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What is Calculus II · Integration

an Improper Integral?

An integral that breaks the rules — either with infinite limits of integration or a function that blows up somewhere in the interval.

Simple Definition

An improper integral is an integral where either:

The interval of integration is infinite (e.g., $\int_1^\infty \frac{1}{x^2} dx$), or
The integrand has a discontinuity (vertical asymptote) within the interval (e.g., $\int_0^1 \frac{1}{\sqrt{x}} dx$)

In both cases, the standard definition of the definite integral (Riemann sum over a finite interval with a bounded function) doesn't apply. Instead, we use limits to define these integrals.

Improper integrals are convergent if the limit exists (finite) and divergent if the limit is infinite or does not exist.

Type 1

$\displaystyle\int_a^\infty f(x)\,dx$

$\displaystyle\int_{-\infty}^b f(x)\,dx$

Type 2

$\displaystyle\int_a^b f(x)\,dx$

with vertical asymptote

→

Evaluate with

Limits

Intuition

🏃

Running to infinity

Imagine you're running an infinite race. The improper integral asks: "If you keep running forever, is there a finite total distance you cover, or do you run infinitely far?" For $\int_1^\infty \frac{1}{x^2} dx$, the answer is finite (the area under the curve is 1). For $\int_1^\infty \frac{1}{x} dx$, the answer is infinite — you run forever and cover infinite distance.

💥

Blowing up

Now imagine walking toward a cliff. The function $\frac{1}{\sqrt{x}}$ blows up to infinity as you approach $x=0$. The improper integral $\int_0^1 \frac{1}{\sqrt{x}} dx$ asks: "Is the area under this curve finite even though the function is infinite at the endpoint?" Surprisingly, it is finite (the area is 2). But $\int_0^1 \frac{1}{x} dx$ is infinite — it blows up too fast.

Formal Definition

Type 1 — Infinite Intervals:

Infinite upper limit

$$\int_a^\infty f(x)\,dx = \lim_{b \to \infty} \int_a^b f(x)\,dx$$

Infinite lower limit

$$\int_{-\infty}^b f(x)\,dx = \lim_{a \to -\infty} \int_a^b f(x)\,dx$$

Both limits infinite

$$\int_{-\infty}^{\infty} f(x)\,dx = \lim_{a \to -\infty} \int_a^c f(x)\,dx + \lim_{b \to \infty} \int_c^b f(x)\,dx$$

Type 2 — Discontinuities:

Vertical asymptote at $x = a$

$$\int_a^b f(x)\,dx = \lim_{t \to a^+} \int_t^b f(x)\,dx$$

Vertical asymptote at $x = b$

$$\int_a^b f(x)\,dx = \lim_{t \to b^-} \int_a^t f(x)\,dx$$

Vertical asymptote at $x = c$ (inside interval)

$$\int_a^b f(x)\,dx = \lim_{t \to c^-} \int_a^t f(x)\,dx + \lim_{s \to c^+} \int_s^b f(x)\,dx$$

Types of Improper Integrals

Type 1A

$\int_a^\infty f(x)\,dx$

Infinite upper limit. Integrate to $b$, then take $b \to \infty$.

Type 1B

$\int_{-\infty}^b f(x)\,dx$

Infinite lower limit. Integrate from $a$ to $b$, then take $a \to -\infty$.

Type 2A

Asymptote at $a$

$\lim_{x \to a^+} f(x) = \pm\infty$. Integrate from $t$ to $b$, then $t \to a^+$.

Type 2B

Asymptote at $b$

$\lim_{x \to b^-} f(x) = \pm\infty$. Integrate from $a$ to $t$, then $t \to b^-$.

Mixed Types

$\int_{-\infty}^{\infty} \frac{1}{x^2+1}\,dx$

Both limits infinite, integrand continuous everywhere. Split at a convenient point (e.g., 0) and evaluate each part separately.

Convergence Tests

Sometimes it's hard to evaluate an improper integral exactly. Use these tests to determine if it converges (finite) or diverges (infinite):

Comparison Test	If $0 \le f(x) \le g(x)$ and $\int g$ converges, then $\int f$ converges. If $\int f$ diverges, then $\int g$ diverges.
Limit Comparison	If $\lim_{x \to \infty} \frac{f(x)}{g(x)} = L$ where $0 < L < \infty$, then $\int f$ and $\int g$ converge or diverge together.
$p$-Test (Type 1)	$\int_1^\infty \frac{1}{x^p}\,dx$ converges if $p > 1$, diverges if $p \le 1$.
$p$-Test (Type 2)	$\int_0^1 \frac{1}{x^p}\,dx$ converges if $p < 1$, diverges if $p \ge 1$.

Quick Examples

Example 1 — Convergent Type 1

$\displaystyle\int_1^\infty \frac{1}{x^2}\,dx$

$= \lim_{b \to \infty} \left[-\frac{1}{x}\right]_1^b$
$= \lim_{b \to \infty} \!\left(1 - \frac{1}{b}\right) = 1$

✅ Converges to 1

Example 2 — Divergent Type 1

$\displaystyle\int_1^\infty \frac{1}{x}\,dx$

$= \lim_{b \to \infty} [\ln x]_1^b$
$= \lim_{b \to \infty} \ln b = \infty$

❌ Diverges to $\infty$

Example 3 — Convergent Type 2

$\displaystyle\int_0^1 \frac{1}{\sqrt{x}}\,dx$

$= \lim_{t \to 0^+} \bigl[2\sqrt{x}\bigr]_t^1$
$= \lim_{t \to 0^+} (2 - 2\sqrt{t}) = 2$

✅ Converges to 2

Example 4 — Divergent Type 2

$\displaystyle\int_0^1 \frac{1}{x}\,dx$

$= \lim_{t \to 0^+} [\ln x]_t^1$
$= \lim_{t \to 0^+} (0 - \ln t) = \infty$

❌ Diverges to $\infty$

Example 5 — Full real line

$\displaystyle\int_{-\infty}^{\infty} \frac{1}{x^2+1}\,dx$

$= \lim_{a \to -\infty} [\tan^{-1}x]_a^0 + \lim_{b \to \infty} [\tan^{-1}x]_0^b$
$= \dfrac{\pi}{2} + \dfrac{\pi}{2} = \pi$

✅ Converges to $\pi$

Common Mistakes

✗ Treating ∞ as a number

You cannot "plug in" $\infty$ directly. Always use limits: $\int_1^\infty f(x)\,dx = \lim_{b \to \infty} \int_1^b f(x)\,dx$. The limit must be evaluated properly.

✗ Ignoring discontinuities inside the interval

If the integrand has a vertical asymptote at $x = c$ where $a < c < b$, you must split at $c$: $\int_a^b = \int_a^c + \int_c^b$. Both parts must converge for the whole integral to converge.

✗ Mixing up the p-test conditions

Type 1 (upper limit $\infty$): $\int_1^\infty x^{-p}\,dx$ converges if $p > 1$. Type 2 (near 0): $\int_0^1 x^{-p}\,dx$ converges if $p < 1$. These are opposite conditions — the boundary $p=1$ always diverges in both cases.

✗ Applying the Comparison Test in the wrong direction

If $f \le g$ and $\int g$ converges → $\int f$ converges. If $f \le g$ and $\int f$ diverges → $\int g$ diverges. No conclusions are possible if $\int g$ diverges while trying to bound $\int f$ above.

Related Concepts

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Improper integrals appear in probability, physics, and advanced calculus. Practice with our quizzes.

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Comparison Test	If \(0 \le f(x) \le g(x)\) and \(\int g\) converges, then \(\int f\) converges. If \(\int f\) diverges, then \(\int g\) diverges.
Limit Comparison	If \(\lim_{x \to \infty} \frac{f(x)}{g(x)} = L\) where \(0 < L < \infty\), then \(\int f\) and \(\int g\) converge or diverge together.
\(p\)-Test (Type 1)	\(\int_1^\infty \frac{1}{x^p}\,dx\) converges if \(p > 1\), diverges if \(p \le 1\).
\(p\)-Test (Type 2)	\(\int_0^1 \frac{1}{x^p}\,dx\) converges if \(p < 1\), diverges if \(p \ge 1\).