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Splitting Integrals At Several Interior Singularities

Use an ordered partition of the interval to determine whether several interior singularities allow an ordinary improper integral to exist.

Intermediate 10 min read

What You'll Learn

  • Order several interior singularities and split the interval at each one.
  • Define convergence through separate one-sided limits at every singular point.
  • Determine whether the full improper integral converges by testing every piece.
  • Evaluate a convergent integral with two distinct square-root singularities.
  • Identify why cancellation between divergent pieces does not establish convergence.

Why Several Singularities Require Several Tests

An integrand can be unbounded at more than one point inside a finite interval. In that situation, splitting the interval is not merely a convenient way to organize a calculation: it is part of the definition of the ordinary improper integral. Each singular point has a left-hand and a right-hand side, and every resulting one-sided integral must converge.

For example, if an integrand is singular at two distinct points \(c_1\) and \(c_2\) in \((a,b)\), where \(c_1<c_2\), the natural partition is \(a<c_1<c_2<b\). It produces three intervals. The middle interval has singular behavior at both ends, so its convergence requires a separate check at each end.

Key principle: An ordinary improper integral across several interior singularities converges only if every one-sided integral adjacent to every singularity converges to a finite value. Divergence on one side cannot be repaired by another side.

Definition by Splitting at Every Singular Point

Let \(f\) be defined on \([a,b]\) except at distinct interior points \(c_1,\ldots,c_k\), ordered so that \(a<c_1<\cdots<c_k<b\). Assume \(f\) is continuous on each open interval between consecutive points, and that any portions away from the \(c_i\) are proper integrals. Split the interval at all of the singular points.

$$ a<c_1<c_2<\cdots<c_k<b. $$

The resulting pieces are \([a,c_1]\), \([c_1,c_2]\), and so on, ending with \([c_k,b]\). At each singular endpoint, define the corresponding integral by a one-sided limit. For example, the piece between \(c_i\) and \(c_{i+1}\) is required to have both of these finite limits:

$$ \lim_{u\to c_i^+}\int_u^d f(x)\,dx \qquad\text{and}\qquad \lim_{v\to c_{i+1}^-}\int_d^v f(x)\,dx, $$

where \(d\) is any fixed point strictly between \(c_i\) and \(c_{i+1}\). The first limit tests the right-hand side of \(c_i\); the second tests the left-hand side of \(c_{i+1}\). If both exist and are finite, their sum defines the integral over that middle piece. Equivalently, one can test the two endpoint limits of the truncated integral independently.

Definition: The ordinary improper integral \(\int_a^b f(x)\,dx\) across the interior singularities \(c_1,\ldots,c_k\) converges if every one-sided integral at every \(c_i\) converges to a finite value. When it does, the value is the sum of the \(k+1\) integrals over the partitioned intervals, each interpreted with its required one-sided limits.

In particular, for two interior singularities \(c_1<c_2\), the definition requires four tests: approach \(c_1\) from the left and from the right, and approach \(c_2\) from the left and from the right. The integral over the middle interval \((c_1,c_2)\) is not accepted just because a calculation with one common cutoff produces a finite-looking expression.

Why the Tests Must Be Separate

On an interval containing no singularity, additivity of definite integrals lets us break the integral into pieces. For an improper integral, the same idea applies only after each required limit exists. The limits at different singularities are independent: a cutoff approaching one singularity cannot be chosen to cancel a divergence at another.

1
List and order the singularities.
Write them as \(c_1<\cdots<c_k\), and partition \([a,b]\) at every \(c_i\).
2
Identify each one-sided test.
At each \(c_i\), inspect the integrals from the left and from the right. Do not combine these tests.
3
Decide convergence before adding.
If any one-sided integral diverges, the ordinary improper integral does not converge. If all are finite, add the resulting piece integrals.

This criterion follows directly from the definition. The integral over the whole interval is defined as a finite sum of the separately defined piece integrals. If one required piece has no finite value, that sum is not an ordinary improper integral. Conversely, if every piece has a finite value, their finite sum exists and defines the integral across the full partition.

Worked Example: Two Integrable Square-Root Singularities

Worked Example: Two Integrable Square-Root Singularities

Determine and evaluate the ordinary improper integral

$$ \int_0^4\left(\frac{1}{\sqrt{|x-1|}}+\frac{1}{\sqrt{|x-3|}}\right)\,dx. $$

Step 1: Split at both singularities. The singular points are \(x=1\) and \(x=3\), so use the intervals \((0,1)\), \((1,3)\), and \((3,4)\). The integral on each piece is improper at any endpoint equal to 1 or 3.

Step 2: Check the piece from 0 to 1. For the first term, let \(u=1-x\). For the second, let \(u=3-x\). Taking the endpoint limits gives

$$ \int_0^1\frac{dx}{\sqrt{1-x}} =\lim_{\varepsilon\to0^+}\int_0^{1-\varepsilon}\frac{dx}{\sqrt{1-x}} =\lim_{\varepsilon\to0^+}2(1-\sqrt{\varepsilon})=2, $$ $$ \int_0^1\frac{dx}{\sqrt{3-x}} =\int_2^3u^{-1/2}\,du =2(\sqrt{3}-\sqrt{2}). $$

Thus the first piece contributes \(2+2(\sqrt{3}-\sqrt{2})\), a finite value.

Step 3: Check the middle piece. At \(x=1\), the first term is singular; at \(x=3\), the second is singular. Each one-sided integral is finite:

$$ \int_1^3\frac{dx}{\sqrt{x-1}} =\lim_{\varepsilon\to0^+}2\bigl(\sqrt{2}-\sqrt{\varepsilon}\bigr) =2\sqrt{2}, $$ $$ \int_1^3\frac{dx}{\sqrt{3-x}} =\lim_{\varepsilon\to0^+}2\bigl(\sqrt{2}-\sqrt{\varepsilon}\bigr) =2\sqrt{2}. $$

The middle piece therefore contributes \(4\sqrt{2}\).

Step 4: Check the piece from 3 to 4. For the first term, set \(u=x-1\); for the second, set \(u=x-3\). Then

$$ \int_3^4\frac{dx}{\sqrt{x-1}} =\int_2^3u^{-1/2}\,du =2(\sqrt{3}-\sqrt{2}), $$ $$ \int_3^4\frac{dx}{\sqrt{x-3}} =\lim_{\varepsilon\to0^+}\int_{3+\varepsilon}^4\frac{dx}{\sqrt{x-3}} =\lim_{\varepsilon\to0^+}2(1-\sqrt{\varepsilon}) =2. $$

This last piece contributes \(2(\sqrt{3}-\sqrt{2})+2\). All required one-sided integrals are finite, so the ordinary improper integral converges. Adding the three contributions gives

$$ \begin{aligned} \int_0^4\left(\frac{1}{\sqrt{|x-1|}}+\frac{1}{\sqrt{|x-3|}}\right)\,dx &=2+2(\sqrt{3}-\sqrt{2})+4\sqrt{2}\\ &\quad+2(\sqrt{3}-\sqrt{2})+2\\ &=4+4\sqrt{3}. \end{aligned} $$

Check: On each side of each singularity, an antiderivative for the singular term is \(2\sqrt{|x-c|}\), with the appropriate sign on that side. Differentiating it gives \(1/\sqrt{|x-c|}\). The finite endpoint limits above therefore match the integrands and confirm the value.

A Divergent Side Makes the Whole Integral Diverge

A singularity may be integrable from one side but not from the other; more generally, another singularity elsewhere cannot compensate for a divergent side. To establish convergence, test the actual behavior on every adjacent interval. Positivity makes the failure especially clear, because no cancellation between positive pieces is possible.

Worked Example: One Nonintegrable Singularity Among Two

Decide whether the ordinary improper integral

$$ \int_0^4\left(\frac{1}{\sqrt{|x-1|}}+\frac{1}{|x-3|}\right)\,dx $$

converges.

Step 1: Partition the interval. The singularities are \(x=1\) and \(x=3\). The pieces are \((0,1)\), \((1,3)\), and \((3,4)\).

Step 2: Check the sides of \(x=1\). The square-root term has finite integral on both sides: its contributions are \(2\) on \((0,1)\) and \(2\sqrt{2}\) on \((1,3)\). The other term is bounded near \(x=1\), since \(1/|x-3|\) approaches \(1/2\). Thus neither side of \(x=1\) causes divergence.

Step 3: Check each side of \(x=3\). On \((1,3)\), the term \(1/|x-3|=1/(3-x)\) gives

$$ \int_1^{3-\varepsilon}\frac{dx}{3-x} =\ln 2-\ln\varepsilon \longrightarrow+\infty \qquad\text{as }\varepsilon\to0^+. $$

On \((3,4)\), it gives

$$ \int_{3+\varepsilon}^4\frac{dx}{x-3} =-\ln\varepsilon \longrightarrow+\infty \qquad\text{as }\varepsilon\to0^+. $$

The square-root term is positive, so it cannot make either of these divergent integrals finite. Both one-sided tests at \(x=3\) fail. Therefore the ordinary improper integral diverges, regardless of the convergence of the sides at \(x=1\).

Worked Example: Two Logarithmic Singularities

Worked Example: Two Logarithmic Singularities

Evaluate

$$ \int_0^4\bigl(\ln|x-1|+\ln|x-3|\bigr)\,dx $$

as an ordinary improper integral.

Step 1: Split at both singularities. Use the intervals \((0,1)\), \((1,3)\), and \((3,4)\). A primitive for \(\ln t\), for \(t>0\), is \(H(t)=t\ln t-t\). In endpoint limits, \(t\ln t\to0\) as \(t\to0^+\). Consequently, the integral of \(\ln t\) from 0 to 1 is \(-1\), and from 0 to 2 it is \(2\ln2-2\). These finite values show that the logarithmic singularities are integrable from the relevant sides.

Step 2: Evaluate the first piece. For \(\ln|x-1|\), the integral from 0 to 1 is \(-1\). For \(\ln|x-3|\), put \(t=3-x\), so the limits for \(t\) are 3 and 2. Hence

$$ \int_0^1\ln|x-3|\,dx =\int_2^3\ln t\,dt =(3\ln3-3)-(2\ln2-2) =3\ln3-2\ln2-1. $$

The first piece is therefore \(3\ln3-2\ln2-2\).

Step 3: Evaluate the middle piece. The two terms have equal integrals, by the substitutions \(t=x-1\) and \(t=3-x\). Each gives

$$ \int_1^3\ln|x-1|\,dx =\int_0^2\ln t\,dt =2\ln2-2. $$

Thus the middle piece contributes \(4\ln2-4\).

Step 4: Evaluate the last piece. For \(\ln|x-1|\), put \(t=x-1\); the resulting integral is \(\int_2^3\ln t\,dt=3\ln3-2\ln2-1\). For \(\ln|x-3|\), put \(t=x-3\); the result is \(\int_0^1\ln t\,dt=-1\). The last piece contributes \(3\ln3-2\ln2-2\).

All six one-sided contributions are finite, so the integral converges. Adding the pieces yields

$$ (3\ln3-2\ln2-2)+(4\ln2-4)+(3\ln3-2\ln2-2) =6\ln3-8. $$

Check: For \(t>0\), \(H'(t)=\ln t\), so differentiating the primitives used on each side recovers the logarithmic terms. The boundary term \(t\ln t\) tends to zero at every singular endpoint, which confirms the finite limits used in the calculation.

A Common Mistake: Combining Divergent Pieces

It is tempting to use one antiderivative across several pieces and hope that infinite terms cancel. This does not test the one-sided integrals separately. For example, on \((0,2)\), the function \(1/(x-1)\) has an interior singularity. The left and right integrals, with cutoff \(\varepsilon\), are

$$ \int_0^{1-\varepsilon}\frac{dx}{x-1}=\ln\varepsilon, \qquad \int_{1+\varepsilon}^2\frac{dx}{x-1}=-\ln\varepsilon. $$

Their sum is zero for every \(\varepsilon>0\), but the left limit is \(-\infty\) and the right limit is \(+\infty\). Neither side has a finite improper integral, so the ordinary improper integral does not converge. A finite result from forcing the two cutoffs to match is not a substitute for the separate one-sided tests.

Takeaway: First establish that every one-sided integral converges. Only then add the values. Cancellation between expressions that diverge is not ordinary convergence.

Check Your Understanding

For each question, begin by marking every singular point and listing the required one-sided tests.

  1. If \(f\) has singularities at \(c_1<c_2<c_3\) inside \([a,b]\), how many one-sided tests are required at the interior points?
  2. Why must the middle piece between two singularities be checked at both of its endpoints?
  3. If all one-sided integrals at one singularity converge but one side at a different singularity diverges, does the full ordinary improper integral converge?
  4. For \(\int_0^4\bigl(|x-1|^{-1/2}+|x-3|^{-1/2}\bigr)\,dx\), write the three intervals in the partition before attempting an evaluation.
  5. Why does adding two cutoff expressions that cancel not prove that the separate improper integrals converge?

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