Improper Double Integrals Examples 1

Recall from the Improper Double Integrals page that if $f$ is a two variable function that does not change signs and either the region $D$ of integration is unbounded or the integrand is unbounded at a boundary or interior point of $D$, then the double integral of $f$ over $D$ can be computed as iterated integrals, one or both of which are improper integrals themselves.

We also made note that improper integral that change signs on their region of integration $D$ cannot be computed this way, and are much more complex to evaluate, so we will not consider any of those examples here.

Let's instead look at some examples of improper integral that we can actually evaluate.

Example 1

Evaluate the double integral $\iint_D \frac{1}{1 + x + y} \: dA$ where $D = \{ (x, y) \in \mathbb{R} : 0 ≤ x < \infty, 0 < y < 1 \}$.

We immediately rewrite the double integral above as iterated integrals to get that:

(1)

\begin{align} \quad \iint_D \frac{1}{1 + x + y} \: dA = \int_0^1 \int_0^{\infty} \frac{1}{1 + x + y} \: dx \: dy = \int_0^1 \lim_{b \to \infty} \int_0^{b} \frac{1}{1 + x + y} \: dx \: dy = \int_0^1 \lim_{b \to \infty} \left [ \ln \mid 1 + x + y \mid \right ]_{x=0}^{x=b} \: dy \\ = \int_0^1 \lim_{b \to \infty} \ln \mid 1 + b + y \mid - \ln \mid 1 + y \mid \: dy = \infty \end{align}

Therefore $\iint_D \frac{1}{1 + x + y} \: dA$ diverges.

Example 2

Evaluate the double integral $\iint_D 1 + \sin x \cos y \: dA$ where $D$ is the first quadrant.

If $D$ represents the first quadrant, then $D = \{ (x, y) \in \mathbb{R}^2 : 0 ≤ x < \infty, 0 ≤ y < \infty \}$ and so:

(2)

\begin{align} \quad \iint_D 1 + \sin x \cos y \: dA = \int_0^{\infty} \int_0^{\infty} 1 + \sin x \cos y \: dy \: dx = \int_0^{\infty} \lim_{b \to \infty} \int_0^b 1 + \sin x \cos y \: dy \: dx = \int_0^{\infty} \lim_{b \to \infty} \left [b + \sin x \sin b \right ]_{x=0}^{x=b} \: dx \\ = \int_0^{\infty} \lim_{b \to \infty} [b + \sin x \sin b] \: dx \end{align}

The interval above diverges, and so $\iint_D 1 + \sin x \cos y \: dA$ diverges.