Why the Singularities Must Be Separated
An integral over a finite interval can still be improper when the integrand is unbounded. If one singularity occurs at an endpoint and another occurs inside the interval, there is more than one limiting process to examine. Each must be handled on its own.
The key rule is simple: an ordinary improper integral converges only when every required one-sided integral has a finite limit. A finite contribution from one part cannot repair a divergent contribution from another, and opposite infinite contributions cannot be combined to produce ordinary convergence.
The first limit treats the unbounded endpoint. The second approaches the interior singularity from the left, and the third approaches it from the right. The cutoffs \(\varepsilon\), \(\delta\), and \(\eta\) are independent: each limit is evaluated separately. A convenient regular splitting point \(r\) lets the endpoint contribution be isolated from the singularity at \(c\).
Why This Definition Uses Separate Limits
On intervals that do not contain a singularity, ordinary additivity of definite integrals allows us to split an integral at a regular point. The improper version keeps that separation: first form integrals on intervals where the integrand is continuous, then take the one-sided limits. The resulting value, if it exists, is the sum of the finite limits.
The choice of regular splitting point does not change the result. If \(r\) and \(s\) are both between \(a\) and \(c\), then the integrand is continuous on the interval between them, so \(\int_r^s f(x)\,dx\) is finite. Moving the split from \(r\) to \(s\) subtracts that same finite integral from one piece and adds it to the adjacent piece. Thus the total is unchanged. This argument also shows why a regular split cannot remove a divergence at either singularity.
Identify the unbounded endpoint and the interior point where the integrand is unbounded.
Place a point between the endpoint and interior singularity where the integrand is continuous.
Use separate cutoffs for the endpoint and the two sides of the interior singularity.
Only if every limit is finite may the pieces be added to give the ordinary integral.
Worked Example: Both Singularities Are Integrable
Worked Example: An Endpoint and Interior Blow-Up with Finite Contributions
Evaluate the ordinary improper integral $$ \int_0^2\left(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{|x-1|}}\right)\,dx. $$ The endpoint singularity is at \(x=0\), and the interior singularity is at \(x=1\). Choose \(r=\tfrac12\). The required pieces are the limits on \([0,\tfrac12]\), \([\tfrac12,1)\), and \((1,2]\).
For \(0<x<1\), an antiderivative is $$ G(x)=2\sqrt{x}-2\sqrt{1-x}, \qquad G'(x)=\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{1-x}}. $$ Since \(G(\tfrac12)=0\), the endpoint piece is $$ \lim_{\varepsilon\to0^+}\int_\varepsilon^{1/2}\left(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{1-x}}\right)\,dx = \lim_{\varepsilon\to0^+}\bigl(0-G(\varepsilon)\bigr) = \lim_{\varepsilon\to0^+}\left(-2\sqrt{\varepsilon}+2\sqrt{1-\varepsilon}\right) =2. $$
The left-hand contribution at the interior singularity is $$ \lim_{\delta\to0^+}\int_{1/2}^{1-\delta}\left(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{1-x}}\right)\,dx = \lim_{\delta\to0^+}\bigl(G(1-\delta)-G(1/2)\bigr) = \lim_{\delta\to0^+}\left(2\sqrt{1-\delta}-2\sqrt{\delta}\right) =2. $$ For \(x>1\), an antiderivative is $$ H(x)=2\sqrt{x}+2\sqrt{x-1}, \qquad H'(x)=\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{x-1}}. $$ Therefore the right-hand contribution is $$ \lim_{\eta\to0^+}\int_{1+\eta}^{2}\left(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{x-1}}\right)\,dx = \lim_{\eta\to0^+}\left(2\sqrt{2}+2-2\sqrt{1+\eta}-2\sqrt{\eta}\right) =2\sqrt{2}. $$
All three limits are finite, so the ordinary improper integral converges, with value $$ 2+2+2\sqrt{2}=4+2\sqrt{2}. $$ The derivative checks above verify the antiderivatives used on each regular interval. The separate limits also check that neither singularity contributes an infinite value.
What Happens If One Side Diverges?
For the ordinary integral to converge, both one-sided approaches to the interior point must work—not just one of them. Likewise, convergence at the endpoint does not compensate for divergence at the interior point. The next example makes this explicit with an integrand whose pieces are nonnegative, so no cancellation is possible.
Worked Example: A Finite Endpoint Piece but a Divergent Interior Singularity
Consider $$ \int_0^2\left(\frac{1}{\sqrt{x}}+\frac{1}{(x-1)^2}\right)\,dx. $$ Again use \(r=\tfrac12\). Away from \(x=1\), an antiderivative is $$ K(x)=2\sqrt{x}+\frac{1}{1-x}, \qquad K'(x)=\frac{1}{\sqrt{x}}+\frac{1}{(1-x)^2}. $$ Since \((1-x)^2=(x-1)^2\), this derivative is the integrand.
The endpoint contribution is finite: $$ \lim_{\varepsilon\to0^+}\int_\varepsilon^{1/2}\left(\frac{1}{\sqrt{x}}+\frac{1}{(x-1)^2}\right)\,dx = \lim_{\varepsilon\to0^+}\left(\sqrt{2}+2-2\sqrt{\varepsilon}-\frac{1}{1-\varepsilon}\right) = \sqrt{2}+1. $$
Approaching \(1\) from the left gives $$ \lim_{\delta\to0^+}\int_{1/2}^{1-\delta}\left(\frac{1}{\sqrt{x}}+\frac{1}{(x-1)^2}\right)\,dx = \lim_{\delta\to0^+}\left(2\sqrt{1-\delta}+\frac{1}{\delta}-\sqrt{2}-2\right) =+\infty. $$ From the right, an antiderivative is \(J(x)=2\sqrt{x}-\frac{1}{x-1}\), since \(J'(x)=\frac{1}{\sqrt{x}}+\frac{1}{(x-1)^2}\). Thus $$ \lim_{\eta\to0^+}\int_{1+\eta}^{2}\left(\frac{1}{\sqrt{x}}+\frac{1}{(x-1)^2}\right)\,dx = \lim_{\eta\to0^+}\left(2\sqrt{2}-1-2\sqrt{1+\eta}+\frac{1}{\eta}\right) =+\infty. $$
The endpoint piece is finite, but both one-sided interior pieces diverge. Therefore the ordinary improper integral diverges. The antiderivatives have been checked by differentiation, and the displayed limits show directly that the divergence comes from the terms \(1/\delta\) and \(1/\eta\).
Worked Example: Endpoint Divergence Despite an Integrable Interior Singularity
Worked Example: The Endpoint Alone Prevents Convergence
Now consider $$ \int_0^2\left(\frac{1}{x}+\frac{1}{\sqrt{|x-1|}}\right)\,dx. $$ The interior singularity has finite contributions on both sides, but the endpoint term must be tested separately. On \(0<x<1\), use $$ L(x)=\ln x-2\sqrt{1-x}, \qquad L'(x)=\frac{1}{x}+\frac{1}{\sqrt{1-x}}. $$
The endpoint piece is $$ \lim_{\varepsilon\to0^+}\int_\varepsilon^{1/2}\left(\frac{1}{x}+\frac{1}{\sqrt{1-x}}\right)\,dx = \lim_{\varepsilon\to0^+}\left(-\ln 2-\sqrt{2}-\ln\varepsilon+2\sqrt{1-\varepsilon}\right) =+\infty, $$ because \(-\ln\varepsilon\to+\infty\). The contribution from the left of \(x=1\) is $$ \lim_{\delta\to0^+}\int_{1/2}^{1-\delta}\left(\frac{1}{x}+\frac{1}{\sqrt{1-x}}\right)\,dx = \lim_{\delta\to0^+}\left(\ln(1-\delta)-2\sqrt{\delta}+\ln 2+\sqrt{2}\right) = \ln 2+\sqrt{2}. $$
On \(x>1\), an antiderivative is \(M(x)=\ln x+2\sqrt{x-1}\), with $$ M'(x)=\frac{1}{x}+\frac{1}{\sqrt{x-1}}. $$ Consequently, $$ \lim_{\eta\to0^+}\int_{1+\eta}^{2}\left(\frac{1}{x}+\frac{1}{\sqrt{x-1}}\right)\,dx = \lim_{\eta\to0^+}\left(\ln 2+2-\ln(1+\eta)-2\sqrt{\eta}\right) = \ln 2+2. $$ The antiderivatives have the stated derivatives, and the two interior limits are finite. However, the endpoint limit is infinite, so the ordinary integral diverges. Finite interior contributions cannot make the endpoint limit finite.
Common Pitfall: Cancelling Opposite Interior Divergences
A tempting but incorrect shortcut is to approach an interior singularity from both sides with the same cutoff and combine the resulting expressions before taking a limit. That procedure can produce a finite symmetric value even though neither one-sided integral converges. A symmetric cancellation does not satisfy the definition of ordinary convergence.
For example, consider the interior term \(1/(x-1)\) near \(x=1\), splitting at \(r=\tfrac12\) and taking the right endpoint of the local interval to be \(\tfrac32\). The separate one-sided integrals are $$ \int_{1/2}^{1-\delta}\frac{1}{x-1}\,dx = \ln\delta-\ln(1/2) = \ln(2\delta)\longrightarrow-\infty $$ and $$ \int_{1+\delta}^{3/2}\frac{1}{x-1}\,dx = \ln(1/2)-\ln\delta = \ln\left(\frac{1}{2\delta}\right)\longrightarrow+\infty. $$ If these expressions are added first with the same \(\delta\), they sum to zero. But the left-hand limit is not finite, and the right-hand limit is not finite. Therefore the ordinary integral diverges. This cancellation is associated with a symmetric principal-value procedure, not ordinary convergence.
Check Your Understanding
For each question, keep the endpoint limit and the two interior one-sided limits distinct.
- If an endpoint contribution is finite but the left-hand limit at an interior singularity is \(+\infty\), does the ordinary improper integral converge?
- Why is a regular splitting point chosen between the endpoint singularity and the interior singularity?
- What must be true of all three limits in the definition for the ordinary integral to converge?
- Why does a finite symmetric cancellation of two interior expressions not establish ordinary convergence?
- In the first worked example, which derivative checks confirm the antiderivatives on the two sides of \(x=1\)?
