EXERCISE 7.8 – Genesismath

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

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

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

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

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

(1) becomes

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

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

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

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

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

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

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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