Why Linearity Needs a Convergence Condition
For ordinary integrals over a finite interval, adding integrands or multiplying an integrand by a constant gives familiar rules for the integral. Improper integrals are defined by limits, however, so we must also establish that the relevant limits exist. The linearity rules hold when the integrals being combined converge.
The useful feature is that the same cutoff can be used for each integrand. At every finite cutoff, ordinary integral linearity applies. Taking the limit then preserves the addition or constant multiple, provided the individual cutoff integrals have finite limits.
Cutoff Definitions
For a function \(f\) integrable on every finite interval \([a,R]\), the right-infinite improper integral is defined by a limit:
provided the limit exists as a finite real number. If it does, the improper integral converges; otherwise it diverges. The cutoff integral \(\int_a^R f(x)\,dx\) is an ordinary finite-interval integral.
For a left-infinite interval, the corresponding definition is
again requiring a finite limit. These definitions make the role of linearity precise: first apply finite-interval linearity at a fixed cutoff, then take the limit.
Linearity Theorem
This result includes constant multiples by taking one coefficient to be zero, and sums by taking both coefficients to be one. A finite number of convergent improper integrals can be combined by applying the two-function result repeatedly.
Why the Theorem Works
Here is the proof for a right-infinite interval. Define the cutoff integrals
By hypothesis, there are finite real numbers \(I\) and \(J\) such that \(F(R)\to I\) and \(G(R)\to J\) as \(R\to\infty\). For every finite \(R\), linearity on the finite interval gives the exact identity
The limit laws for sums and constant multiples apply because both limits on the right exist and are finite. Therefore,
The right-hand side is finite, so the combined improper integral converges, and its value is the claimed linear combination. For a left-infinite integral, use the same argument with cutoff integrals \(\int_{-R}^b f(x)\,dx\) and \(\int_{-R}^b g(x)\,dx\).
Worked Examples on Half-Lines
Worked Example: Combining Exponential Tails
Evaluate \(\int_0^\infty \bigl(2e^{-x}-3e^{-2x}\bigr)\,dx\), and find the part of this integral from \(R\) to infinity, where \(R\ge 0\).
First evaluate the two separate integrals by using a finite cutoff \(B\):
Both improper integrals converge. Linearity therefore gives
To calculate the tail, use a finite upper cutoff \(B>R\), then let \(B\to\infty\):
The antiderivative checks the integrand: its derivative is \(2e^{-x}-3e^{-2x}\). At \(R=0\), the tail formula gives \(2-\frac32=\frac12\), matching the total integral. As \(R\to\infty\), both exponential terms tend to zero, as a tail should.
Worked Example: A Polynomial Factor in the Combined Tail
Evaluate \(\int_0^\infty \bigl(3xe^{-x}-2e^{-x}\bigr)\,dx\), and calculate its tail from \(R\) to infinity.
For the first individual integral, use the antiderivative \(-(x+1)e^{-x}\), whose derivative is \(xe^{-x}\):
The last limit follows because \((B+1)e^{-B}\to 0\). The other integral is \(\int_0^\infty e^{-x}\,dx=1\), as calculated in the previous example. Thus both converge, and
An antiderivative of the combined integrand is \(-(3x+1)e^{-x}\), since
Therefore the tail is
At \(R=0\), this is \(1\), agreeing with the full integral. It tends to zero as \(R\to\infty\). The calculation also shows how to combine tails: the separate tails are \((R+1)e^{-R}\) and \(e^{-R}\), so \(3(R+1)e^{-R}-2e^{-R}=(3R+1)e^{-R}\).
Worked Example: Linearity on a Left-Infinite Interval
Evaluate \(\int_{-\infty}^0 \bigl(-2e^x+4e^{2x}\bigr)\,dx\), and calculate the portion from \(-\infty\) to \(-R\), for \(R\ge 0\).
Use the left cutoff \(-B\). The two individual cutoff integrals are
Both integrals converge, so the combined value is
For the left tail, first integrate over \([-B,-R]\), where \(B>R\), and then let \(B\to\infty\):
Differentiating \(-2e^x+2e^{2x}\) gives \(-2e^x+4e^{2x}\), verifying the antiderivative. At \(R=0\), the tail is zero, matching the full integral. As \(R\to\infty\), the tail tends to zero. Its value can also be obtained by combining the individual left tails: \(-2e^{-R}+4\left(\frac12e^{-2R}\right)\).
Whole-Real-Line Integrals
An integral over the whole real line is defined by splitting at a finite point \(c\). It converges only when both one-sided improper integrals converge:
Consequently, linearity on the whole real line follows by applying the half-line theorem separately to the left and right pieces. For each function being combined, both pieces must converge. The same splitting also gives linearity for a two-sided tail, such as the parts outside \([-R,R]\).
Worked Example: Combining Tails on the Whole Real Line
Consider \(3e^{-|x|}-2e^{-2|x|}\). Find its integral over the whole real line and its two-sided tail outside \([-R,R]\), where \(R\ge 0\).
The integrands are even, so each whole-line integral is twice its right-half-line integral. For \(k>0\), \(\int_0^\infty e^{-kx}\,dx=1/k\). Thus
Each is convergent on both sides. Linearity gives
On the right, the tail from \(R\) to infinity is \(3e^{-R}-e^{-2R}\), because the separate tails are \(e^{-R}\) and \(\frac12e^{-2R}\). The left tail has the same value by evenness. Therefore the two-sided tail is
At \(R=0\), the expression is \(6-2=4\), the full integral. As \(R\to\infty\), it tends to zero. This verifies the tail calculation against both limiting cases.
A Common Mistake: Finite Cutoffs Do Not Guarantee Convergence
At a finite cutoff, an identity such as
is valid whenever the finite-interval integrals exist. But it does not, by itself, show that the limit as \(R\to\infty\) exists. For example, take \(f(x)=1\) and \(g(x)=0\). Then
which has no finite limit. The mistake would be to infer convergence merely from finite-cutoff linearity. The theorem instead assumes that the individual improper integrals converge before taking their linear combination. This condition is what allows the limit laws used in the proof.
Using Linearity to Find a Tail
If \(\int_a^\infty f(x)\,dx\) converges, then for any finite \(R\ge a\), the tail from \(R\) onward also converges. Indeed, for \(B>R\),
Taking \(B\to\infty\) shows that the tail equals the full integral minus the finite initial part. For two convergent integrals, applying the same cutoff identity to each gives
This is often the most efficient way to calculate a combined tail: find the individual tail formulas, multiply by the coefficients, and add. The result follows from cutoff limits, not from treating infinity as an ordinary endpoint.
Check Your Understanding
Use the convergence hypotheses carefully in each question.
- State the definition of convergence for \(\int_a^\infty f(x)\,dx\).
- Why does the proof of linearity use the same cutoff for both functions?
- If \(\int_a^\infty f(x)\,dx=5\) and \(\int_a^\infty g(x)\,dx=-2\), what is \(\int_a^\infty (3f(x)-g(x))\,dx\), assuming both converge?
- What must be true of the two one-sided pieces for an integral over the whole real line to converge?
- Why does finite-cutoff linearity alone not establish convergence of an improper integral?
