Type 1: Infinite Limits
Type 1: Infinite Limits of Integration
Textbook Reference
| Primary source | OpenStax Calculus Volume 2, Section 3.7: “Improper Integrals” |
| Direct link | https://openstax.org/books/calculus-volume-2/pages/3-7-improper-integrals |
| Textbook used in class | Stewart, Calculus, Section 7.8: “Improper Integrals” |
Opening Scenario
The definite integral $\int_a^b f(x)\,dx$ was defined for finite $a$ and $b$ with $f$ continuous on $[a,b]$. What if the upper limit is $\infty$? The function $1/x^2$ approaches zero as $x \to \infty$, and geometrically the area under the curve on $[1, \infty)$ should be finite (the rectangles shrink fast enough). But the definition does not allow $b = \infty$. The fix: define the integral over an infinite interval as a limit of ordinary integrals over $[a, t]$ as $t \to \infty$.
Quick Reference
Type 1 improper integrals (infinite limits):
$$\int_a^{\infty} f(x)\,dx = \lim_{t\to\infty} \int_a^t f(x)\,dx \quad \text{(if the limit exists)}.$$
$$\int_{-\infty}^b f(x)\,dx = \lim_{t\to -\infty} \int_t^b f(x)\,dx.$$
$$\int_{-\infty}^{\infty} f(x)\,dx = \int_{-\infty}^c f(x)\,dx + \int_c^{\infty} f(x)\,dx \quad \text{(split at any finite } c\text{)}.$$
Convergence/divergence: An improper integral converges if the limit exists and is finite. It diverges if the limit is infinite or does not exist.
$p$-test (important special case): $$\int_1^{\infty}\frac{1}{x^p}\,dx \text{ converges if } p > 1, \text{ diverges if } p \leq 1.$$
Key Concepts
1. The Definition
$\displaystyle\int_1^{\infty}\frac{1}{x^2}\,dx$ is not an ordinary integral -- the upper limit is not a number. Define it as:
$$\int_1^{\infty}\frac{1}{x^2}\,dx = \lim_{t\to\infty}\int_1^t\frac{1}{x^2}\,dx = \lim_{t\to\infty}\left[-\frac{1}{x}\right]_1^t = \lim_{t\to\infty}\left(-\frac{1}{t}+1\right) = 1.$$
The integral converges to 1. Geometrically, the infinite region under $y = 1/x^2$ has finite area 1.
2. The $p$-Test
For power functions on $[1, \infty)$:
$$\int_1^{\infty}\frac{dx}{x^p} = \lim_{t\to\infty}\left[\frac{x^{1-p}}{1-p}\right]_1^t \quad (p \neq 1).$$
- $p > 1$: $\lim_{t\to\infty} t^{1-p} = 0$ (since $1-p < 0$), so the limit is $\dfrac{0 - 1}{1-p} = \dfrac{1}{p-1}$. Converges.
- $p < 1$: $\lim_{t\to\infty} t^{1-p} = \infty$. Diverges.
- $p = 1$: $\int_1^t \frac{dx}{x} = \ln t \to \infty$. Diverges.
Key borderline case: $\displaystyle\int_1^{\infty}\frac{1}{x}\,dx$ diverges, even though $1/x \to 0$ as $x \to \infty$.
3. Both Limits at Infinity
$$\int_{-\infty}^{\infty} f(x)\,dx = \int_{-\infty}^0 f(x)\,dx + \int_0^{\infty} f(x)\,dx.$$
Both halves must converge separately. If either diverges, the whole integral diverges. Do not use the “cancellation” shortcut $\lim_{t\to\infty}\int_{-t}^t f(x)\,dx$ as the definition -- this is called the Cauchy principal value and differs from the standard improper integral when both halves diverge.
“$\int_{-\infty}^{\infty} f(x)\,dx = 0$ when $f$ is odd.” For an odd function, $\int_{-t}^t f(x)\,dx = 0$ for all $t$. But if both $\int_{-\infty}^0 f$ and $\int_0^{\infty} f$ diverge individually, the integral $\int_{-\infty}^{\infty} f\,dx$ diverges. Example: $\int_{-\infty}^{\infty} x\,dx$ diverges (both $\int_0^{\infty} x\,dx$ and $\int_{-\infty}^0 x\,dx$ diverge).
Worked Example
Evaluate $\displaystyle\int_0^{\infty} e^{-x}\,dx$.
$\displaystyle\int_0^{\infty} e^{-x}\,dx = \lim_{t\to\infty}\int_0^t e^{-x}\,dx = \lim_{t\to\infty}\left[-e^{-x}\right]_0^t = \lim_{t\to\infty}(-e^{-t}+1) = 0 + 1 = 1$.
Converges to 1. This is the simplest case of the Gamma function: $\Gamma(1) = \int_0^{\infty} e^{-x}\,dx = 1$.
Evaluate $\displaystyle\int_1^{\infty}\frac{dx}{\sqrt{x}}$.
$\displaystyle\lim_{t\to\infty}\int_1^t x^{-1/2}\,dx = \lim_{t\to\infty}\left[2x^{1/2}\right]_1^t = \lim_{t\to\infty}(2\sqrt{t}-2) = \infty$.
Diverges. ($p = 1/2 < 1$ in the $p$-test.)
Common Errors Summary
| Error | Example | Correction |
|---|---|---|
| Evaluating at $\infty$ directly | Writing $\left[-\frac{1}{x}\right]_1^{\infty} = -\frac{1}{\infty}+\frac{1}{1}$ | Use the limit: $\lim_{t\to\infty}\left[-\frac{1}{x}\right]_1^t = \lim_{t\to\infty}(-\frac{1}{t}+1) = 1$ |
| Using cancellation for $\int_{-\infty}^{\infty}$ | $\int_{-\infty}^{\infty}x\,dx = 0$ by symmetry | Split at 0; each half diverges, so the integral diverges |
| Applying the $p$-test to $\int_0^1$ instead of $\int_1^{\infty}$ | Claiming $\int_0^1\frac{dx}{x^2}$ converges because $p=2>1$ | The $p$-test for $\int_1^{\infty}$ says $p>1$ converges. For $\int_0^1$, the rule reverses: $p<1$ converges, $p\geq 1$ diverges (Type 2 improper integral) |
Leveled Practice
Level 1 -- Basic Convergence
Problem 1. Determine whether $\displaystyle\int_2^{\infty}\frac{1}{x^3}\,dx$ converges, and if so, find its value.
Show answer
$p = 3 > 1$: converges. $\displaystyle\lim_{t\to\infty}\left[-\frac{1}{2x^2}\right]_2^t = \lim_{t\to\infty}\left(-\frac{1}{2t^2}+\frac{1}{8}\right) = \frac{1}{8}$.
Level 2 -- Divergence
Problem 2. Show that $\displaystyle\int_1^{\infty}\frac{1}{x}\,dx$ diverges.
Show answer
$\displaystyle\lim_{t\to\infty}\int_1^t\frac{dx}{x} = \lim_{t\to\infty}\ln t = \infty$. Diverges.
Level 3 -- Both Limits
Problem 3. Evaluate $\displaystyle\int_{-\infty}^{\infty}\frac{1}{1+x^2}\,dx$.
Show answer
Split at 0:
$\displaystyle\int_{-\infty}^0\frac{dx}{1+x^2} = \lim_{t\to-\infty}\arctan(x)\Big|_t^0 = 0 - (-\pi/2) = \pi/2$.
$\displaystyle\int_0^{\infty}\frac{dx}{1+x^2} = \lim_{t\to\infty}\arctan(x)\Big|_0^t = \pi/2 - 0 = \pi/2$.
Total: $\pi/2 + \pi/2 = \pi$.
Mastery Checklist
Mental Model
An improper integral with an infinite upper limit is like a hotel with infinitely many rooms. Some integrals accumulate finite total area as $t \to \infty$ (the room sizes shrink fast enough); others accumulate infinite total area (the rooms shrink too slowly). The limit is the formal way to ask: “Does the accumulation ever stabilize?”
The $p$-test gives the threshold: $p > 1$ means the function decays fast enough, $p \leq 1$ means it does not.
Connections
Looking back
- Limits at infinity (Section 3.4): The limit $\lim_{t\to\infty} F(t)$ is evaluated using the techniques of Chapter 3.
- FTC Part 2 (Section 4.3): Used to evaluate $\int_a^t f\,dx = F(t) - F(a)$ before taking the limit.
Looking ahead
- Discontinuous integrands (Section 7.8): Type 2 improper integrals, handled with the same limit approach.
- Comparison test (Section 7.8): When an antiderivative is unavailable, compare with a known convergent or divergent integral.
Back to Techniques of Integration | Next: Type 2 -- Discontinuous Integrands