Improper Integrals

Foundation refresher - §4.1 review, Kaplan

1-D Improper Integrals - The Refresher You Need

Single-variable improper integrals come in two flavors: the function blows up (e.g. $\int_0^1 \frac{1}{\sqrt{x}}\,dx$) or the interval is infinite (e.g. $\int_1^\infty \frac{1}{x^2}\,dx$). In both cases, the integral is defined as a limit:

Singular integrand at $a$:

$$\int_a^b f(x)\,dx \;=\; \lim_{c\to a^+}\int_c^b f(x)\,dx.$$

Infinite interval:

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

If the limit exists and is finite, the improper integral converges; otherwise it diverges. Both cases are reviewed thoroughly on the dedicated chapter page §4.8 Improper Multiple Integrals, which opens with this 1-D case before moving to higher dimensions. The two cards below cover what doesn't show up there: the critical $p$-pattern and two integrals worth memorizing.

The Critical $p$ Pattern

Let's put the two results side by side:

Near a singularity ($x \to 0^+$)
$$\int_0^1 \frac{dx}{x^p} \begin{cases} = \frac{1}{1-p} & p < 1 \\[4pt] \text{diverges} & p \ge 1 \end{cases}$$
At infinity ($x \to \infty$)
$$\int_1^\infty \frac{dx}{x^p} \begin{cases} = \frac{1}{p-1} & p > 1 \\[4pt] \text{diverges} & p \le 1 \end{cases}$$

The critical exponent for 1D is $p = 1$. This is worth memorizing. But when you move to double integrals (§4.8), something shifts: $\int\!\int \frac{dA}{r^p}$ over a disk has critical exponent $p = 2$. In $n$ dimensions, the critical exponent is $p = n$. The extra dimensions contribute enough "room" that the integrand can decay more slowly and still converge.

Two Famous Improper Integrals

These show up everywhere in probability, physics, and the rest of this course. Both are Type 2 (infinite limits), and both converge even though it's not obvious at first glance.

The arctangent integral:

$$\int_0^\infty \frac{dx}{1 + x^2} = \lim_{b \to \infty} \arctan b = \frac{\pi}{2}$$

We know $\frac{d}{dx}\arctan x = \frac{1}{1+x^2}$, and $\arctan b \to \pi/2$ as $b \to \infty$. Clean and simple.

The Gaussian integral:

$$\int_{-\infty}^{\infty} e^{-x^2}\,dx = \sqrt{\pi}$$

This one is remarkable because $e^{-x^2}$ has no elementary antiderivative. The trick - which you'll see in §4.8 - is to square the integral, convert to a double integral, and switch to polar coordinates: $I^2 = \int\!\int e^{-(x^2+y^2)}\,dx\,dy = \int_0^{2\pi}\!\int_0^\infty e^{-r^2}\,r\,dr\,d\theta = \pi$. So $I = \sqrt{\pi}$.

Below, both integrals are computed numerically over $[0, b]$ as $b$ grows. Watch them converge to their exact values.

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