Improper Integrals

Improper Integrals

Suppose that we have a function $f$ containing an infinite discontinuity, and we want to calculate the area of the integral containing this continuity. For example, suppose that we want to calculate the area of the function $f(x) = \frac{1}{x^2}$ on the interval $\int_a^{\infty} f(x) \: dx$ where $a > 0$, or more appropriately, the following shaded region:


We might suspect that this area is infinite, but in fact, we will see that the area from $x = a$ to $\infty$ is actually finite. We can evaluate this integral with limits in the following manner:

\begin{align} \int_a^{\infty} \frac{1}{x^2} \: dx = \lim_{b \to \infty} \int_a^b \frac{1}{x^2} \: dx \\ \int_a^{\infty} \frac{1}{x^2} \: dx = \lim_{b \to \infty} - \frac{1}{x} \big |_{a}^{b} \\ \int_a^{\infty} \frac{1}{x^2} \: dx = - \lim_{b \to \infty} \frac{1}{x} \big |_{a}^{b} \\ \int_a^{\infty} \frac{1}{x^2} \: dx = - \lim_{b \to \infty} \frac{1}{b} - \frac{1}{a} \\ \int_a^{\infty} \frac{1}{x^2} \: dx = \frac{1}{a} \\ \end{align}

So in fact, the area under bounded by the curve $f(x) = \frac{1}{x^2}$ on the interval $1$ to $\infty$ is finite, precisely $\frac{1}{a}$. This is not necessarily always the case though. Let's look at the very similar curve, $g(x) = \frac{1}{x}$ on the interval $[a, \infty)$, where $a > 0$. The math is very similar:

\begin{align} \int_a^{\infty} \frac{1}{x} \: dx = \lim_{b \to \infty} \int_a^b \frac{1}{x} \: dx \\ \int_a^{\infty} \frac{1}{x} \: dx = \lim_{b \to \infty} \ln x \big |_a^b \\ \int_a^{\infty} \frac{1}{x} \: dx = \lim_{b \to \infty} ( \ln b - \ln a ) \\ \int_a^{\infty} \frac{1}{x} \: dx = \infty \end{align}

In this case though, it turns out the area bounded by the curve and our limits of integration is infinity, which is rather interesting since $f$ and $g$ look very similar geometrically.

Convergence and Divergence

We will now introduce some definitions regarding improper integrals in this form:

Definition: Suppose that $\int_a^b f(x) \: dx$ is defined for $a ≤ b$. It thus follows that $\int_a^{\infty} f(x) \: dx = \lim_{b \to \infty} \int_a^b f(x) \: dx$. If $a ≥ b$, it thus follows that $\int_{-\infty}^b f(x) \: dx = \lim_{a \to -\infty} \int_a^b f(x) \: dx$. These improper integrals $\int_a^{\infty} f(x) \: dx$ and $\int_{-\infty}^b f(x) \: dx$ are said to be Convergent if their limits exist as a finite number, and called Divergent if their limits do not exist.

In the examples we looked at before, we can say that $\int_a^{\infty} \frac{1}{x^2} \: dx = \frac{1}{a}$ is convergent since for $a > 0$, the corresponding limit is defined as a finite value. Additionally, we can say that $\int_a^{\infty} \frac{1}{x} \: dx = \infty$ is divergent since the limit does not exist, that is, there is no defined finite value for this limit.

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