}\,dx = \lim_{r\to1^-}\int_0^r\frac{dx}{\sqrt{1-x}} + \lim_{s\to1^+}\int_s^2\frac{dx}{\sqrt{x-1}}. $$
For the left side, use \(x=1-u^2\), with \(u>0\). Then \(dx=-2u\,du\) and \(\sqrt{1-x}=u\). When \(x=0\), \(u=1\); when \(x=r\), \(u=\sqrt{1-r}\). Therefore the truncated integral is $$ \int_0^r\frac{dx}{\sqrt{1-x}} = \int_1^{\sqrt{1-r}}\frac{-2u}{u}\,du = \int_{\sqrt{1-r}}^1 2\,du = 2-2\sqrt{1-r}. $$
As \(r\to1^-\), \(\sqrt{1-r}\to0\), so the left-hand limit is \(2\).
For the right side, use \(x=1+u^2\), with \(u>0\). Now \(dx=2u\,du\) and \(\sqrt{x-1}=u\). When \(x=s\), \(u=\sqrt{s-1}\); when \(x=2\), \(u=1\). Hence $$ \int_s^2\frac{dx}{\sqrt{x-1}} = \int_{\sqrt{s-1}}^1\frac{2u}{u}\,du = 2-2\sqrt{s-1}. $$
As \(s\to1^+\), this limit is also \(2\). The two finite one-sided values can therefore be added, giving $$ \int_0^2 \frac{1}{\sqrt{|x-1|}}\,dx=2+2=4. $$
Check: On the left, an antiderivative is \(-2\sqrt{1-x}\), whose derivative is \(1/\sqrt{1-x}\). Its truncated value from \(0\) to \(r\) is \(-2\sqrt{1-r}+2\), as found above. On the right, an antiderivative is \(2\sqrt{x-1}\), whose derivative is \(1/\sqrt{x-1}\). Its value from \(s\) to \(2\) is \(2-2\sqrt{s-1}\), also matching the substitution. Both sides tend to \(2\), confirming the total \(4\).
Rational Denominators: Track the Limits, Not Just the Formula
A substitution involving a rational expression may send an approach to the singular point to an unbounded parameter value. That is acceptable: it simply changes how the one-sided limit is expressed. The transformed bounds must still be attached to the correct side and kept in the correct order.
Worked Example: A Simple Pole with Opposite One-Sided Divergences
Consider $$ \int_0^2 \frac{1}{x-1}\,dx. $$ The denominator vanishes at \(x=1\), so split first: $$ \lim_{r\to1^-}\int_0^r\frac{dx}{x-1} + \lim_{s\to1^+}\int_s^2\frac{dx}{x-1}. $$
For the left integral, let \(u=1-x\). Then \(dx=-du\), \(x-1=-u\), and \(dx/(x-1)=du/u\). The bounds are \(u=1\) at \(x=0\), and \(u=1-r\) at \(x=r\). Thus $$ \int_0^r\frac{dx}{x-1} = \int_1^{1-r}\frac{du}{u} = \ln(1-r). $$ As \(r\to1^-\), this tends to \(-\infty\).
For the right integral, let \(u=x-1\). Then \(dx=du\) and \(x-1=u\). The bounds are \(u=s-1\) and \(u=1\), so $$ \int_s^2\frac{dx}{x-1} = \int_{s-1}^1\frac{du}{u} = -\ln(s-1). $$ As \(s\to1^+\), this tends to \(+\infty\). Since neither one-sided limit is finite, the ordinary improper integral diverges.
Check: An antiderivative away from \(x=1\) is \(\ln|x-1|\), whose derivative is \(1/(x-1)\). The left truncated value is \(\ln(1-r)-\ln(1)=\ln(1-r)\); the right truncated value is \(\ln(1)-\ln(s-1)=-\ln(s-1)\). These agree with the substitutions and confirm the divergent limits.
The opposite signs of these divergences do not make the ordinary improper integral converge. Its definition requires two finite one-sided limits. Combining the truncated integrals before taking their limits would be a different procedure, and cannot replace that requirement.
When a Rational Substitution Sends the Singularity to Infinity
Worked Example: A Reciprocal Substitution at a Double Pole
Examine the one-sided behavior of $$ \int_0^2\frac{1}{(x-1)^2}\,dx. $$ Again, \(x=1\) is an interior singularity, so consider the left and right integrals separately. On the left, set \(u=1/(1-x)\), so \(x=1-1/u\), \(dx=du/u^2\), and \((x-1)^2=1/u^2\). For the truncated interval from \(x=0\) to \(x=1-\varepsilon\), the bounds are \(u=1\) and \(u=1/\varepsilon\). Hence $$ \int_0^{1-\varepsilon}\frac{dx}{(x-1)^2} = \int_1^{1/\varepsilon} 1\,du = \frac{1}{\varepsilon}-1. $$ As \(\varepsilon\to0^+\), this tends to \(+\infty\).
On the right, set \(u=1/(x-1)\), so \(x=1+1/u\), \(dx=-du/u^2\), and \((x-1)^2=1/u^2\). For \(x\) from \(1+\varepsilon\) to \(2\), \(u\) runs from \(1/\varepsilon\) down to \(1\). Therefore $$ \int_{1+\varepsilon}^2\frac{dx}{(x-1)^2} = \int_{1/\varepsilon}^1 (-1)\,du = \frac{1}{\varepsilon}-1. $$ This also tends to \(+\infty\). The ordinary improper integral diverges.
Check: An antiderivative is \(-1/(x-1)\), since its derivative is \(1/(x-1)^2\). On the left, its value from \(0\) to \(1-\varepsilon\) is \(1/\varepsilon-1\). On the right, its value from \(1+\varepsilon\) to \(2\) is \(1/\varepsilon-1\). Both agree with the transformed calculations.
A Common Mistake: Substituting Across the Singular Point
A tempting but incorrect approach is to apply the ordinary substitution formula to the full interval \([0,2]\) in the first example, as though the integrand were continuous throughout. It is not: at \(x=1\), the expression \(1/\sqrt{|x-1|}\) is undefined and unbounded. The proper substitution theorem applies to truncated intervals that avoid that point, not to an interval crossing it.
Another error is to omit the derivative factor. For example, on the right side of the root example, \(x=1+u^2\) gives \(dx=2u\,du\). Replacing \(dx\) by \(du\) would incorrectly transform \(1/\sqrt{x-1}\) into \(1/u\), rather than the correct product \((1/u)(2u)=2\). The factor \(2u\) is essential both to the value and to the behavior near the limit.
Why This Method Matters
Substitution can reveal the local structure of a singularity. A square-root substitution can turn an unbounded integrand into a bounded constant, as in the first example. A reciprocal substitution can turn a pole into an integral over an interval whose endpoint moves to infinity, as in the double-pole example. In each case, the substitution clarifies the limiting behavior; it does not alter the definition of convergence.
Always distinguish three tasks: defining the integral by its one-sided limits, transforming each truncated integral correctly, and deciding whether each transformed limit is finite. Keeping these tasks separate prevents orientation errors and prevents divergent contributions from being mistaken for a convergent improper integral.
Check Your Understanding
For each question, keep the two one-sided limits distinct whenever the interval contains an interior singularity.
- Why must an integral with a singularity at \(c\) be split at \(c\) before applying a substitution?
- In the substitution \(x=1+u^2\), what happens to \(dx\), and what is the transformed lower bound when \(x=s>1\)?
- For \(\int_0^2 1/(x-1)\,dx\), what are the limits of the left and right truncated integrals as their endpoints approach \(1\)?
- Under \(u=1/(x-1)\), what limit does \(u\) approach as \(x\to1^+\)?
- What additional issue must be addressed if the integrand is also singular at an outer endpoint?
