Integrals Across the Whole Real Line

Introduction

An integral over the whole real line has no finite endpoints at which to begin and end an ordinary definite integral. To define it, we must account for both directions in which the interval extends without bound: toward negative infinity and toward positive infinity.

The key requirement is that the two tails converge separately. Their values need not be equal, and one tail cannot make up for a divergent other tail. Once both limits exist as finite numbers, the integral over the whole real line is their sum.

Central idea: An integral across the whole real line is defined by two independent one-sided improper integrals, not by sending two endpoints to infinity in a coordinated way.

Definition of a Whole-Line Integral

Let \(f\) be integrable on every finite interval \([a,b]\) with \(a<b\). The left-hand improper integral is defined, when the limit exists and is finite, by

\int_{-\infty}^{0} f(x)\,dx = \lim_{a\to-\infty}\int_a^0 f(x)\,dx.

The right-hand improper integral is defined, when the limit exists and is finite, by

\int_{0}^{\infty} f(x)\,dx = \lim_{b\to\infty}\int_0^b f(x)\,dx.

The choice of \(0\) as the finite dividing point gives a precise definition for this tutorial. The integral over the whole real line exists if and only if both one-sided integrals above converge to finite values. When they do, define

\int_{-\infty}^{\infty} f(x)\,dx = \int_{-\infty}^{0} f(x)\,dx + \int_{0}^{\infty} f(x)\,dx.

Definition: The whole-line improper integral \(\int_{-\infty}^{\infty} f(x)\,dx\) converges when both \(\lim_{a\to-\infty}\int_a^0 f(x)\,dx\) and \(\lim_{b\to\infty}\int_0^b f(x)\,dx\) exist as finite real numbers. Its value is the sum of those limits.

For signed functions, convergence here means that each one-sided limit is finite. The definition does not require either tail to be positive. It does, however, require the limits to exist separately. The value of one tail cannot be used to cancel a divergence in the other.

Why the Two Limits Are Added

On any finite interval \([a,b]\) with \(a<0<b\), additivity of definite integrals gives

\int_a^b f(x)\,dx = \int_a^0 f(x)\,dx + \int_0^b f(x)\,dx.

This finite-interval identity explains the definition. As the left endpoint moves toward negative infinity and the right endpoint moves toward positive infinity, the two terms on the right approach their respective one-sided improper integrals. If both limits exist as finite numbers, the limit of their sum is the sum of their limits.

Cut off the left tail.
Use a finite endpoint \(a<0\) and form \(\int_a^0 f(x)\,dx\).

Cut off the right tail.
Use a finite endpoint \(b>0\) and form \(\int_0^b f(x)\,dx\).

Take the limits independently.
Check \(a\to-\infty\) and \(b\to\infty\) as two separate convergence questions.

Add only if both converge.
The whole-line integral is the sum of the two finite one-sided values.

Worked Examples

Worked Example: Exponential Tails with Different Values

Define \(f\) by \(f(x)=e^x\) for \(x<0\), and \(f(x)=3e^{-2x}\) for \(x\geq 0\). Evaluate \(\int_{-\infty}^{\infty} f(x)\,dx\), if it converges.

First calculate the left tail, using an antiderivative of \(e^x\):

\int_a^0 e^x\,dx = \left[e^x\right]_a^0 = e^0-e^a = 1-e^a.

As \(a\to-\infty\), \(e^a\to0\), so the left tail converges to \(1\):

\int_{-\infty}^0 f(x)\,dx = \lim_{a\to-\infty}(1-e^a) = 1.

For the right tail, an antiderivative of \(3e^{-2x}\) is \(-\frac{3}{2}e^{-2x}\). Therefore,

\int_0^b 3e^{-2x}\,dx = \left[-\frac{3}{2}e^{-2x}\right]_0^b = -\frac{3}{2}e^{-2b}+\frac{3}{2}.

Since \(e^{-2b}\to0\) as \(b\to\infty\), the right tail converges to \(\frac{3}{2}\). Both tails are finite, so the whole-line integral exists and equals

\int_{-\infty}^{\infty} f(x)\,dx = 1+\frac{3}{2} = \frac{5}{2}.

Check: Differentiating \(e^x\) gives \(e^x\), and differentiating \(-\frac{3}{2}e^{-2x}\) gives \(3e^{-2x}\). The separate tail values are \(1\) and \(\frac{3}{2}\); their sum is \(\frac{5}{2}\), so the two unequal contributions have been included.

Worked Example: Rational Tails with Unequal Contributions

Let \(g(x)=\frac{1}{(1-x)^2}\) for \(x<0\), and \(g(x)=\frac{2}{(1+x)^2}\) for \(x\geq0\). Evaluate the whole-line integral.

For the left tail, an antiderivative of \((1-x)^{-2}\) is \((1-x)^{-1}\), since its derivative is \((1-x)^{-2}\). Thus

\int_a^0 \frac{1}{(1-x)^2}\,dx = \left[\frac{1}{1-x}\right]_a^0 = 1-\frac{1}{1-a}.

As \(a\to-\infty\), \(1-a\to\infty\), so \(\frac{1}{1-a}\to0\). Hence the left tail equals \(1\).

For the right tail, an antiderivative of \(2(1+x)^{-2}\) is \(-\frac{2}{1+x}\). Therefore,

\int_0^b \frac{2}{(1+x)^2}\,dx = \left[-\frac{2}{1+x}\right]_0^b = -\frac{2}{1+b}+2.

As \(b\to\infty\), \(\frac{2}{1+b}\to0\), so the right tail equals \(2\). Both one-sided integrals converge, and therefore

\int_{-\infty}^{\infty} g(x)\,dx = 1+2 = 3.

Check: The derivative of \(\frac{1}{1-x}\) is \(\frac{1}{(1-x)^2}\), and the derivative of \(-\frac{2}{1+x}\) is \(\frac{2}{(1+x)^2}\). The computed tail values \(1\) and \(2\) are unequal, and their sum is \(3\).

Worked Example: One Divergent Tail Prevents Convergence

Define \(h(x)=\frac{1}{1-x}\) for \(x<0\), and \(h(x)=e^{-x}\) for \(x\geq0\). The right tail converges, but determine whether the whole-line integral exists.

An antiderivative of \(\frac{1}{1-x}\) is \(-\ln(1-x)\). For the left tail,

\int_a^0 \frac{1}{1-x}\,dx = \left[-\ln(1-x)\right]_a^0 = 0+\ln(1-a) = \ln(1-a).

As \(a\to-\infty\), \(1-a\to\infty\), and \(\ln(1-a)\to\infty\). The left tail does not converge to a finite value.

For comparison, the right tail is finite:

\int_0^b e^{-x}\,dx = \left[-e^{-x}\right]_0^b = 1-e^{-b} \longrightarrow 1 \quad\text{as }b\to\infty.

Although the right tail converges to \(1\), the whole-line integral does not exist under the definition: its left tail diverges. The two sides must each pass their own convergence test.

Check: Differentiating \(-\ln(1-x)\) gives \(\frac{1}{1-x}\), and differentiating \(-e^{-x}\) gives \(e^{-x}\). The left limit is infinite while the right limit is \(1\), so there is no pair of finite tail values to add.

A Common Mistake: Using the Same Cutoff to Force Cancellation

A tempting but invalid shortcut is to integrate from \(-R\) to \(R\), take \(R\to\infty\), and treat the resulting limit as the whole-line integral. That procedure can hide the behavior of the two tails. For example, take \(f(x)=x\). With equal cutoffs,

\int_{-R}^{R}x\,dx = \left[\frac{x^2}{2}\right]_{-R}^{R} = \frac{R^2}{2}-\frac{R^2}{2} = 0.

It would be wrong to conclude from this calculation that the whole-line improper integral is \(0\). The separate tail calculations give

\int_a^0 x\,dx = \left[\frac{x^2}{2}\right]_a^0 = -\frac{a^2}{2} \longrightarrow -\infty,

and

\int_0^b x\,dx = \left[\frac{x^2}{2}\right]_0^b = \frac{b^2}{2} \longrightarrow \infty.

Neither one-sided integral converges to a finite number. Therefore the whole-line integral does not exist by the definition, even though the equal-cutoff calculation is always \(0\). The correction is to take and test the two one-sided limits separately before adding anything.

Takeaway: Matching cutoffs can make two divergent quantities appear to cancel. A whole-line improper integral exists only when the left and right one-sided integrals each converge independently.

How to Evaluate a Whole-Line Integral Reliably

For a new integral, write down the two one-sided limits before doing any limiting calculation. Evaluate the finite integrals on each side, determine each limit, and record whether it is finite. Only after both limits have been shown to exist should you add their values.

This order matters even when the integrand is simple or symmetric. A familiar antiderivative on finite intervals does not by itself establish convergence at either infinite endpoint. Likewise, a finite answer from a combined cutoff calculation does not establish that the two one-sided integrals converge.

Examples with unequal tails are useful because they make clear that there is no requirement for the contributions to match. In the rational example, one side contributed \(1\) and the other contributed \(2\); the integral existed because both were finite, and its value was their sum. The decisive question is not whether the tails are equal, but whether each has a finite limit.

Check Your Understanding

Answer each question by using the separate one-sided definition.

State the two limits that must converge for \(\int_{-\infty}^{\infty} f(x)\,dx\) to exist.
If the left tail converges to \(-2\) and the right tail converges to \(5\), what is the whole-line integral?
For the function \(f(x)=x\), why does \(\int_{-R}^{R}x\,dx=0\) for every \(R>0\) not establish convergence of the whole-line integral?
Suppose the right tail converges to a finite value but the left tail diverges. Does the whole-line integral exist under this definition?
Must the left and right tail values be equal when the whole-line integral converges?