Improper Integrals: Basics

Definition

In general, an Improper Integral is a definite integral over an infinite area. It's a definite integral that doesn't meet the criteria for definite integrals.

Types of Improper Integrals

An integral is improper if it has any of the following boundary conditions (remember: a definite integral is the bounded area between the curve of $f(x)$ and the x-axis):

Type I: Infinite Integration Limits

• $b = \infty$
• $a = -\infty$
• $a = -\infty$ and $b = \infty$

Type II: Discontinuous Integrands

$f(x)$ isn't bounded in $[a, b]$

Typically, this means that $f(x)$ has one or more vertical asymptotes in $[a, b]$

How to Handle Improper Integrals

The general approach to integrating improper integrals uses limits to work around the discontinuities in the integrand's boundary.

In general, you follow these steps:

1. Define a small margin $h$ near the discontinuity at $x = c$ in the interval $[a, b]$. Now you have a window, $c + h$, between the discontinuity and the area under the curve.
2. Sneak up on the discontinuity by letting $c + h$ shrink towards $c$.
3. Take the limit as $h\to 0$

Specific techniques for integrating improper integrals are described in the next note,
7.8 Improper Integration Techniques.

Convergence & Divergence

When you insert a margin between the area under the $f(x)$ curve and the discontinuity, and then take the limit as the width of the margin shrinks to zero, one of two things will happen:

1. The limit exists, so the integral converges.
2. The limit does not exist, so the integral diverges.

Only convergent integrals have a finite area. All divergent integrals have infinite area, which cannot be measured.

The problem [convergence vs divergence] centers around whether $h\to 0$ faster than $f(c\pm h)\to\infty$.

--Prof Herbert Gross, Calculus Revisited

MIT OpenCourseware

Integration Techniques for Improper Integrals

Infinite Integration Limits (Type I)

To handle an integrand $f(x)$ which is continuous over the entire interval of integration, but has $\pm \infty$ as one of both of the integration limits:

1. Choose a massive number $N$ and integrate over $[a, N]$
2. Take the limit as $N\to \infty$ and see what happens to the value of the integral.

$$\int_a^{\infty} f(x)dx = \lim_{a\to N} \int_a^N f(x)dx$$

Discontinuous Integrands (Type II)

To handle an integral which has a discontinuity at either end, $x = a$ or $x = b$ (assume $\int_a^bf(x)dx$ has a vertical asymptate at $a$ here:

1. Choose a small number $h$ and integrate over $[a + h, b]$
2. Take the limit as $h\to 0$ and see what happens to the value of the integral.

If $f(x)$ is unbounded for $x$ near $a$ only: $$\int_a^b f(x) dx = \lim_{h\to 0} \int_{a+h}^b f(x) dx$$

If $f(x)$ is unbounded for $x$ near $b$ only: $$\int_a^b f(x) dx = \lim_{h\to 0} \int_a^{b+h} f(x) dx$$

More comprehensive techniques, including consideration for particular types of functions is in: 7.8 Improper Integration Techniques.

Examples

Divergence

Consider the integral $$\int_0^1 \frac{1}{x}dx$$

It is improper because it has a vertical asymptote at $x = 0$. So, use the algorithm for discontinuous integrands:

$$\int_0^1 \frac{1}{x} dx = \lim_{\varepsilon \to 0^+} \int_\varepsilon^1 \frac{1}{x} dx = \lim_{\varepsilon\to 0^+} \left\{\ln|x|\right\}_{\varepsilon}^1$$$$= \lim_{\varepsilon\to 0^+} \left[ \ln(1) - \ln(\varepsilon) \right] = \infty$$