# Improper Integrals

Mathematics I · Integral Calculus · https://tryals.app/learn/mathematics-i/improper-integrals

## Integrating to Infinity

An ordinary definite integral has finite limits. An **improper integral** relaxes this: either the **interval is infinite** ($\pm\infty$) or the **integrand is unbounded** (a vertical asymptote).

Improper integrals are defined as **limits** of proper ones. For an infinite upper limit:

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

If the limit exists and is finite, the integral **converges**; otherwise it **diverges**.

| Term | Condition | Result |
| --- | --- | --- |
| **Convergent** | Limit is a finite number | Finite area/value |
| **Divergent** | Limit is $\pm\infty$ or DNE | Infinite/undefined |

**Common pitfall:** assuming that because the integrand goes to zero, the integral must **converge**. Decay to zero is necessary but not sufficient.

## The $p$-Test & Comparison

Behavior depends delicately on decay speed. Consider the family $\int_1^{\infty} \frac{1}{x^p}\, dx$:

| Parameter $p$ | Integral Behavior | Reason |
| --- | --- | --- |
| $p > 1$ | **Converges** | Tail shrinks fast enough |
| $p \le 1$ | **Diverges** | Tail shrinks too slowly |

For example, $\int_1^{\infty} \frac{1}{x^2}\, dx = 1$ (finite), but $\int_1^{\infty} \frac{1}{x}\, dx = \infty$, even though both integrands approach zero.

When exact evaluation fails, use **comparison**: if $0 \le f \le g$ and $\int g$ converges, $\int f$ converges too. If $\int f$ diverges, $\int g$ diverges.

## Practice questions

5 of this lesson's 11 practice questions, with answers. The full set is in the app.

### 1. Complete the statement about decay and convergence.

**Answer:** For an improper integral over an infinite interval, the integrand tending to zero is **necessary but not sufficient** for convergence.

**Why:** 1/x tends to zero, yet its integral from 1 to infinity diverges, so decay is not sufficient. It is necessary in the sense that a non-decaying tail cannot give a finite total. What matters is _how fast_ the decay happens.

Page: https://tryals.app/practice/mathematics-i/improper-integrals/complete-the-statement-about-decay-and-convergence

### 2. ∫₀^∞ e^(−x²) dx converges to √π/2, a finite answer involving π, for a function with no elementary antiderivative. Why does the impossibility of the indefinite integral not block the definite one?

A. Because the antiderivative does exist but has not yet been discovered
B. Because the definite integral is a limit of areas, not an antiderivative
C. Because the limits being 0 and infinity makes the integrand elementary
D. Because e^(−x²) is elementary, so its antiderivative must be too

**Answer:** B. Because the definite integral is a limit of areas, not an antiderivative

**Why:** The integral is defined as a limit of sums and exists whether or not a formula does, the Fundamental Theorem is a convenience, not the definition. This one is evaluated by a polar-coordinate trick in two dimensions, and its value is why the normal distribution's constant contains a √π.

Page: https://tryals.app/practice/mathematics-i/improper-integrals/e-x-dx-converges-to-2-a-finite-answer-involving-for-a

### 3. ∫₁^∞ (1/x²) dx converges to 1 while ∫₁^∞ (1/x) dx diverges, although both integrands shrink to zero. Why does going to zero not settle the question?

A. Because 1/x diverges only at the lower end, and its behaviour at infinity is irrelevant.
B. Because the two integrals have different lower limits, which changes convergence.
C. Because 1/x is undefined at zero while 1/x² is not, which is what causes the divergence.
D. Because what matters is how fast the integrand shrinks.

**Answer:** D. Because what matters is how fast the integrand shrinks.

**Why:** Evaluate both to b: 1 − 1/b approaches 1, and ln b grows without bound, slowly and without stopping. Shrinking to zero is necessary and not sufficient, and the boundary is sharp, ∫x^(−p) converges exactly when p > 1. It is the harmonic series again, connected directly by the integral test, and it is why Gabriel's horn has infinite area and finite volume.

Page: https://tryals.app/practice/mathematics-i/improper-integrals/1-x-dx-converges-to-1-while-1-x-dx-diverges-although

### 4. Select every integral that is improper.

A. x^2 dx from 0 to 1
B. 1/x dx from 1 to 2
C. e^(-x) dx from 1 to infinity
D. e^x dx from minus infinity to 0
E. 1/x dx from 0 to 1

**Answer:** C. e^(-x) dx from 1 to infinity; D. e^x dx from minus infinity to 0; E. 1/x dx from 0 to 1

**Why:** 1/x blows up at the left endpoint 0, so the first is improper; the last is perfectly ordinary because 1/x is bounded on [1,2]. Options 3 and 4 have infinite limits. A bounded integrand on a bounded interval is never improper.

Page: https://tryals.app/practice/mathematics-i/improper-integrals/select-every-integral-that-is-improper

### 5. If f(x) tends to 0 as x tends to infinity, then the integral of f from 1 to infinity must converge.

**Answer:** False

**Why:** False. 1/x tends to zero yet its integral from 1 to infinity is infinite. Convergence depends on the _rate_ of decay, not on the mere fact of it, this is the pitfall the lesson singles out, and the same one that catches students in the theory of series.

Page: https://tryals.app/practice/mathematics-i/improper-integrals/if-f-x-tends-to-0-as-x-tends-to-infinity-then-the-integral-of-f