Improper Integrals for MATH 122
Exam Relevance for MATH 122
Appears in the later part of the course. Convergence/divergence and evaluation.
This skill appears on:
Understanding Improper Integrals
So far, all your integrals have had nice, finite bounds and well-behaved functions. But what happens when:
- The interval extends to infinity? (like $\int_1^{\infty} \frac{1}{x^2} \, dx$)
- The function blows up somewhere in the interval? (like $\int_0^{1} \frac{1}{\sqrt{x}} \, dx$)
These are improper integrals. They require special handling because we can't just plug infinity into our antiderivative!
The key idea: Replace the "problem" with a limit, then evaluate.
Two Types of Improper Integrals
Type 1: Infinite Limits of Integration
The interval extends to $\infty$ or $-\infty$.
$$\int_a^{\infty} f(x) \, dx = \lim_{t \to \infty} \int_a^{t} f(x) \, dx$$
$$\int_{-\infty}^{b} f(x) \, dx = \lim_{t \to -\infty} \int_t^{b} f(x) \, dx$$
Type 2: Discontinuous Integrand
The function has a vertical asymptote at $x = c$ in $[a, b]$.
If the discontinuity is at $x = b$:
$$\int_a^{b} f(x) \, dx = \lim_{t \to b^-} \int_a^{t} f(x) \, dx$$
If the discontinuity is at $x = a$:
$$\int_a^{b} f(x) \, dx = \lim_{t \to a^+} \int_t^{b} f(x) \, dx$$
Convergent vs Divergent
- Convergent: The limit exists and equals a finite number
- Divergent: The limit is $\pm\infty$ or doesn't exist
Problem: Evaluate $\int_1^{\infty} \frac{1}{x^2} \, dx$
Step 1: Replace $\infty$ with a limit
$$\int_1^{\infty} \frac{1}{x^2} \, dx = \lim_{t \to \infty} \int_1^{t} \frac{1}{x^2} \, dx$$
Step 2: Evaluate the definite integral
$$= \lim_{t \to \infty} \left[ -\frac{1}{x} \right]_1^{t}$$
$$= \lim_{t \to \infty} \left( -\frac{1}{t} - \left(-\frac{1}{1}\right) \right)$$
$$= \lim_{t \to \infty} \left( -\frac{1}{t} + 1 \right)$$
Step 3: Evaluate the limit
As $t \to \infty$, $\frac{1}{t} \to 0$
$$= 0 + 1 = \boxed{1}$$
The integral converges to 1.
Problem: Evaluate $\int_1^{\infty} \frac{1}{x} \, dx$
Step 1: Replace $\infty$ with a limit
$$\int_1^{\infty} \frac{1}{x} \, dx = \lim_{t \to \infty} \int_1^{t} \frac{1}{x} \, dx$$
Step 2: Evaluate the definite integral
$$= \lim_{t \to \infty} \left[ \ln|x| \right]_1^{t}$$
$$= \lim_{t \to \infty} \left( \ln t - \ln 1 \right)$$
$$= \lim_{t \to \infty} \ln t$$
Step 3: Evaluate the limit
As $t \to \infty$, $\ln t \to \infty$
$$= \boxed{\text{Diverges}}$$
Problem: Evaluate $\int_{-\infty}^{\infty} \frac{1}{1+x^2} \, dx$
Step 1: Split at any convenient point
$$\int_{-\infty}^{\infty} \frac{1}{1+x^2} \, dx = \int_{-\infty}^{0} \frac{1}{1+x^2} \, dx + \int_{0}^{\infty} \frac{1}{1+x^2} \, dx$$
Step 2: Evaluate each piece
For the right piece:
$$\int_{0}^{\infty} \frac{1}{1+x^2} \, dx = \lim_{t \to \infty} \left[ \arctan x \right]_0^{t}$$
$$= \lim_{t \to \infty} (\arctan t - \arctan 0)$$
$$= \frac{\pi}{2} - 0 = \frac{\pi}{2}$$
For the left piece:
$$\int_{-\infty}^{0} \frac{1}{1+x^2} \, dx = \lim_{t \to -\infty} \left[ \arctan x \right]_t^{0}$$
$$= \arctan 0 - \lim_{t \to -\infty} \arctan t$$
$$= 0 - \left(-\frac{\pi}{2}\right) = \frac{\pi}{2}$$
Step 3: Add the pieces
$$= \frac{\pi}{2} + \frac{\pi}{2} = \boxed{\pi}$$
Problem: Evaluate $\int_0^{1} \frac{1}{\sqrt{x}} \, dx$
Step 1: Identify the problem
$f(x) = \frac{1}{\sqrt{x}}$ is undefined at $x = 0$ (blows up to $\infty$).
The discontinuity is at the lower limit.
Step 2: Replace with a limit
$$\int_0^{1} \frac{1}{\sqrt{x}} \, dx = \lim_{t \to 0^+} \int_t^{1} \frac{1}{\sqrt{x}} \, dx$$
Step 3: Evaluate the integral
$$= \lim_{t \to 0^+} \left[ 2\sqrt{x} \right]_t^{1}$$
$$= \lim_{t \to 0^+} \left( 2\sqrt{1} - 2\sqrt{t} \right)$$
$$= \lim_{t \to 0^+} \left( 2 - 2\sqrt{t} \right)$$
Step 4: Evaluate the limit
As $t \to 0^+$, $\sqrt{t} \to 0$
$$= 2 - 0 = \boxed{2}$$
Problem: Evaluate $\int_{-1}^{1} \frac{1}{x^2} \, dx$
Step 1: Identify the problem
$f(x) = \frac{1}{x^2}$ has a vertical asymptote at $x = 0$, which is inside $[-1, 1]$.
Step 2: Split at the discontinuity
$$\int_{-1}^{1} \frac{1}{x^2} \, dx = \int_{-1}^{0} \frac{1}{x^2} \, dx + \int_{0}^{1} \frac{1}{x^2} \, dx$$
Step 3: Evaluate each piece
For the right piece:
$$\int_{0}^{1} \frac{1}{x^2} \, dx = \lim_{t \to 0^+} \int_t^{1} \frac{1}{x^2} \, dx$$
$$= \lim_{t \to 0^+} \left[ -\frac{1}{x} \right]_t^{1}$$
$$= \lim_{t \to 0^+} \left( -1 + \frac{1}{t} \right)$$
As $t \to 0^+$, $\frac{1}{t} \to +\infty$
This piece diverges!
Step 4: Conclusion
If either piece diverges, the whole integral diverges.
$$\boxed{\text{Diverges}}$$
⚠️ Warning: You cannot just compute $\left[-\frac{1}{x}\right]_{-1}^{1} = -1 - 1 = -2$. This ignores the discontinuity and gives a wrong answer!
The p-Test for $\int_1^{\infty} \frac{1}{x^p} \, dx$
This integral comes up often:
$$\int_1^{\infty} \frac{1}{x^p} \, dx$$
Rule:
- Converges if $p > 1$
- Diverges if $p \leq 1$
Examples:
- $\int_1^{\infty} \frac{1}{x^2} \, dx$ → converges ($p = 2 > 1$)
- $\int_1^{\infty} \frac{1}{x} \, dx$ → diverges ($p = 1$)
- $\int_1^{\infty} \frac{1}{\sqrt{x}} \, dx$ → diverges ($p = 0.5 < 1$)
Common Mistakes and Misunderstandings
❌ Mistake: Ignoring discontinuities inside the interval
Wrong: $\int_{-1}^{1} \frac{1}{x^2} \, dx = \left[-\frac{1}{x}\right]_{-1}^{1} = -2$
Why it's wrong: There's a vertical asymptote at $x = 0$. You must split and check each piece.
Correct: Split at $x = 0$, evaluate both improper integrals. (This one diverges.)
❌ Mistake: Forgetting the limit notation
Wrong: Writing $\int_1^{\infty} \frac{1}{x^2} \, dx = \left[-\frac{1}{x}\right]_1^{\infty} = 0 + 1 = 1$
Why it's wrong: You can't plug $\infty$ directly into an expression.
Correct: Write $\lim_{t \to \infty} \left[-\frac{1}{x}\right]_1^{t}$, then evaluate the limit.
❌ Mistake: Not splitting $\int_{-\infty}^{\infty}$
Wrong: Treating $\int_{-\infty}^{\infty} f(x) \, dx$ as a single limit
Why it's wrong: Both ends need separate limits. If either diverges, the whole thing diverges.
Correct: Split into $\int_{-\infty}^{0} + \int_{0}^{\infty}$ and evaluate each with its own limit.
Improper Integral: Infinite Upper Limit
Replace the infinite upper limit with a variable t, then take the limit as t approaches infinity.
Variables:
- $a$:
- the finite lower limit
- $t$:
- temporary upper limit that approaches infinity
- $f(x)$:
- the function being integrated
Improper Integral: Infinite Lower Limit
Replace the infinite lower limit with a variable t, then take the limit as t approaches negative infinity.
Variables:
- $b$:
- the finite upper limit
- $t$:
- temporary lower limit that approaches negative infinity
- $f(x)$:
- the function being integrated
Improper Integral: Discontinuity at Endpoint
If f(x) has a vertical asymptote at x = b, approach it from the left with a limit. Similarly for discontinuity at x = a, approach from the right.
Variables:
- $a$:
- lower limit
- $b$:
- upper limit (where discontinuity occurs in this example)
- $t$:
- approaches the discontinuity from inside the interval
The p-Test
Converges if p > 1, diverges if p ≤ 1. Quick way to determine convergence without computing.
Variables:
- $p$:
- the exponent (p > 1 means convergent, p ≤ 1 means divergent)
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