Improper integrals

Integrate all the way to infinity

An important generalization of definite integral (the Riemann integral) which we have learned in basic calculus is the special kind of integral that involves infinity. As strange as it may sound, it actually has broad applications in real-world problems.

The "wrong" kind of integrals

By now, we should have a solid understanding of definite integrals $\int_a^b f(x) dx$. We defined this rigorously via Riemann sum. The focus of today is the "wrong" kind of integrals like \[ \int_1^\infty \frac{1}{x} dx \quad \int_1^\infty \frac{1}{x^2} dx \quad \int_{-\infty}^\infty \frac{1}{x^2} dx \quad \int_{0}^1 \frac{1}{\sqrt{x}} dx \] It is not even clear they make any sense at all. Our goal is to make sense of them and determine their value rigorously.

Hello everyone! Welcome back to another calculus lesson! The topic today is improper integrals: It's all about doing integrals, but doing it wrong. Yet, we will develop a mathematical solution that will help us to make sense of these wrong kind of integrals.

By now, you should have a solid understanding of definite integrals of the form $\int_a^b f(x) dx$ which gives us the total signed area bounded by the graph of the function $y=f(x)$ and the $x$-axis from $x = a$ to $x=b$. We defined this rigorously via Riemann sum (this is Riemann integral). The focus of today's discussion is the ``wrong'' kind of integrals like these ones.

By the way, can you see what's wrong with the last one? Pay close attention to where the integral starts.

It is not even clear they make any sense at all. Think about it, does it really make sense to compute the area of an infinitely long shape? Or infinitely tall shape? Our goal is to make sense of this type of integrals and determine their values rigorously.

Integrals of this kind are quite common in applications. Probably more than you think. For example, in probability theory, continuous probability distributions and expected values are defined in terms of integrals to infinity. In control theory or signal processing, the Laplace transform is also defined in terms of such integrals to infinity.

Integration over infinite intervals

The first kind of integrals we are going to look at involve interval that are infinitely long. E.g. \[ \begin{aligned} &\int_a^{\infty} f(x) dx, & &\int_{-\infty}^b f(x) dx, & &\int_{-\infty}^{\infty} f(x) dx \end{aligned} \] They are known as improper integrals.

These integrals cannot be understood from the view point of Riemann sums: the intervals are infinitely long.

The right approach is to define them in terms of limits.

The first kind of integrals we are going to look at is the kind in which the interval of integration is infinitely long, like these ones. They are integrals from somewhere to infinity, or from negative infinity to somewhere, or from negative infinity to positive infinity, which is essentially over the entire real line.

Integrals of these forms are known as improper integrals.

It is important to see that these integrals cannot be understood from the view point of Riemann sums. Can you see why? So to make sense of these, we will need something else. In typical calculus fashion, the right approach is to define them in terms of limits.

Definitions

(Improper integral) For a continuous function $f(x)$, we define \[ \int_a^{\infty} f(x) dx = \lim_{t \to \infty} \int_a^t f(x) dx \] if the limit exists (and finite), in which case we say the improper integral converges (or is convergent). Otherwise, we say it diverges (or is divergent).

For a continuous function $f(x)$, we define the integral from $a$ to infinity as the limit of of the integral from $a$ to $t$ as $t$ goes to infinity, IF this limit exists. And that's a big IF, of course. As you probably know by now, there are many situations in which limits can fail to exist. So that's a very big IF. But IF the limit does exist, we say the improper integral CONVERGES (or the integral is CONVERGENT). Otherwise, we say it diverges (or is divergent).

A side note here: For consistency, we do not consider $\infty$ or $-\infty$ to be valid limits in this part of the discussion. Since infinity is not a real number. We will, however, discuss more the general concept of ``infinite limits'' in later discussions.

You can see this is really the classic calculus solution to everything problem we have: IF we cannot know what happens AT infinity, we take the limit as we go toward infinity.

If we understand that, we can certainly go the other direction. We can define the integral from negative infinity to $b$ to be the limit of the integral from $t$ to $b$ as $t$ goes to negative infinity, IF this limit exists. Again, if the limit exists, we say the improper integral converges. Otherwise, we say the it diverges.

Following the exact same recipe, we define \[ \int_{-\infty}^b f(x) dx = \lim_{t \to -\infty} \int_t^b f(x) dx. \]

From $-\infty$ to $\infty$

Based on these, we can make sense of integrals from $-\infty$ to $\infty$ by splitting it into two separate integrals.

For a function $f(x)$ continuous on $(-\infty,\infty)$, we define \[ \int_{-\infty}^{\infty} f(x) dx = \int_{-\infty}^a f(x) \, dx + \int_a^{\infty} f(x) \, dx \] for any (finite) $a$, if both integrals on the right hand side are convergent. In this case, we say the integral is convergent. Otherwise, we say this integral is divergent.

Note that the choice of $a$ here is arbitrary, and we can make any choice to make the calculation easier.

For a function f(x) that is continuous on the entire real line, we define the improper integral of f(x) from negative infinity to positive infinity to be the sum of the two improper integrals, one from negative infinity to a number a, and the other one from a to infinity, if both of these integrals on the right hand side are convergent. If either of these integrals diverge, then we say the original integral also diverge.

Here, A is an arbitrary choice of number. It can be shown that the above definition is actually independent from the value of a, and you can pick any number you like to make the calculation easier.

Determine if the improper integral $\int_1^\infty \frac{1}{x} \, dx$ is convergent. If it is, compute its value.

Using the definition given above, \begin{align*} \int_1^\infty \frac{1}{x} \, dx &= \lim_{t \to \infty} \int_1^t \frac{1}{x} \, dx = \lim_{t \to \infty} [ \ln|x| ]_1^t \\ &= \lim_{t \to \infty} [ \ln |t| - \ln 1 ] = \lim_{t \to \infty} \ln |t| \end{align*} which does not exist. Therefore, the improper integral $\int_1^\infty \frac{1}{x} \, dx$ is divergent.

Determine if the improper integral $\int_1^\infty \frac{1}{x^2} \, dx$ is convergent. If it is, compute its value.

Using the definition given above, \begin{align*} \int_1^\infty \frac{1}{x^2} \, dx &= \lim_{t \to \infty} \int_1^t \frac{1}{x^2} \, dx = \lim_{t \to \infty} \left[ -\frac{1}{x} \right]_1^t \\ &= \lim_{t \to \infty} -\frac{1}{t} + 1 = 1. \end{align*} Therefore, the improper integral $\int_1^\infty \frac{1}{x} \, dx$ is convergent, and its value is 1.

For any given rational (or real) number $p>0$, determine if \[ \int_1^\infty \frac{1}{x^p} \, dx \] is convergent, and compute its value if it is convergent. (Hint: the result, of course, depends on $p$)

Integrating over discontinuities

The second kind of improper integrals involves discontinuities. For example, even though \[ \int_0^1 \frac{1}{\sqrt{x}} \, dx \] only involves a finite interval, the integrand $1 / \sqrt{x}$ "goes to infinity" as $x \to 0$ and hence not defined (or finite) over the entire interval $[0,1]$.

It is still meaningful to compute this type of integrals, and "limits" is going to get us out of this trouble, again.

If $f(x)$ is continuous in $[a,b)$, then \[ \int_a^b f(x) dx = \lim_{t \to b^-} \int_a^t f(x) dx \quad\text{if this limit exists}. \] If $f(x)$ is continuous in $(a,b]$, then \[ \int_a^b f(x) dx = \lim_{t \to a^+} \int_t^b f(x) dx. \quad\text{if this limit exists}. \] In both cases, if the limits exist, we say the integrals converge. Otherwise we say the integrals diverge.

Determine if the integral $\int_0^1 1 / \sqrt{x} \, dx$ is convergent, and compute its value if it is convergent.

The integrand $1 / \sqrt{x}$ is continuous on $(0,1]$ but grows unboundedly as $x \to 0^+$. By the definition given above, \begin{align*} \int_0^1 \frac{1}{\sqrt{x}} \, dx &= \lim_{t \to 0+} \int_t^1 \frac{1}{\sqrt{x}} \, dx \\ &= \lim_{t \to 0+} [2 \sqrt{x} ]_t^1 = \lim_{t \to 0+} [2 - 2\sqrt{t} ] = 2. \end{align*} Therefore, this improper integral converges to 2.

The same idea applies to integrals for which the integrand has a discontinuity in the middle of the interval.

If $f(x)$ is continuous in $[a,c)$ and $(c,b]$, then \[ \int_a^b f(x) dx = \int_a^c f(x) dx + \int_c^b f(x) dx \] if both of the integrals on the right hand side converge. In this case, we say the integral converges.

Determine if the integral $\int_{-1}^1 1 / x \, dx$ is convergent, and compute its value if it is convergent.

The integrand $1 / x$ has a vertical asymptote at $x=0$, and hence not continuous on $[-1,1]$. By the definition above, \begin{align*} \int_{-1}^1 \frac{1}{x} \, dx &= \int_{-1}^0 \frac{1}{x} \, dx + \int_0^1 \frac{1}{x} \, dx \\ &= \lim_{t \to 0^-} \int_{-1}^t \frac{1}{x} \, dx + \lim_{t \to 0^+} \int_{t}^1 \frac{1}{x} \, dx \\ &= \lim_{t \to 0^-} [ \ln|t| - \ln 1] + \lim_{t \to 0^+} [ \ln 1 - \ln|t|]. \end{align*} Both limits do not exist. Therefore this integral is divergent.

For any given rational (or real) number $p>0$, determine if the improper integral \[ \int_0^1 \frac{1}{x^p} \, dx \] is convergent, and compute its value if it is convergent.