Improper Multiple Integrals
§4.8 - Convergence over unbounded regions and near singularities
The surface $z = \dfrac{1}{x^2 + y^2}$ explodes to infinity at the origin - yet the volume it encloses over an expanding disk is finite. How?
1D Review: Convergent vs. Divergent
Recall from single-variable calculus: $\displaystyle\int_1^\infty \frac{1}{x^p}\,dx$ converges iff $p > 1$. Watch the area accumulate as we push the upper limit $R \to \infty$.
Unbounded Regions
For double integrals over all of $\mathbb{R}^2$, we integrate over a growing disk of radius $R$ and take the limit. The Gaussian integral is the classic example:
Watch the volume converge to $\pi$ as $R$ grows:
Going Up a Dimension - When Harder Becomes Easier
We just saw that $\iint_{\mathbb{R}^2}e^{-(x^2+y^2)}\,dA = \pi$. But how do we know that? The function $e^{-x^2}$ has no elementary antiderivative - yet we can evaluate $\int_0^\infty e^{-x^2}\,dx$ exactly.
The strategy is called dimensional promotion - deliberately going up a dimension to exploit symmetry that doesn't exist in the original problem. The 1D integral $\int_0^\infty e^{-x^2}\,dx$ has no useful symmetry. But if we could somehow get to the 2D integral $\iint e^{-(x^2+y^2)}\,dx\,dy$, the exponent $x^2 + y^2 = r^2$ has circular symmetry - and polar coordinates would crack it open.
The question is: how do we get from a 1D integral to a 2D one? Think about that for a moment before reading on. What operation on $I = \int_0^\infty e^{-x^2}\,dx$ would naturally produce a double integral?
And doubling gives the full-line Gaussian: $\int_{-\infty}^{\infty}e^{-x^2}\,dx = \sqrt{\pi}$. That $\frac{\sqrt{\pi}}{2}\approx 0.886$ governs every normal distribution in statistics - the bell curve's area is finite precisely because of this result.
The key lesson: Going up a dimension made the problem easier, not harder. That's rare and beautiful. The Jacobian factor $r$ in $r\,dr\,d\theta$ is what saves the day - without it, the polar integral would be just as impossible as the original. That extra $r$ turns the integrand into something with a simple u-substitution. Work through the full derivation yourself - the steps are short and satisfying.
Singularities
When the integrand blows up inside the domain, we cut out a small hole of radius $\varepsilon$ and take $\varepsilon \to 0$. A key result in polar coordinates on the unit disk:
This converges iff $p < 2$. Compare $p = 1$ (converges) vs $p = 2$ (diverges):
Value $=$ 0 Converges to $2\pi$
Value $=$ 0 Diverges
Comparison Test
If $0 \le f(x,y) \le g(x,y)$ on a region $R$, then:
Playground: Explore Convergence
Enter a function and watch the integral accumulate over an expanding domain.
Practice Problems (§4.8)
From Kaplan, problems after §4.8
Step 1: Convert to spherical coordinates
In spherical coordinates, $dV = r^2\sin\phi\,dr\,d\phi\,d\theta$ and the integrand is $r^{-p}$:
$$\iiint_R r^{-p}\,dV = \int_0^{2\pi}\!\int_0^{\pi}\!\int_0^1 r^{-p}\cdot r^2\sin\phi\,dr\,d\phi\,d\theta$$Step 2: Separate the integrals
The angular part factors out:
$$= \underbrace{\int_0^{2\pi}d\theta}_{2\pi}\cdot\underbrace{\int_0^{\pi}\sin\phi\,d\phi}_{2}\cdot\int_0^1 r^{2-p}\,dr = 4\pi\int_0^1 r^{2-p}\,dr$$The radial integral $\int_0^1 r^{2-p}\,dr$ converges iff $2-p > -1$, i.e., $p \lt 3$.
Step 3: Evaluate
For $p \lt 3$:
$$\int_0^1 r^{2-p}\,dr = \frac{r^{3-p}}{3-p}\bigg|_0^1 = \frac{1}{3-p}$$Step 1: Convert to polar
Set $x = r\cos\theta$, $y = r\sin\theta$. Then $\log(x^2+y^2) = \log(r^2) = 2\log r$ and $\sqrt{x^2+y^2} = r$:
$$\iint_R \frac{\log(x^2+y^2)}{\sqrt{x^2+y^2}}\,dA = \int_0^{2\pi}\!\int_0^1 \frac{2\log r}{r}\cdot r\,dr\,d\theta = 4\pi\int_0^1 \log r\,dr$$Step 2: Evaluate the radial integral
Integration by parts with $u = \log r$, $dv = dr$:
$$\int_0^1 \log r\,dr = \bigl[r\log r - r\bigr]_0^1$$At $r=1$: $1\cdot 0 - 1 = -1$. At $r \to 0^+$: $r\log r \to 0$ (by L'Hopital) and $r \to 0$. So the integral equals $-1$.