Why Split an Infinite Integral?
An integral over an infinite interval is defined by a limit, so it is natural to ask whether its value depends on where we begin measuring the tail. If the integral converges, it does not: we can cut the interval at any finite point, evaluate the finite part and the remaining tail separately, and add them.
This fact is useful when one part is easier to evaluate directly than the whole. The split point may be chosen to simplify an antiderivative, create a familiar integral, or make the two pieces straightforward to calculate. The essential requirement is that the original improper integral converges.
Definition of a Right-Infinite Integral
Let \(f\) be continuous on \([a,\infty)\). The improper integral from \(a\) to infinity is defined by taking the limit of ordinary integrals with a finite upper endpoint.
Continuity ensures that each integral over a finite interval, such as \([a,c]\), is an ordinary finite-interval integral. We will split at a fixed finite point \(c>a\). The word “fixed” matters: \(c\) is chosen first, and then the cutoff \(R\) tends to infinity.
The Splitting Theorem
The theorem says more than a rule for rearranging notation. It guarantees that the tail beginning at \(c\) is itself a convergent improper integral. The equality is justified by finite-cutoff integrals and their limits.
For \(R>c\), split the ordinary integral at \(c\): \(\int_a^R f(x)\,dx=\int_a^c f(x)\,dx+\int_c^R f(x)\,dx\).
As \(R\to\infty\), the left side tends to the assumed value of \(\int_a^\infty f(x)\,dx\).
The finite integral \(\int_a^c f(x)\,dx\) is constant, so subtracting it shows that \(\int_c^R f(x)\,dx\) has a finite limit.
Why the Theorem Works
Here is the complete limit argument. Fix \(c>a\). For every \(R>c\), additivity of ordinary definite integrals gives
Let \(L=\int_a^\infty f(x)\,dx\), which exists by assumption. Rearranging the finite-cutoff identity gives
The first term on the right tends to \(L\) as \(R\to\infty\), and the second is a fixed finite number. Therefore the right side tends to \(L-\int_a^c f(x)\,dx\). This proves that the tail integral converges and that
Adding \(\int_a^c f(x)\,dx\) to both sides gives the splitting identity. Thus the identity follows from an ordinary finite-interval splitting followed by a limit; no informal manipulation of an infinite endpoint is needed.
Worked Examples
Worked Example: Splitting an Exponential Tail
Evaluate \(\int_0^\infty e^{-x}\,dx\) by splitting at \(c=2\).
First evaluate the finite piece. An antiderivative is \(-e^{-x}\), since its derivative is \(e^{-x}\):
For the tail, keep a finite upper cutoff \(R>2\) until evaluating the limit:
The split pieces add to
Therefore \(\int_0^\infty e^{-x}\,dx=1\). As a check, the cutoff calculation for the whole interval gives \(\lim_{R\to\infty}[-e^{-x}]_0^R=\lim_{R\to\infty}(1-e^{-R})=1\), agreeing with the split result.
Worked Example: Splitting an Arctangent Integral
Evaluate \(\int_0^\infty \frac{1}{1+x^2}\,dx\) by splitting at \(c=1\).
An antiderivative of \(1/(1+x^2)\) is \(\arctan x\), because \(\frac{d}{dx}\arctan x=1/(1+x^2)\). The finite piece is
For the tail, evaluate with cutoff \(R>1\):
Here we use the standard limit \(\lim_{R\to\infty}\arctan R=\pi/2\). Adding the pieces gives \(\pi/4+\pi/4=\pi/2\). As a check, evaluating from \(0\) to \(R\) directly gives \(\arctan R-\arctan 0\), whose limit is also \(\pi/2\).
Worked Example: Choosing a Split for a Rational Tail
Evaluate \(\int_0^\infty \frac{1}{(x+1)(x+3)}\,dx\), splitting at \(c=1\). This split separates the finite interval \([0,1]\) from a tail whose antiderivative has a simple limit.
Use partial fractions:
Multiplying by \((x+1)(x+3)\) gives \(1=A(x+3)+B(x+1)\). Setting \(x=-1\) gives \(1=2A\), so \(A=1/2\). Setting \(x=-3\) gives \(1=-2B\), so \(B=-1/2\). Hence the integrand is \(\frac12\left(\frac1{x+1}-\frac1{x+3}\right)\).
For the finite piece,
For the tail, using \(R>1\),
Indeed, \((R+1)/(R+3)\to1\), so its logarithm tends to \(\ln1=0\). The sum of the two pieces is \(\frac12\ln(3/2)+\frac12\ln2=\frac12\ln3\). To check, the whole cutoff integral is \(\frac12\ln\left(\frac{3(R+1)}{R+3}\right)\), which tends to \(\frac12\ln3\) as \(R\to\infty\).
Choosing and Changing the Split Point
The theorem allows any finite \(c>a\), but some choices are more useful than others. In the exponential example, the split at \(2\) produced simple expressions for both pieces. In the rational example, the split at \(1\) made the finite endpoint evaluations especially direct. A split is a tool for organizing an evaluation, not a change to the value of the convergent integral.
In particular, if \(c\) and \(d\) are two finite points greater than \(a\), the full integral can be represented using either point. The finite segment between \(c\) and \(d\) is accounted for in one representation as part of the finite piece and in the other as part of the tail. The theorem guarantees that both representations sum to the same total, provided the original integral converges.
For practical calculations, make sure the split point is in the domain of \(f\) and that the finite piece is well-defined. Here the continuity assumption supplies that condition. Then evaluate each ordinary or improper piece with the appropriate endpoint procedure, and only afterward combine the results.
A Common Sign Error
A frequent mistake is reversing the endpoint subtraction when evaluating the finite piece. For example, a wrong calculation of \(\int_0^2 e^{-x}\,dx\) might write
This subtracts the value at the upper endpoint from the value at the lower endpoint. The correct convention is \(F(2)-F(0)\), so
The incorrect value is negative, even though \(e^{-x}>0\) throughout \([0,2]\). That sign check immediately signals a problem. For a split integral, verify each piece’s endpoint order before adding the results.
Why This Identity Matters
Splitting is a basic way to make an infinite-interval calculation manageable while preserving its meaning. It permits a convenient finite interval to be evaluated separately and isolates the part whose convergence is controlled by a limit. The proof also gives a useful logical check: if the full integral converges, every tail starting at a finite point must converge as well.
The safe procedure is always the same: choose a finite split point, establish the finite-cutoff identity, and take the cutoff limit only after the pieces have been evaluated or their limits justified. This keeps the role of the improper limit visible and prevents endpoint notation from hiding a missing convergence argument.
Check Your Understanding
Answer each question using the cutoff-limit definition of an improper integral.
- State the splitting identity for a convergent integral \(\int_a^\infty f(x)\,dx\) at a finite point \(c>a\).
- Why does convergence of \(\int_a^\infty f(x)\,dx\) imply convergence of \(\int_c^\infty f(x)\,dx\)?
- Evaluate \(\int_0^\infty e^{-x}\,dx\) by splitting at \(c=1\).
- When evaluating \(\int_c^\infty f(x)\,dx\), why must the upper cutoff remain finite until the limit is taken?
- What endpoint-order check can help detect a sign error in a finite piece?
