1. \(\begin{aligned}\int_1^\infty \frac{1}{x^2}\ dx\ =\ \lim_{b\to\infty} \int_1^b\frac1{x^2}\ dx\ \amp =\ \lim_{b\to\infty} \frac{-1}{x}\Big|_1^b \\ \amp = \lim_{b\to\infty} \frac{-1}{b} + 1\\ \amp = 1. \end{aligned}\) A graph of the area defined by this integral is given in Figure 6.8.4.

2. \(\begin{aligned}\int_1^\infty \frac1x\ dx \amp = \lim_{b\to\infty}\int_1^b\frac1x\ dx \\ \amp = \lim_{b\to\infty} \ln\abs{x}\Big|_1^b \\ \amp = \lim_{b\to\infty} \ln(b)\\ \amp = \infty. \end{aligned}\) The limit does not exist, hence the improper integral \(\ds\int_1^\infty\frac1x\ dx\) diverges. Compare the graphs in Figures 6.8.4 and 6.8.5; notice how the graph of \(f(x) = 1/x\) is noticeably larger. This difference is enough to cause the improper integral to diverge.

3. \(\begin{aligned}\int_{-\infty}^0 e^x \ dx \amp = \lim_{a\to-\infty} \int_a^0e^x\ dx \\ \amp = \lim_{a\to-\infty} e^x\Big|_a^0 \\ \amp = \lim_{a\to-\infty} e^0-e^a \\ \amp = 1. \end{aligned}\) A graph of the area defined by this integral is given in Figure 6.8.6.

4. We will need to break this into two improper integrals and choose a value of \(c\) as in part 3 of Definition 6.8.2. Any value of \(c\) is fine; we choose \(c=0\text{.}\) \begin{align*} \int_{-\infty}^\infty \frac1{1+x^2}\ dx \amp = \lim_{a\to-\infty} \int_a^0\frac{1}{1+x^2}\ dx + \lim_{b\to\infty} \int_0^b\frac{1}{1+x^2}\ dx\\ \amp = \lim_{a\to-\infty} \tan^{-1}(x) \Big|_a^0 + \lim_{b\to\infty} \tan^{-1}(x) \Big|_0^b\\ \amp = \lim_{a\to-\infty} \left(\tan^{-1}(0) -\tan^{-1}(a) \right) + \lim_{b\to\infty} \left(\tan^{-1}(b) -\tan^{-1}(0) \right)\\ \amp = \left(0-\frac{-\pi}2\right) + \left(\frac{\pi}2-0\right).\\ \end{align*} Each limit exists, hence the original integral converges and has value: \begin{align*} \amp = \pi. \end{align*} A graph of the area defined by this integral is given in Figure 6.8.7.

This integral will require the use of Integration by Parts. Let \(u = \ln(x)\) and \(dv = 1/x^2\ dx\text{.}\) Then

1. A graph of \(f(x) = 1/\sqrt{x}\) is given in Figure 6.8.12. Notice that \(f\) has a vertical asymptote at \(x=0\text{;}\) in some sense, we are trying to compute the area of a region that has no “top.” Could this have a finite value? \begin{align*} \int_0^1 \frac{1}{\sqrt{x}}\ dx \amp = \lim_{a\to0^+}\int_a^1 \frac1{\sqrt{x}}\ dx\\ \amp = \lim_{a\to0^+} 2\sqrt{x}\Big|_a^1\\ \amp = \lim_{a\to0^+} 2\left(\sqrt{1}-\sqrt{a}\right)\\ \amp = 2. \end{align*} It turns out that the region does have a finite area even though it has no upper bound (strange things can occur in mathematics when considering the infinite).

   In Definition 6.8.10, \(c\) can be one of the endpoints (\(a\) or \(b\)). In that case, there is only one limit to consider as part of the definition.

2. The function \(f(x) = 1/x^2\) has a vertical asymptote at \(x=0\text{,}\) as shown in Figure 6.8.13, so this integral is an improper integral. Let's eschew using limits for a moment and proceed without recognizing the improper nature of the integral. This leads to: \begin{align*} \int_{-1}^1\frac1{x^2}\ dx \amp = -\frac1x\Big|_{-1}^1\\ \amp = -1 - (1)\\ \amp =-2 ! \end{align*}

   Clearly the area in question is above the \(x\)-axis, yet the area is supposedly negative! Why does our answer not match our intuition? To answer this, evaluate the integral using Definition 6.8.10. \begin{align*} \int_{-1}^1\frac1{x^2}\ dx \amp = \lim_{t\to0^-}\int_{-1}^t \frac1{x^2}\ dx + \lim_{t\to0^+}\int_t^1\frac1{x^2}\ dx\\ \amp = \lim_{t\to0^-}-\frac1x\Big|_{-1}^t + \lim_{t\to0^+}-\frac1x\Big|_t^1\\ \amp = \lim_{t\to0^-}-\frac1t-1 + \lim_{t\to0^+} -1+\frac1t\\ \amp \Rightarrow \Big(\infty-1\Big)\ + \ \Big(- 1+\infty\Big). \end{align*} Neither limit converges hence the original improper integral diverges. The nonsensical answer we obtained by ignoring the improper nature of the integral is just that: nonsensical.

We begin by integrating and then evaluating the limit. \begin{align*} \int_1^\infty \frac1{x^p}\ dx \amp = \lim_{b\to\infty}\int_1^b\frac1{x^p}\ dx\\ \amp = \lim_{b\to\infty}\int_1^b x^{-p}\ dx \qquad \text{ (assume \(p\neq 1\)) }\\ \amp = \lim_{b\to\infty} \frac{1}{-p+1}x^{-p+1}\Big|_1^b\\ \amp = \lim_{b\to\infty} \frac{1}{1-p}\big(b^{1-p}-1^{1-p}\big). \end{align*}

When does this limit converge — i.e., when is this limit not \(\infty\text{?}\) This limit converges precisely when the power of \(b\) is less than 0: when \(1-p\lt 0 \Rightarrow 1\lt p\text{.}\)

Our analysis shows that if \(p>1\text{,}\) then \(\ds\int_1^\infty \frac1{x^p}\ dx\) converges. When \(p\lt 1\) the improper integral diverges; we showed in Example 6.8.3 that when \(p=1\) the integral also diverges.

Figure 6.8.15 graphs \(y=1/x\) with a dashed line, along with graphs of \(y=1/x^p\text{,}\) \(p\lt 1\text{,}\) and \(y=1/x^q\text{,}\) \(q>1\text{.}\) Somehow the dashed line forms a dividing line between convergence and divergence.

1. The function \(f(x) = e^{-x^2}\) does not have an antiderivative expressible in terms of elementary functions, so we cannot integrate directly. It is comparable to \(g(x)=1/x^2\text{,}\) and as demonstrated in Figure 6.8.19, \(e^{-x^2} \lt 1/x^2\) on \([1,\infty)\text{.}\) We know from Key Idea 6.8.16 that \(\ds \int_1^\infty \frac{1}{x^2}\ dx\) converges, hence \(\ds\int_1^\infty e^{-x^2}\ dx\) also converges.

2. Note that for large values of \(x\text{,}\) \(\ds \frac{1}{\sqrt{x^2-x}} \approx \frac{1}{\sqrt{x^2}} =\frac{1}{x}\text{.}\) We know from Key Idea 6.8.16 and the subsequent note that \(\ds \int_3^\infty \frac1x\ dx\) diverges, so we seek to compare the original integrand to \(1/x\text{.}\) It is easy to see that when \(x>0\text{,}\) we have \(x = \sqrt{x^2} > \sqrt{x^2-x}\text{.}\) Taking reciprocals reverses the inequality, giving \begin{equation*} \frac1x \lt \frac1{\sqrt{x^2-x}}. \end{equation*} Using Theorem 6.8.17, we conclude that since \(\ds\int_3^\infty\frac1x\ dx\) diverges, \(\ds\int_3^\infty\frac1{\sqrt{x^2-x}}\ dx\) diverges as well. Figure 6.8.20 illustrates this.

As \(x\) gets large, the quadratic inside the square root function will begin to behave much like \(y=x\text{.}\) So we compare \small\(\ds\frac{1}{\sqrt{x^2+2x+5}}\) to \small\(\ds\frac1x\) with the Limit Comparison Test: \begin{equation*} \lim_{x\to\infty} \frac{1/\sqrt{x^2+2x+5}}{1/x} = \lim_{x\to\infty}\frac{x}{\sqrt{x^2+2x+5}}. \end{equation*}

The immediate evaluation of this limit returns \(\infty/\infty\text{,}\) an indeterminate form. Using l'Hôpital's Rule seems appropriate, but in this situation, it does not lead to useful results. (We encourage the reader to employ l'Hôpital's Rule at least once to verify this.)

The trouble is the square root function. To get rid of it, we employ the following fact: If \(\lim\limits_{x\to c} f(x) = L\text{,}\) then \(\lim\limits_{x\to c} f(x)^2 = L^2\text{.}\) (This is true when either \(c\) or \(L\) is \(\infty\text{.}\)) So we consider now the limit \begin{equation*} \lim_{x\to\infty} \frac{x^2}{x^2+2x+5}. \end{equation*}

This converges to 1, meaning the original limit also converged to 1. As \(x\) gets very large, the function \small\(\ds\frac{1}{\sqrt{x^2+2x+5}}\) looks very much like \small\(\ds\frac1x\text{.}\)\ Since we know that \small\(\ds\int_3^{\infty} \frac1x\ dx\) diverges, by the Limit Comparison Test we know that \small\(\ds\int_3^\infty\frac{1}{\sqrt{x^2+2x+5}}\ dx\) also diverges. Figure 6.8.23 graphs \(f(x)=1/\sqrt{x^2+2x+5}\) and \(f(x)=1/x\text{,}\) illustrating that as \(x\) gets large, the functions become indistinguishable.