Improper Integrals

The standard Riemann integral requires the integrand to be bounded and the interval of integration to be finite. When either of these conditions is not satisfied, the usual definition of the Riemann integral cannot be applied directly. However, by extending the standard definition to handle unbounded limits of integration or functions, we obtain the concept of improper integral.

Unbounded Integrand

Let’s start by studying the case when the limits of integration are bounded but the integrand is unbounded.

Let $f : [a, b) \to \mathbb{R}$ be a Riemann-integrable function on every $[a, b- \epsilon]$ interval, for all $\epsilon$ such that $0 < \epsilon < b - a$. Then, we can write:

$$ \int_a^b f(x) \, dx = \lim_{\epsilon \to 0^+} \int_a^{b-\epsilon} f(x) \, dx $$

By doing this, we stay away from the problematic endpoint $b$.

Likewise, given a function $f : (a, b] \to \mathbb{R}$ that is Riemann-integrable on every interval $[a + \epsilon, b]$, for all $\epsilon$ such that $0 < \epsilon < b-a$, we can write:

$$ \int_a^b f(x) \, dx = \lim_{\epsilon \to 0^+} \int_{a + \epsilon}^b f(x) \, dx $$

In both cases, if the limit exists and is finite, we say that the improper integral is convergent. Otherwise, we say that it is divergent.

Let’s look at an example. Let $a > 0$ and let’s consider a function $$ f(x) = \frac{1}{x^{\alpha}}, \quad \forall \alpha \in \mathbb{R}, \, x \in (0,1] $$

Our goal is to determine the values of $\alpha$ for which the following improper integral converges:

$$ \int_0^1 \frac{1}{x^{\alpha}} \, dx $$

We notice that $f$ is continuous on $(0,1]$ and therefore Riemann-integrable on every interval $[\epsilon, 1]$ with $0 < \epsilon < 1$. Let’s now determine a primitive:

$$ \begin{aligned} \int_a^b \frac{1}{x^{\alpha}} \, dx &= \int_a^b x^{-\alpha} \, dx\\ &= \left[ \frac{x^{1 - \alpha}}{1 - \alpha} \right]_a^b, \quad \alpha \neq 1\\ &= \frac{b^{1-\alpha}}{1-\alpha} - \frac{a^{1-\alpha}}{1 - \alpha}\\ &= \frac{b^{1-\alpha} - a^{1-\alpha}}{1 - \alpha} \end{aligned} $$

In our case, this corresponds to:

$$ \begin{aligned} \int_{\epsilon}^1 f(x) \, dx &= \int_{\epsilon}^1 x^{-\alpha} \, dx\\ &= \begin{cases} \frac{1 - \epsilon^{1 - \alpha}}{1- \alpha}, &\alpha \neq 1\\ -\log(\epsilon), &\alpha = 1 \end{cases} \end{aligned} $$

Therefore, we conclude that:

$$ \int_0^1 \frac{1}{x^{\alpha}} \, dx = \begin{cases} \text{converges to } \frac{1}{1 - \alpha}, &\alpha < 1\\ \text{diverges to } +\infty, & \alpha \geq 1 \end{cases} $$

Direct Comparison

Instead of determining whether an improper integral converges by finding a primitive and evaluating the corresponding limit, we can use a class of criteria known as convergence tests. The first one that we will consider is called direct comparison.

Let $f, g : (a, b] \to \mathbb{R}$ be continuous and nonnegative functions such that: $$ 0 \leq f(x) \leq g(x), \quad \forall x \in (a, b] $$ Then:

1. If $\int_a^b g(x) \, dx$ converges, then $\int_a^b f(x) \, dx$ also converges;
2. If $\int_a^b f(x) \, dx$ diverges, then $\int_a^b g(x) \, dx$ also diverges.

Naturally, the same convergence criteria apply when the improper integral has a singularity at the right endpoint $b$.

Let’s now look at an example. We want to determine the convergence of the following integral:

$$ \int_0^1 \frac{1}{\sqrt{x}} \, dx $$

We notice that $f(x)$ tends to $+\infty$ as $x\to 0^+$. Therefore, the integral is improper and we interpret it as:

$$ \int_0^1 \frac{1}{\sqrt{x}} \, dx = \lim_{\epsilon \to 0^+} \int_{\epsilon}^1 \frac{1}{\sqrt{x}} \, dx $$

To apply the direct comparison criterion, let’s choose the following function:

$$ g(x) = \frac{1}{x^{3/4}} $$

Then, for $0 < x < 1$, we have:

$$ \frac{1}{\sqrt{x}} = \frac{1}{x^{1/2}} \leq \frac{1}{x^{3/4}} $$

Let’s now consider the following integral:

$$ \int_0^1 \frac{1}{x^{3/4}} \, dx $$

We already know that integrals of this kind converge to $\frac{1}{1-\alpha}$ when $0 < \alpha < 1$, that is:

$$ \frac{1}{1 - \frac{3}{4}} = 4 $$

In fact:

$$ \begin{aligned} \lim_{a \to 0^+} \int_a^1 x^{-3/4} \, dx &= \lim_{a \to 0^+} \left[ 4x^{1/4} \right]_a^1\\ &= \lim_{a \to 0^+} (4 - 4a^{1/4})\\ &= 4 \end{aligned} $$

Since the comparison function converges and since the following inequality holds:

$$ 0 \leq \frac{1}{\sqrt{x}} \leq \frac{1}{x^{3/4}} $$

By the direct comparison criterion, $\int_0^1 \frac{1}{\sqrt{x}} \, dx$ converges as well. In fact:

$$ \begin{aligned} \int_0^1 \frac{1}{\sqrt{x}} \, dx &= \lim_{\epsilon \to 0^+} \left[ 2 \sqrt{x} \right]_{\epsilon}^1\\ &= \lim_{\epsilon \to 0^+} \left( 2 \sqrt{1} - 2 \sqrt{\epsilon} \right)\\ &= 2 \end{aligned} $$

Limit Comparison Test

As a corollary of the criterion studied in the previous section, we can introduce another comparison test known as the limit comparison test. This method can be used to determine the convergence of an improper integral by comparing its integrand to a simpler function.

Let $f, g : (a, b] \to \mathbb{R}$ be continuous and strictly positive functions. Suppose that the following limit exists: $$ L = \lim_{x \to a^+} \frac{f(x)}{g(x)}, \quad 0 < L < +\infty $$

Then the following two integrals:

$$ \int_a^b f(x) \, dx, \quad \int_a^b g(x) \, dx $$

have the same nature; that is, either both converge or both diverge.

Let’s try to determine the convergence of the following integral using this criterion:

$$ \int_0^1 \frac{\sin x}{x^{3/2}} \, dx $$

We notice that both the numerator and the denominator tend to zero as $x \to 0^+$, it is therefore an improper integral.

Let’s begin by recalling that Taylor expansion of the sine function around zero gives:

$$ \sin x \sim x, \quad x \to 0^+ $$

If we substitute this approximation into the expression for $f(x)$, we get:

$$ f(x) = \frac{\sin(x)}{x^{3/2}} \sim \frac{x}{x^{3/2}} = \frac{1}{x^{3/2 - 1}} = \frac{1}{x^{1/2}} = \frac{1}{\sqrt{x}} $$

Therefore, the comparison function will be: $$ g(x) = \frac{1}{\sqrt{x}} $$

Let’s verify that the limit of the ratio of the two functions has a finite limit:

$$ \begin{aligned} \lim_{x \to 0^+} \frac{f(x)}{g(x)} &= \lim_{x \to 0^+} \frac{\frac{\sin x}{x^{3/2}}}{\frac{1}{\sqrt{x}}}\\ &= \lim_{x \to 0^+} \frac{\sin x}{x^{3/2}} \cdot \sqrt{x}\\ &= \lim_{x \to 0^+} \frac{\sin x}{x} = 1 \end{aligned} $$

In particular, the last result is justified on this lesson about the squeeze theorem. Since the limit of the ratio of the two functions converges to a finite value that is not zero, by the limit comparison test, the following two integrals have the same convergence behavior:

$$ \int_0^1 \frac{\sin x}{x^{3/2}} \, dx, \quad \int_0^1 \frac{1}{\sqrt{x}} \, dx $$

Since we know that the following integral converges:

$$ \int_0^1 \frac{1}{\sqrt{x}} \, dx $$

The limit comparison test implies that the improper integral $\int_0^1 \frac{\sin x}{x^{3/2}}$ converges as well.

Unbounded Limits of Integration

The other kind of improper integral arises when the interval of integration is unbounded. For example:

$$ [a, +\infty), \, (-\infty, b], \, (a, +\infty), \, (-\infty, b) $$

Let $f : [a, +\infty) \to \mathbb{R}$ be a function that is Riemann-integrable on every compact interval $[a, c]$, for all $(c > a)$. Then, we define:

$$ \lim_{c \to +\infty} \int_a^c f(x) \, dx $$

Likewise, let $f : (-\infty, b] \to \mathbb{R}$ be a function that is Riemann-integrable on every compact interval $[c, b]$, for all $c < b$. Then, we define:

$$ \lim_{c \to -\infty} \int_c^b f(x) \, dx $$

In both cases, if the corresponding limit exists and is finite, we say that the improper integral is convergent. Otherwise, we say that it is divergent.

Just like for the previous kind of improper integrals, we can define particular convergence criteria to establish whether an integral of this kind converges or not.

Direct Comparison

To apply this convergence test to this kind of integral, we just need to slightly modify the statement of the theorem.

Let $f, g : (-\infty, b] \to \mathbb{R}$ be continuous and nonnegative functions such that: $$ 0 \leq f(x) \leq g(x), \quad \forall x \in (-\infty, b] $$ Then:

1. If $\int_{-\infty}^b g(x) \, dx$ converges, then $\int_{-\infty}^b f(x) \, dx$ also converges;
2. If $\int_{-\infty}^b f(x) \, dx$ diverges, then $\int_{-\infty}^b g(x) \, dx$ also diverges.

Before trying to apply this method to a practical problem, let’s first define the p-integral. Let $a > 0$ and let’s consider the following function defined on $[a, +\infty)$:

$$ f(x) = \frac{1}{x^{\alpha}}, \quad \forall \alpha \in \mathbb{R} $$

We notice that $f$ is continuous on $[a, +\infty)$ and therefore integrable in any interval of the type $[a, b]$ with $b > a$. Therefore, we have:

$$ \int_a^b f(x) \, dx = \int_a^b \frac{1}{x^{\alpha}} \, dx = \begin{cases} \frac{b^{1-\alpha} - a^{1-\alpha}}{1 - \alpha}, &\alpha \neq 1\\ \log(b) - \log(a), &\alpha = 1 \end{cases} $$

Thus, for $a > 0$, we have:

$$ \int_a^{+\infty} \frac{1}{x^{\alpha}} \, dx = \begin{cases} \text{converges to } \frac{a^{1-\alpha}}{\alpha - 1}, &\alpha > 1\\ \text{diverges to } +\infty, &\alpha \leq 1 \end{cases} $$

Now, let’s try to apply the direct comparison principle to the following integral:

$$ \int_{-\infty}^{-1} \frac{1}{x^2 + 1} \, dx $$

We notice that the integrand is continuous on $(-\infty, -1]$ and that the integral is improper since the lower limit of integration is unbounded. Let’s determine a comparison function. we observe that:

$$ \forall x \leq -1, \, x^2 + 1 \geq x^2 $$

Taking the reciprocals of both quantities, gives us:

$$ 0 < \frac{1}{x^2 + 1} \leq \frac{1}{x^2} $$

Therefore, we can choose $g(x) = \frac{1}{x^2}$. Let’s study the following integral:

$$ \int_{-\infty}^{-1} \frac{1}{x^2} \, dx $$

Since we already know that it converges since $\alpha = 2 > 1$, by the direct comparison criterion, we can conclude that $\int_{-\infty}^{-1} \frac{1}{x^2 + 1} \, dx$ converges as well.

Limit Comparison Test

Finally, let’s try to apply the limit comparison criterion to improper integrals with unbounded limits of integration.

Let $f, g : (-\infty, b] \to \mathbb{R}$ be continuous and strictly positive functions. Suppose that the following limit exists: $$ L = \lim_{x \to -\infty} \frac{f(x)}{g(x)}, \quad 0 < L < +\infty $$

Then the following two integrals:

$$ \int_{-\infty}^b f(x) \, dx, \quad \int_{-\infty}^b g(x) \, dx $$

have the same nature; that is, either both converge or both diverge.

Let’s look at a final example. Consider the following integral:

$$ \int_1^{+\infty} \frac{x^2 + 3}{x^4 + x} \, dx $$

We notice that this integral is improper since the upper limit of integration is unbounded. Before trying to apply the limit comparison test, let’s verify whether the function is continuous and positive on the interval $[1, +\infty)$.

Additionally, for $x \geq 1$, both the numerator and the denominator are positive; thus, $f(x) > 0$ $\forall x \in [1, +\infty)$ and we can apply the criterion. We notice that, as $x \to +\infty$, we have:

$$ f(x) = \frac{x^2 + 3}{x^4 + x} \sim \frac{x^2}{x^4} = \frac{1}{x^2} = g(x) $$

Therefore, our comparison function will be $g(x) = \frac{1}{x^2}$. Let’s begin by evaluating the limit of the ratio of the two functions:

$$ \begin{aligned} \lim_{x \to +\infty} \frac{f(x)}{g(x)} &= \lim_{x \to +\infty} \frac{\frac{x^2+3}{x^4+x}}{\frac{1}{x^2}}\\ &= \lim_{x \to +\infty} \frac{x^2(x^2 + 3)}{x^4 + x}\\ &= \lim_{x \to +\infty} \frac{x^4 + 3x^2}{x^4 + x}\\ &= \lim_{x \to +\infty} \frac{x^4(1 + \frac{3}{x^2})}{x^4(1 + \frac{1}{x^3})}\\ &= \frac{1 + 0}{1 + 0}\\ &= 1 \end{aligned} $$

Since this limit converges to a finite value that is not zero, by the limit comparison test, the following two integrals behaves identically:

$$ \int_1^{+\infty} \frac{x^2 + 3}{x^4 + x} \, dx, \quad \int_1^{+\infty} \frac{1}{x^2} \, dx $$

Since we know that the following integral converges:

$$ \int_1^{+\infty} \frac{1}{x^2} \, dx $$

The limit comparison test implies that the improper integral $\int_1^{+\infty} \frac{x^2 + 3}{x^4 + x} \, dx$ converges as well.