EXERCISE 7.8

3.

$$
\begin{aligned}
\int_{0}^{+\infty} e^{-2 x} d x & =\lim _{b \rightarrow+\infty} \int_{0}^{b} e^{-2 x} d x \\ \\
& =\lim_{b \rightarrow+\infty}\left[\frac{e^{-2 x}}{-2}\right]_{0}^{b} \\ \\
& =\lim _{b \rightarrow+\infty}\left[-\frac{e^{-2 b}}{2}+\frac{e^{0}}{2}\right] \\ \\
& =0+\frac{1}{2}=\frac{1}{2}
\end{aligned}
$$
4.
$$
\begin{aligned}
\int_{-1}^{+\infty} \frac{x}{1+x^{2}} d x & =\lim _{b \rightarrow+\infty} \int_{-1}^{b} \frac{x}{1+x^{2}} d x \\ \\
& =\lim _{b \rightarrow+\infty} \frac{1}{2} \int_{-1}^{b} \frac{2 x}{1+x^{2}} d x \\ \\
& =\lim _{b \rightarrow+\infty} \frac{1}{2}\left[\ln \left(1+x^{2}\right)\right]_{-1}^{b} \\ \\
& =\lim _{b \rightarrow+\infty} \frac{1}{2}\left[\ln \left(1+b^{2}\right)-\ln (1+1)\right] \\ \\
& =\lim _{b \rightarrow+\infty} \frac{1}{2}\left[\ln \left(1+b^{2}\right)-\ln 2\right]=+\infty
\end{aligned}
$$

5.
$$
\begin{aligned}
\int_{3}^{+\infty} \frac {2}{x^2-1}dx =
\lim _{b \rightarrow+\infty} \int_{3}^{b} \frac {2}{x^2-1}d x \quad \text { (1) } \\ \\
\text {Let} \quad x=\operatorname{coth} \theta \Rightarrow \theta=\operatorname{coth}^{-1}(x)
\end{aligned}
$$

$$
\begin{gathered}
d x=-\operatorname{csch}^{2} \theta d \theta \\ \\
x^{2}-1=\operatorname{coth}^{2} \theta-1=+\operatorname{csch}^{2} \theta
\end{gathered}
$$

$$
\begin{aligned}
& \int \frac{2}{\operatorname{coth}^{2} \theta-1}\left(-\operatorname{csch}^{2} \theta d \theta\right) \\ \\
& =2\int \frac{-\operatorname{csch}^{2} \theta}{\operatorname{csch}^{2} \theta} d \theta \\ \\
& = 2 \int -d \theta=-2[\theta] \\ \\
& =-2 \operatorname{coth}^{-1} x
\end{aligned}
$$

(1) becomes

$$
\begin{aligned}
& \lim _{b \rightarrow+\infty} \int_{3}^{b} \frac{2}{x^{2}-1} d x=\lim _{b \rightarrow+\infty}-2\left[\operatorname{coth}^{-1} x\right]_{3}^{b} \\ \\
& =\lim _{b \rightarrow+\infty}-2\left[\operatorname{coth}^{-1} b-\operatorname{coth}^{-1}(3)\right] \\ \\
& =-2\left[0-\operatorname{coth}^{-1} 3\right]=2 \operatorname{coth}^{-1} 3
\end{aligned}
$$

6.

$$
\begin{aligned}
\int_{0}^{+\infty} x e^{-x^{2}} d x & =\lim _{b \rightarrow+\infty} \int_{0}^{b} x e^{-x^{2}} d x \\ \\
& =\lim _{b \rightarrow+\infty}-\frac{1}{2} \int_{0}^{b}-2 x e^{-x^{2}} d x \\ \\
& =\lim _{b \rightarrow+\infty}-\frac{1}{2} \int_{0} \frac{d}{d x}\left(e^{-x^{2}}\right) d x \\ \\
& =\lim _{b \rightarrow+\infty}-\frac{1}{2}\left[e^{-x^{2}}\right]_{0}^{b} \\ \\
& =\lim _{b \rightarrow+\infty}-\frac{1}{2}\left[e^{-b^{2}}-e^{0}\right] \\ \\
& =-\frac{1}{2}[0-1]=\frac{1}{2}
\end{aligned}
$$

7.

$$
\begin{aligned}
\int_{e}^{+\infty} \frac{1}{x \ln ^{3} x} d x & =\lim _{b \rightarrow+\infty} \int_{e}^{b} \frac{1}{x(\ln x)^{3}} d x \\ \\
& =\lim _{b \rightarrow+\infty} \int_{e}^{b} \frac{(\ln x)^{-3}}{x} d x \\ \\
& =\lim _{b \rightarrow+\infty}\left[\frac{{(\ln x)}^{-3+1}}{-3+1}\right]_{e}^{b} \\ \\
& \left.=\lim _{b \rightarrow+\infty} \left [\frac{1}{2(\ln x)^{2}}\right]_{e}^{b} \\ \\
& =\lim _{b \rightarrow+\infty}\left[\frac{-1}{2(\ln b)^{2}}+\frac{1}{2(\ln e)^{2}}\right]=0+\frac{1}{2}=\frac{1}{2}
\end{aligned}
$$

8.

$$
\begin{aligned}
\int_{2}^{+\infty} \frac{1}{x \sqrt{\ln x}} d x & =\lim _{b \rightarrow+\infty} \int_{2}^{b} \frac{1}{x(\ln x)^{1 / 2}} d x \\ \\
& =\lim _{b \rightarrow+\infty} \int_{2}^{b} \frac{(\ln x)^{-1 / 2}}{x} d x \\ \\
& =\lim _{b \rightarrow+\infty}\left[\frac{(\ln x)^{-\frac{1}{2}+1}}{\frac{-1}{2}+1}\right]_{2}^{b} \\ \\
& =\lim _{b \rightarrow+\infty}\left[\frac{(\ln x)^{1 / 2}}{1 / 2}\right]_{2}^{b} \\ \\
& =\lim _{b \rightarrow+\infty}\left[2(\ln b)^{1 / 2}-2(\ln 2)^{1/2}\right]=+\infty
\end{aligned}
$$

9.

$$
\begin{aligned}
\int_{-\infty}^{0} \frac{d x}{(2 x-1)^{3}} & =\lim _{b \rightarrow-\infty} \int_{b}^{0}(2 x-1)^{-3} d x \\ \\
& =\lim _{b \rightarrow-\infty} \frac{1}{2} \int_{b}^{0} 2(2 x-1)^{-3} d x \\ \\
& =\lim _{b \rightarrow-\infty} \frac{1}{2}\left[\frac{(2 x-1)^{-3+1}}{-3+1}\right]_{b}^{0} \\ \\
& =\lim _{b \rightarrow-\infty} \frac{1}{2}\left[\frac{(2 x-1)^{2}}{-2}\right]_{b}^{0} \\ \\
& =\lim _{b \rightarrow-\infty}-\frac{1}{4}\left[\frac{1}{(2 x-1)^{2}}\right]_{b}^{0}
\end{aligned}
$$

$$
\begin{aligned}
& =\lim _{b \rightarrow-\infty}-\frac{1}{4}\left[\frac{1}{(0-1)^{2}}-\frac{1}{(2 b-1)^{2}}\right] \\ \\
& =-\frac{1}{4}[1-0]=-\frac{1}{4}
\end{aligned}
$$

10.

$$
\begin{aligned}
\int_{-\infty}^{3} \frac{d x}{x^{2}+9} & =\lim _{b \rightarrow-\infty} \int_{b}^{3} \frac{d x}{x^{2}+3^{2}} \\ \\
& =\lim _{b \rightarrow-\infty}\left[\frac{1}{3} \tan ^{-1}\left(\frac{x}{3}\right)\right]_{b}^{3} \\ \\
& =\lim _{b \rightarrow-\infty}\left[\frac{1}{3} \tan ^{-1}\left(\frac{3}{3}\right)-\frac{1}{3} \tan ^{-1}(b / 3)\right] \\ \\
& =\frac{1}{3} \tan ^{-1}(1)-\frac{1}{3} \tan ^{-1}(-\infty) \\ \\
& =\frac{1}{3}\left(\frac{\pi}{4}\right)-\frac{1}{3}\left(\frac{-\pi}{2}\right) \\ \\
& =\frac{\pi}{12}+\frac{\pi}{6}=\frac{\pi+2 \pi}{12}=\frac{3 \pi}{12} \\ \\
& =\frac{\pi}{4}
\end{aligned}
$$

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