Truncations and Definitions of Improper Integrals

Truncation Turns an Improper Integral into a Limit

A proper Riemann integral is taken over a bounded interval where the function is Riemann integrable. When an interval is unbounded, or a function is singular at an endpoint, the notation for an integral over the whole interval does not by itself define a value. The definition is a limit of proper integrals over truncated intervals. The required limits must be finite, and when there is more than one problematic endpoint, each endpoint is handled separately.

In this tutorial, functions are real-valued on the interiors of the intervals under discussion. Assume they are Riemann integrable on every compact interval used for a truncation; continuity on those intervals is a sufficient condition. The integrals in the definitions below are proper Riemann integrals. The ambient Lebesgue space for the real-line examples is $(\mathbb{R},\mathcal{L},m)$, but defining an improper Riemann integral does not assert that it equals a Lebesgue integral.

Definition: Let $a\in\mathbb{R}$, and suppose $f$ is Riemann integrable on every $[a,R]$ with $R>a$. The improper Riemann integral $\int_a^\infty f(x)\,dx$ converges to the finite real number $I$ if $$ \lim_{R\to\infty}\int_a^R f(x)\,dx=I. $$ If this finite limit does not exist, the improper integral diverges. For an interval unbounded to the left, define $\int_{-\infty}^b f(x)\,dx$ by the finite limit $\lim_{A\to\infty}\int_{-A}^b f(x)\,dx$, provided $f$ is Riemann integrable on every $[-A,b]$.

The endpoints are treated in the direction that approaches them: $R\to\infty$ on the right and $A\to\infty$ on the left. If a function is defined only beyond some finite starting point, the definition uses that point as $a$ or $b$. Changing a finite starting point adds or removes only a proper integral; it does not remove the need for a finite limit at infinity.

Worked Example: An Integral on a Half-Line

Evaluate $\int_0^\infty e^{-2x}\,dx$ from its truncations. The function is continuous and Riemann integrable on every $[0,R]$, so the definition applies.

For finite $R>0$, an antiderivative is $-\tfrac12e^{-2x}$. Therefore $$ \int_0^R e^{-2x}\,dx =\left[-\frac12e^{-2x}\right]_0^R =-\frac12e^{-2R}+\frac12 =\frac{1-e^{-2R}}{2}. $$ As $R\to\infty$, $e^{-2R}\to0$, so the limit is $\tfrac12$. Hence the improper integral converges to $\tfrac12$.

Check: at $R=1$, the truncated value is $(1-e^{-2})/2$, which is less than $1/2$ because $e^{-2}>0$. The difference between the limiting value and this truncation is $e^{-2}/2$, exactly the integral $\int_1^\infty e^{-2x}\,dx$ computed by the same limit. This independently checks the tail calculation.

Finite Singular Endpoints

Definition: Let $a<b$, and suppose $f$ is Riemann integrable on every $[a+\varepsilon,b]$ for $0<\varepsilon<b-a$. The improper integral at the left endpoint, written $\int_a^b f(x)\,dx$, converges if the finite real limit $$ \lim_{\varepsilon\downarrow0}\int_{a+\varepsilon}^b f(x)\,dx $$ exists. For a singular right endpoint, truncate at $b-\delta$ and require the finite limit $\lim_{\delta\downarrow0}\int_a^{b-\delta}f(x)\,dx$. If both endpoints are singular, require both one-sided limits separately to exist as finite real numbers; the improper integral is the sum of those limits.

The notation $\varepsilon\downarrow0$ means that the positive truncation distance tends to zero. It does not permit assigning an infinite value as the integral. For a singularity at an interior point $c\in(a,b)$, split the interval at $c$. The integral converges only if both $\lim_{\varepsilon\downarrow0}\int_a^{c-\varepsilon}f(x)\,dx$ and $\lim_{\delta\downarrow0}\int_{c+\delta}^b f(x)\,dx$ exist as finite real numbers; its value is their sum. This is also how a finite interval with singularities at both endpoints and at interior points is defined: split at every singular point and require every resulting one-sided limit separately.

Worked Example: A Finite Singular Endpoint

Compute the improper integral of $f(x)=x^{-1/2}$ over $(0,1]$. This function is continuous on each truncated interval $[\varepsilon,1]$ for $0<\varepsilon<1$.

For such $\varepsilon$, an antiderivative is $2\sqrt{x}$, and $$ \int_\varepsilon^1 x^{-1/2}\,dx =2\sqrt{1}-2\sqrt{\varepsilon} =2-2\sqrt{\varepsilon}. $$ As $\varepsilon\downarrow0$, $\sqrt{\varepsilon}\to0$. The finite endpoint limit is therefore $2$, so the improper integral converges to $2$.

Check: with $\varepsilon=1/4$, the truncated integral is $2-2(1/2)=1$, and the omitted interval contributes $2\sqrt{1/4}=1$. These add to $2$, the limiting value. The computation uses a limit of proper integrals, not evaluation of an antiderivative at the singular endpoint without justification.

Why Endpoint Limits Must Be Separate

When two endpoints are problematic, truncating both by the same distance traces out just one path through the two-parameter family of proper integrals. That path can hide cancellation. The definition instead asks for the separate one-sided limits, or equivalently for truncations that can approach the endpoints independently.

Proposition: Suppose $f$ is Riemann integrable on every compact subinterval of $(a,b)$, where either or both endpoints may be singular. If the two finite limits $$ L=\lim_{\varepsilon\downarrow0}\int_{a+\varepsilon}^{c}f(x)\,dx, \qquad R=\lim_{\delta\downarrow0}\int_c^{b-\delta}f(x)\,dx $$ exist for an interior point $c\in(a,b)$, then the independently truncated integrals satisfy $$ \lim_{\substack{\varepsilon\downarrow0\\ \delta\downarrow0}} \int_{a+\varepsilon}^{b-\delta}f(x)\,dx=L+R. $$ Here the two-parameter limit means that $\varepsilon$ and $\delta$ can tend to zero independently.

Proof. Take positive $\varepsilon$ and $\delta$ small enough that $a+\varepsilon<c<b-\delta$. Additivity of the proper Riemann integral gives $$ \int_{a+\varepsilon}^{b-\delta}f(x)\,dx =\int_{a+\varepsilon}^{c}f(x)\,dx+\int_c^{b-\delta}f(x)\,dx. $$ Let $\eta>0$. Since the first limit equals $L$, there is $\varepsilon_0>0$ such that $0<\varepsilon<\varepsilon_0$ implies $\left|\int_{a+\varepsilon}^{c}f-L\right|<\eta/2$. Since the second limit equals $R$, there is $\delta_0>0$ such that $0<\delta<\delta_0$ implies $\left|\int_c^{b-\delta}f-R\right|<\eta/2$. Thus, whenever both bounds hold, the triangle inequality gives $$ \left|\int_{a+\varepsilon}^{b-\delta}f-(L+R)\right| \le \left|\int_{a+\varepsilon}^{c}f-L\right| +\left|\int_c^{b-\delta}f-R\right| <\eta. $$ This is precisely the independent two-parameter limit. $\square$

This proposition gives a consequence of two separately convergent endpoint integrals. It does not say that a limit along one prescribed path, such as $\varepsilon=\delta$, implies either separate limit.

Worked Example: Two Endpoint Truncations

Consider $f(x)=x^{-1/2}+(1-x)^{-1/2}$ on $(0,1)$. It has a singularity at each endpoint and is continuous on every truncated interval $[\varepsilon,1-\delta]$, where $0<\varepsilon,\delta<1/2$. Keep the two truncation distances independent.

Integrating the two terms separately on the proper interval gives $$ \int_\varepsilon^{1-\delta}x^{-1/2}\,dx =2\sqrt{1-\delta}-2\sqrt{\varepsilon}, $$ and $$ \int_\varepsilon^{1-\delta}(1-x)^{-1/2}\,dx =2\sqrt{1-\varepsilon}-2\sqrt{\delta}. $$ Thus the truncated integral is $$ 2\sqrt{1-\delta}-2\sqrt{\varepsilon} +2\sqrt{1-\varepsilon}-2\sqrt{\delta}. $$ As $\varepsilon\downarrow0$ and $\delta\downarrow0$ independently, the first and third terms tend to $2$ each, while the second and fourth tend to $0$. The limit is $4$. In particular, the left and right one-sided contributions are each $2$, so their sum is $4$, as required by the definition.

Check: for $\varepsilon=\delta=1/4$, the truncated value is $2\sqrt{3/4}-1+2\sqrt{3/4}-1=2\sqrt{3}-2$. As both truncation distances decrease to zero, this expression approaches $4$; the finite-cutoff value is not itself the improper integral.

A Symmetric Cutoff Can Conceal Divergence

Worked Example: A Pole with Opposite-Signed Contributions

Consider $f(x)=1/x$ on $(-1,1)$, which has an interior singularity at $0$. By definition, split at $0$ and examine the two one-sided integrals independently.

For $0<\varepsilon<1$, the left truncation is $$ \int_{-1}^{-\varepsilon}\frac{dx}{x} =\left[\ln|x|\right]_{-1}^{-\varepsilon} =\ln(\varepsilon)-\ln(1) =\ln(\varepsilon), $$ which tends to $-\infty$, not to a finite real number. The right truncation is $$ \int_{\varepsilon}^{1}\frac{dx}{x} =\left[\ln|x|\right]_{\varepsilon}^{1} =\ln(1)-\ln(\varepsilon) =-\ln(\varepsilon), $$ which tends to $+\infty$. Neither one-sided integral converges as a finite improper integral. But if one adds the two truncated integrals using the same $\varepsilon$, the result is $\ln(\varepsilon)-\ln(\varepsilon)=0$ for every $0<\varepsilon<1$. That cancellation along equal cutoffs does not make the improper integral converge: it does not supply either of the two required finite endpoint limits.

Check: the left and right formulas are exact negatives at each common cutoff, so their sum is indeed $0$. The separate limits, however, are $-\infty$ and $+\infty$; the definition requires each to be finite before they can be added.

The same issue arises at infinity. A whole-line improper integral is defined by splitting at a finite point, for example $0$, and requiring both $\int_{-\infty}^0 f(x)\,dx$ and $\int_0^\infty f(x)\,dx$ to converge separately. A symmetric truncation $\int_{-R}^R f(x)\,dx$ alone is not that definition. It only tests the particular choice that sends the two endpoints outward together.

Cutoff Sequences and a Common Pitfall

Proposition: Let $F(R)=\int_a^R f(x)\,dx$ be defined for all $R>a$. If $F(R)\to I\in\mathbb{R}$ as $R\to\infty$, then for every sequence $R_n\to\infty$ with $R_n>a$, $F(R_n)\to I$. Conversely, if there is a finite $I$ such that $F(R_n)\to I$ for every sequence $R_n\to\infty$, then $F(R)\to I$ as $R\to\infty$.

Proof. For the forward implication, fix $\eta>0$. By the definition of $F(R)\to I$, there is $R_0>a$ such that $R>R_0$ implies $|F(R)-I|<\eta$. Since $R_n\to\infty$, eventually $R_n>R_0$, so eventually $|F(R_n)-I|<\eta$. For the converse, suppose $F(R)$ does not tend to $I$. Then there is an $\eta>0$ such that for every $R_0>a$, some $R>R_0$ satisfies $|F(R)-I|\ge\eta$. Choose such an $R_n>n+a$ for each positive integer $n$. Then $R_n\to\infty$, but $|F(R_n)-I|\ge\eta$ for every $n$, contradicting the assumed convergence along every sequence. Hence $F(R)\to I$. $\square$

A common mistake is to compute a symmetric truncation and treat its limit as the integral over an interval with two problematic endpoints. For $1/x$ on $(-1,1)$, the wrong calculation is $\lim_{\varepsilon\downarrow0}\bigl(\int_{-1}^{-\varepsilon}1/x\,dx+\int_{\varepsilon}^{1}1/x\,dx\bigr)=0$, followed by the claim that the improper integral is $0$. The correction is to apply the definition to each side: the left limit is $-\infty$, the right limit is $+\infty$, and the improper integral therefore does not converge. A finite integral over a truncated interval is not enough; the required endpoint limits must exist independently and be finite.

Check Your Understanding

For each question, identify the proper truncated integrals first, then apply the relevant limit definition.

Write the definition of convergence for $\int_{-\infty}^b f(x)\,dx$, including the condition on the finite truncations.
Use truncation to evaluate $\int_0^\infty e^{-3x}\,dx$. Which finite limit defines the answer?
For $f(x)=x^{-1/3}$ on $(0,1]$, write the truncated integral and decide whether its limit as the cutoff decreases to zero is finite.
Explain why convergence of $\int_{-R}^{R}f(x)\,dx$ as $R\to\infty$ alone does not define the improper integral over the whole real line.
For a singularity at $c\in(a,b)$, state the two separate limits required for convergence of $\int_a^b f(x)\,dx$.