Improper Double Integrals

# Improper Double Integrals

Just like in single variable calculus, we can sometimes encounter improper double integrals if we integrate a two variable real-valued function if our domain of integration is unbounded or if the function we are integrating is unbounded for some point in our domain.

We will only be dealing with improper integrals for which $f(x, y) ≥ 0, \forall (x, y) \in D$, or $f(x, y) ≤ 0, \forall (x, y) \in D$. If $f$ changes sign on $D$, then improper double integrals are much more complicated to evaluate, and their iterated integrals my converge when in actuality their double integral does not.

That said, evaluating improper double integrals of the simpler type specified above is much like evaluating regular improper integrals. For example, suppose that wanted to evaluate the integral $\iint_R \frac{1}{(1 + x^2)(1 + y^2)} \: dA$ where $R = \{ (x, y) : x ≥ 0, y ≥ 0 \}$. If we rewrite this double integral as iterated integrals, we would have that:

(1)
\begin{align} \quad \iint_R \frac{1}{(1 + x^2)(1 + y^2)} \: dA = \int_0^{\infty} \int_0^{\infty} \frac{1}{(1 + x^2)(1 + y^2)} \: dy \: dx \end{align}

If we evaluated in the inner improper integral $\int_0^{\infty} \frac{1}{(1 + x^2)(1 + y^2)} \: dy$ first while treating $x$ as fixed, we would get that:

(2)
\begin{align} \quad \int_0^{\infty} \frac{1}{(1 + x^2)(1 + y^2)} \: dy = \lim_{b \to \infty} \int_0^b \frac{1}{(1 + x^2)(1 + y^2)} \: dy = \lim_{b \to \infty} \left [ \frac{1}{1 + x^2} \tan^{-1}(y) \right ]_0^b = \frac{1}{1 + x^2} \cdot \lim_{b \to \infty} \tan^{-1} (b) = \frac{1}{1 + x^2} \cdot \frac{\pi}{2} \end{align}

Therefore we have that:

(3)
\begin{align} \quad \quad \iint_R \frac{1}{(1 + x^2)(1 + y^2)} \: dA = \int_0^{\infty} \frac{1}{1 + x^2} \cdot \frac{\pi}{2} \: dx = \frac{\pi}{2} \int_0^{\infty} \frac{1}{1 + x^2} \: dx = \frac{\pi}{2} \cdot \lim_{b \to \infty} \left [ \tan^{-1} (x) \right ]_0^b = \frac{\pi}{2} \cdot \lim_{b \to \infty} \tan^{-1} (b) = \frac{\pi}{2} \cdot \frac{\pi}{2} = \frac{\pi^2}{4} \end{align}

Therefore, $\iint_R \frac{1}{(1 + x^2)(1 + y^2)} \: dA$ converges to $\frac{\pi^2}{4}$. The graph of $z = \frac{1}{(1 + x^2)(1 + y^2)}$ in the first quadrant is given below: