Class 17 - Improper integrals (10/12/2017, Thursday)

2nd Gateway

You must pass the second Gateway by Oct 13, Friday. Otherwise, your grade will be damaged a lot!

Webwork

• 8.3 due 10/18 Wed

Review: Polar coordinates

Exercise*. Solve #2 of Exam 2, Winter 2017.

Today we will cover Section 7.6 of your book.

Today: Imporper integrals

We start by an example. You know how to compute

$$\int_{0.001}^{2}\frac{1}{x}dx = \ln(2) - \ln(0.001).$$

So what if we replace $0.001$ by $0$? Then you will have to consider

$$\int_{0}^{2}\frac{1}{x}dx,$$

but you would immediately say "You cannot talk about $1/0$ in Math 116." You are right, but as a notation, we define 

$$\int_{0}^{2}\frac{1}{x}dx := \lim_{\epsilon \rightarrow 0+}\int_{\epsilon}^{2}\frac{1}{x}dx.$$

How do we compute this? For any $\epsilon > 0$, we can compute

$$\int_{\epsilon}^{2}\frac{1}{x}dx = \ln(2) - \ln(\epsilon),$$

so

$$\lim_{\epsilon \rightarrow 0+}\int_{\epsilon}^{2}\frac{1}{x}dx = \lim_{\epsilon \rightarrow 0+}(\ln(2) - \ln(\epsilon)) = \infty,$$

or in other words, the limit diverges to the $\infty$, so the limit does not exists.

In fact, drawing a neat picture of rectangles can tell you that

$$\int_{0}^{\infty}\frac{1}{x}dx \geq 1 + 1/2 + 1/3 + 1/4 + \cdots.$$

However, we also can see that

$$\begin{align*}1 + 1/2 + 1/3 + 1/4 + \cdots &\geq 1 + 1/2 + 1/4 + 1/4 + 1/4 + 1/4 + \cdots \\ &= 1 + 1/2 + 1/2 + 1/2 + \cdots = \infty,\end{align*}$$

so we get

$$\int_{0}^{\infty}\frac{1}{x}dx = \infty,$$

in a geometric fashion.

Exercise. Compute 

$$\int_{1}^{\infty}\frac{1}{x}dx = \lim_{r \rightarrow \infty}\int_{1}^{r}\frac{1}{x}dx.$$

(Hint: You can compute directly, or draw picture to realize that this integral is the same as the previous integral.)

Unlike the above exercise, we have

$$\begin{align*}\int_{1}^{\infty}\frac{1}{x^{2}}dx &= \lim_{r\rightarrow\infty}\left(\frac{-1}{r} + 1\right) \\ &= 1.\end{align*}$$

The integrals that require us to take limits are called improper integrals. We say an improper integral converges/diverges when the associated limit converges/diverges.

Exercise ($p$-test I). Show that when $p > 1,$ the following integral converges

$$\int_{1}^{\infty}\frac{1}{x^{p}} dx.$$

Exercise ($p$-test II). Show that when $p \leq 1,$ the following integral diverges

$$\int_{1}^{\infty}\frac{1}{x^{p}} dx.$$

Exercise ($p$-test III). Show that when $p \geq 1,$ the following integral diverges

$$\int_{1}^{\infty}\frac{1}{x^{p}} dx.$$

(Hint: Use substitution $u = 1/x$ and $p$-test II.)

Exercise ($p$-test IV). Show that when $p < 1,$ the following integral converges

$$\int_{1}^{\infty}\frac{1}{x^{p}} dx.$$

(Hint: Use substitution $u = 1/x$ and $p$-test I.)

Exercise. Compute

$$\int_{1}^{\infty}e^{-4x} dx.$$

Exercise. Find all real numbers $c$ such that

$$\int_{1}^{\infty}e^{cx} dx$$

is convergent.

Warning. Your book (p.390) interprets the following integral to be divergent:

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

Exercise. Using $p$-test, determine whether the following improper integral is convergent:

$$\int_{0}^{\infty}\frac{1}{x^{2}} dx.$$

is convergent.