The integrals we'll work on in this section aren't literally "improper," like maybe you shouldn't do them, they're just a little different. They're basically integrals for which one or both limits aren't strictly defined, or that have a discontinuity somewhere between the limits of integration.

For example, what if we wanted to know the area under the curve $f(x) = \frac{1}{x}$ between the limits of x = 1 and x → ∞ ? We might write that integral like this:

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

but a better way to write it is

$$\lim_{t \rightarrow \infty} \, \int_1^t \, \frac{1}{x} \, dx.$$

As a reminder, the area we're looking for is:

Now this is an integral we can do easily:

$$ \begin{align} \lim_{t \rightarrow \infty} \, &\int_1^t \, \frac{1}{x} \, dx \\[5pt] &= \lim_{t \rightarrow \infty} \, ln|x| \bigg|_1^t \\[7pt] &= \lim_{t \rightarrow \infty} \, ln|t| - 0 \: \rightarrow \: \infty \end{align}$$

When the limit is evaluated, we see that the area under this curve is infinite. That's not too surprising given that we know that y = 0 is an asymptote, and that the function approaches but never quite reaches a value of zero as $x \rightarrow \infty.$ What's interesting is that similar functions *do* bound a finite area. We'll see one in the next example.

We say that this integral **diverges**. If one or both limits of an improper integral is infinite or doesn't exist, the integral diverges. An improper integral that has a finite limit is said to **converge** to that limit.

Calculate the value of the area under $f(x) = \frac{1}{x^2}$ from x = 0 to infinity.

The integral is

$$\lim_{t \rightarrow \infty} \, \int_1^t \, \frac{1}{x^2} \, dx.$$

This again is not a difficult integral to evaluate:

$$ \begin{align} \lim_{t \rightarrow \infty} \, &\int_1^t \, \frac{1}{x^2} \, dx \\[5pt] &= \lim_{t \rightarrow \infty} \; -\frac{1}{x} \, \bigg|_1^t \\[5pt] &= \lim_{t \rightarrow \infty} \; \frac{-1}{t} + 1 = 1 \; unit^2 \end{align}$$

So this curve, very similar to the one in our first example above, encloses a *finite* area, even though the function still approaches the x-axis asymptotically as x grows. Strange.

We say that an improper integral like this **converges** to a finite value. The interpretation of these two results that makes the most sense is that the second function approaches zero more rapidly, and apparently sufficiently fast that it encloses a finite area. You'll find a somewhat more satisfying interpretation in the section on the integral test for infinite series.

The first two examples here show how two similar functions can enclose finite and infinite areas, even though they both behave asymptotically &emdash; that is, they both approach a steady value but never actually reach it. That can be a difficult thing to wrap your mind around.

To help you visualize what's going on, take a look at some sum of our expressions, 1/n and 1/n^{2} as n grows by factors of two. The first row is the first term, the second is the sum of the first 10 terms, and so on:

$n$ | $1/n$ | $1/n^2$ |

1 | 1.00000000 | 1.00000000 |

10 | 2.92896825 | 1.54976773 |

20 | 3.59773966 | 1.59616324 |

40 | 4.27854304 | 1.62024396 |

80 | 4.96547928 | 1.63251187 |

160 | 5.65551122 | 1.63870356 |

320 | 6.34709835 | 1.64181394 |

640 | 7.03946489 | 1.64337279 |

1280 | 7.73222160 | 1.64415312 |

diverges |
converges |

And below is a graph of these growing sums.

You can see that the sum of terms of the form 1/n^{2} grows rapidly at first, then approaches a limit. In fact, the three digits after the decimal will settle in to .644 and never change thereafter. Further terms just refine digits successively to the right of the decimal. In contrast, the sum of terms of the form 1/n, the so-called **harmonic series**, continues to grow. In fact, it can be shown that this sum can be used to represent any number if we just add enough terms. That's the difference between **divergence** and **convergence**.

One way to think of it, if it helps, is that one denominator approaches infinity "faster" than the other, and that makes the difference.

This concept of convergence and divergence is central to many aspects of mathematics, including how computers and calculators compute things like sines, cosines and logarithms.

We can expand on the previous two examples to ask, for what value of p does the integral $\int_1^{\infty} \, \frac{1}{x^p} \, dx$ converge?

The integral is

$$ \begin{align} \lim_{t \rightarrow \infty} \, \int_1^t \, \frac{1}{x^p} \, dx &= \lim_{t \rightarrow \infty} \, \frac{x^{-p + 1}}{1 - p} \, \bigg|_1^t \\[5pt] &= \frac{1}{1 - p} \, \lim_{t \rightarrow \infty} \left[ t^{-p + 1} - 1 \right] \\[5pt] &= \frac{1}{1 - p} \, \lim_{t \rightarrow \infty} \left[ \frac{1}{t^{p - 1}} - 1 \right] \end{align}$$

Now the fraction goes to zero as $t \rightarrow \infty$ as long as $p \gt 1,$ so as long as $p \gt 1,$ the integral has a finite value of $\frac{1}{p - 1}.$

This is a refinement in our understanding of the convergence of the area under functions like $f(x) = 1/x^p.$ As long as $p \gt 1,$ the area is finite and can be calculated.

Now let's consider the integral of a function with a discontinuity. Compute the integral

$$\int_3^7 \, \frac{1}{\sqrt{x - 3}} \, dx$$

The function is infinite at x = 3, so we'll need to rewrite it, approaching the limit x = 3 from above or from inside the area being integrated, like this,

$$\lim_{t \rightarrow 3^+} \, \int_t^7 \, \frac{1}{\sqrt{x - 3}} \, dx$$

Now the integral is done by u-substitution to yield

$$ \begin{align} &= \lim_{t \rightarrow 3^+} \, 2 \sqrt{x - 3} \, \bigg|_t^7 \\[5pt] &= 2(\sqrt{4} - 0) = 4 \end{align}$$

Notice that for this integral, once we found the antiderivative, our discontinuity wasn't a problem any more, and simply performing the integral without regard to it would have worked. Nevertheless, it's important to be aware of discontinuities and make sure that any integral over one converges to a real value.

Compute the integral of $f(x) = ln(x)$ between $x = 0$ and $x = 1.$

This is a tricky problem because ln(0) is undefined. Its domain is (0, ∞). Here's the graph of y = ln(x):

The integral is

$$\lim_{t \rightarrow 0^+} \, \int_t^1 \, ln(x) \, dx$$

You might recall that this integral can be done by parts; the antiderivative of $ln(x)$ is $x \, ln(x) + x + C,$ so our limit is

$$ \begin{align} \lim_{t \rightarrow 0^+} &[x \, ln(x) + x] \, \bigg|_t^1 \\[5pt] &= 1 \cdot ln(1) - 1 - \lim_{t \rightarrow 0^+} \, [t \, ln(t) - t] \\[5pt] &= -1 - \lim_{t \rightarrow 0^+} \, [t \, ln(t) - t] \\[5pt] &= -1 - \lim_{t \rightarrow 0^+} \, t \, ln(t) \end{align}$$

Now the remaining limit can be evaluated as an ∞/∞ limit using L'Hopital's rule:

$$ \begin{align} \lim_{t \rightarrow 0^+} \, t \, ln(t) &= \lim_{t \rightarrow 0^+} \, \frac{ln(t)}{1/t} \\[5pt] &= \frac{1/t}{-1/t^2} = -t \: \rightarrow \: 0 \end{align}$$

So remembering the -1 from our original limit, the result is -1. The shaded area of our integral is 1 square unit, lying below the x axis. The integral *converges* to a finite value.

Consider the integral of a function with a discontinuity within the interval of integration.

Integrate

$$f(x) = \frac{1}{x - 2}$$

between $x = 0$ and $x = 10.$

Here's a graph of the function and the area we're trying to find:

Let's first think about this integral as if we didn't know about the discontinuity. It's pretty simple:

$$ \begin{align} \int_0^{10} \, \frac{1}{x - 2} \, dx &= ln |x - 2| \, \bigg|_0^{10} \\[5pt] &= ln|8| - ln|2| = ln \left| \frac{8}{2} \right| = ln(4) \end{align}$$

But this isn't correct, as we'll see. Now let's do the integral but this time consider the vertical asymptote at x = 2.

$$ \begin{align} &\lim_{t \rightarrow 2^-} \, int_0^t \, \frac{dx}{x - 2} + \lim_{t \rightarrow 2^+} \, int_t^{10} \, \frac{dx}{x - 2} \\[5pt] &= \lim_{t \rightarrow 2^-} ln|x - 2| \bigg|_0^t + \lim_{t \rightarrow 2^+} ln|x - 2| \bigg|_t^{10} \\[5pt] &= \lim_{t \rightarrow 2^-} ln|t - 2| - ln(2) \\[5pt] &\phantom{0000} + ln(8) - \lim_{t \rightarrow 2^+} ln|t - 2| \end{align}$$

Now both of those limits evaluate to ln(0), which is infinite, so this integral **diverges**, and the area it encloses is infinite. So you can see that we need to consider the discontinuity.

Integrals with infinite limits, like

$$\int_a^{\infty} \, f(x) \, dx, \: \: \text{ or } \: \: \int_{-\infty}^{a} \, f(x) \, dx, \: \: \text{ or } \: \: \int_{-\infty}^{\infty} \, f(x) \, dx$$

are sometimes called **Type 1** improper integrals in books and articles. Likewise, those without infinite limits, but with a discontinuity somewhere inside the interval of integration are called **Type 2** improper integrals.

Calculate the value of the following integrals, or show that they diverge.

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

2. |
$$\int_{0}^1 \, \frac{ln(x)}{\sqrt{x}} \, dx$$ |

3. |
$$\int_{0}^{\infty} \, \frac{e^x}{e^{2x} + 1} \, dx$$ |

4. |
$$\int_{0}^5 \, \frac{x}{x - 1} \, dx$$ |

5. |
$$\int_{0}^{\infty} \, x \, e^{-3x} \, dx$$ |

6. |
$$\int_{-\infty}^{\infty} \, \frac{1}{1 + x^2} \, dx$$ |

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