Why a Left-Infinite Integral Needs a Definition
An integral such as \(\int_{-5}^{b} f(x)\,dx\) has finite endpoints. But \(-\infty\) is not a real number, so \(\int_{-\infty}^{b} f(x)\,dx\) cannot be evaluated as an ordinary definite integral with that endpoint. Instead, we first integrate over a finite interval and then move its lower endpoint farther and farther to the left.
This limiting process is essential. Depending on the function, the cutoff integrals may approach a finite number, grow without bound, or fail to approach any single value. The improper integral is defined by the limit; it is not an instruction to substitute negative infinity into an antiderivative.
For example, continuity of \(f\) on \((-\infty,b]\) guarantees that each finite-cutoff integral \(\int_L^b f(x)\,dx\) exists. Continuity is a convenient sufficient condition, though the definition itself only requires that the finite-cutoff integrals be well-defined.
The direction of the limit matters: \(L\) moves toward negative infinity, while \(b\) stays fixed. The integral accumulates area from the distant left up to the fixed endpoint. The cutoff \(L\) is temporary; convergence means that the resulting values settle toward one finite number as \(L\) recedes.
Using an Antiderivative to Read the Limit
Suppose \(F\) is an antiderivative of \(f\) on the interval in question, so \(F'(x)=f(x)\). The Fundamental Theorem of Calculus gives a formula for each finite cutoff. Taking the limit of that formula converts the improper-integral question into a question about \(F(L)\).
For a real number \(L<b\), consider the ordinary integral \(\int_L^b f(x)\,dx\).
Since \(F'=f\), the finite integral equals \(F(b)-F(L)\).
The improper integral is \(\lim_{L\to-\infty}\bigl(F(b)-F(L)\bigr)\), if that limit is finite.
This also explains exactly when an antiderivative that grows without bound causes divergence. The endpoint value \(F(b)\) is fixed. If \(F(L)\to+\infty\) as \(L\to-\infty\), then \(F(b)-F(L)\to-\infty\). If \(F(L)\to-\infty\), then \(F(b)-F(L)\to+\infty\). In either case, there is no finite improper-integral value.
Why the Antiderivative Formula Works
The formula follows from the definition and the Fundamental Theorem of Calculus; no evaluation at an infinite endpoint is involved. For each finite \(L<b\), the theorem applies on the closed interval \([L,b]\). Then the definition asks for the limit of those finite answers.
On this finite interval, \(F'=f\), so the Fundamental Theorem gives \(\int_L^b f(x)\,dx=F(b)-F(L)\).
Substitute this finite-interval identity into \(\lim_{L\to-\infty}\int_L^b f(x)\,dx\).
This yields \(\lim_{L\to-\infty}(F(b)-F(L))\), provided it exists as a finite real number. The conclusion is precisely the defining limit, evaluated using an antiderivative.
Because \(F(b)\) is a constant, convergence requires \(F(L)\) to approach a finite limit as \(L\to-\infty\). Indeed, if \(F(b)-F(L)\) has a finite limit \(A\), then rearranging gives \(F(L)=F(b)-(F(b)-F(L))\), which also has a finite limit. Thus an unbounded \(F(L)\) cannot produce a convergent left-infinite integral. More generally, boundedness alone is not enough: if \(F(L)\) does not approach a limit, the cutoff integrals do not approach one either.
Worked Examples
Worked Example: A Finite Limit at the Left Endpoint
Evaluate \(\int_{-\infty}^{2} e^x\,dx\) from the definition.
First use a finite cutoff \(L<2\). An antiderivative is \(F(x)=e^x\), since \(F'(x)=e^x\). Therefore,
Now take the required limit. Since \(e^L\to 0\) as \(L\to-\infty\),
The limit is finite, so the improper integral converges to \(e^2\). As a check, differentiating \(F(x)=e^x\) gives \(F'(x)=e^x\), and the finite-cutoff expression \(e^2-e^L\) tends to the stated value.
Worked Example: A Rational Integrand with a Finite Limit
Evaluate \(\int_{-\infty}^{0}\frac{1}{(x-1)^2}\,dx\).
The function is continuous on every interval \([L,0]\), because \(x=1\) is outside the interval whenever \(L<0\). An antiderivative is \(F(x)=-\frac{1}{x-1}\), since
For a finite \(L<0\), the cutoff integral is
As \(L\to-\infty\), the fraction \(\frac{1}{L-1}\) tends to zero. Hence
The answer is finite. The derivative check confirms the antiderivative, and the cutoff expression has limit \(1\), as required by the definition.
Worked Example: An Antiderivative That Grows Without Bound
Determine whether \(\int_{-\infty}^{1}2x\,dx\) converges.
The integrand is continuous on each finite cutoff interval \([L,1]\). An antiderivative is \(F(x)=x^2\), because \(F'(x)=2x\). For finite \(L<1\),
As \(L\to-\infty\), \(L^2\to+\infty\), so \(1-L^2\to-\infty\). Thus the cutoff integrals do not approach a finite real number, and the improper integral diverges to negative infinity.
This result is consistent with the antiderivative criterion: \(F(L)=L^2\to+\infty\), so \(F(1)-F(L)\to-\infty\). Differentiating \(x^2\) recovers \(2x\), and the exact cutoff expression verifies the direction of divergence.
Convergence and Divergence: What the Limit Must Do
Convergence is not the same as having an antiderivative, or even as having finite values for every cutoff. It is a condition on the behavior of all the cutoff integrals as \(L\) moves left without bound.
| Behavior as \(L\to-\infty\) | Conclusion for the improper integral |
|---|---|
| \(F(b)-F(L)\) approaches a finite real number | The integral converges to that number. |
| \(F(L)\to+\infty\) | The cutoff integrals tend to \(-\infty\); the integral diverges. |
| \(F(L)\to-\infty\) | The cutoff integrals tend to \(+\infty\); the integral diverges. |
| \(F(L)\) has no limit | The cutoff integrals have no finite limit; the integral diverges. |
The table uses an antiderivative on the whole interval from arbitrarily far left to \(b\). The convergence test is always the limit in the definition. If an antiderivative is unavailable, or is inconvenient, the definition still applies directly to the cutoff integrals.
A Common Sign and Endpoint Mistake
A frequent error is to reverse the finite-interval evaluation and write \(F(L)-F(b)\) instead of \(F(b)-F(L)\). This reverses the sign of every cutoff integral. For example, with \(f(x)=e^x\) and \(b=0\), the incorrect calculation would be
But the finite-interval Fundamental Theorem gives the upper endpoint first and the lower endpoint second:
The correction is to preserve the order \(F(\text{upper endpoint})-F(\text{lower endpoint})\) at every finite cutoff, and only then take \(L\to-\infty\). A related mistake is to insert \(-\infty\) directly into \(F\). The endpoint is not a real input to the antiderivative; the limit of \(F(L)\) is what matters.
Why the Definition Matters
The cutoff definition makes the meaning of an infinite interval precise. It distinguishes a genuine finite integral value from a limit that grows without bound, and from a cutoff process that does not settle at all. The same discipline also prevents a misleading use of antiderivative notation: a symbolic expression at an infinite endpoint is valid only when it represents the defining limit.
When starting a left-infinite integral, keep the order of reasoning clear: identify the finite-cutoff integral, evaluate it on that finite interval if an antiderivative is available, and then examine its limit as the cutoff moves to negative infinity. This is the basic test for whether the improper integral has a finite value.
Check Your Understanding
Answer each question using the cutoff definition and, where useful, the finite-interval Fundamental Theorem of Calculus.
- Write the definition of \(\int_{-\infty}^{b}f(x)\,dx\), including the condition for convergence.
- If \(F'=f\), what finite-cutoff expression must be used before taking \(L\to-\infty\)?
- If \(F(L)\to+\infty\), what happens to \(F(b)-F(L)\), and what does that imply?
- Why is it not valid to substitute \(-\infty\) directly into an antiderivative?
- Does boundedness of \(F(L)\) alone guarantee convergence? Explain what additional behavior is needed.
