Solutions - 0130-05

(a)

(1) Definition of improper integral:

$${\int }_0^1\ln xdx=\underset{R\to 0^{+}}{\lim }{\int }_R^1\ln xdx$$

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(2) Antiderivative and limit:

$$\begin{array}{ccc}& \gg \gg \underset{R\to 0^{+}}{\lim }x\ln x-x{|}_R^1& \\ & & \\ & \gg \gg \underset{R\to 0^{+}}{\lim }0-1-(R\ln R-R)\gg \gg -1\text{(converges)}& \text{(Note A)}\end{array}$$

Note A: Use L’Hopital:

$$R\ln R=\frac{\ln R}{1/R}{\gg \gg }^{d/dR}\frac{1/R}{-1/R^2}=-R\to 0$$

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(b)

(1) Definition of improper integral:

$${\int }_{-\infty }^{+\infty }x{e}^{-x^2}dx=\underset{R\to +\infty }{\lim }{\int }_0^{R}x{e}^{-x^2}dx+\underset{R\to -\infty }{\lim }{\int }_R^{0}x{e}^{-x^2}dx$$

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(2) Antiderivative and limit:

$$\begin{array}{c}\gg \gg \underset{R\to +\infty }{\lim }{-\frac{1}{2}{e}^{-x^2}|}_0^R+\underset{R\to -\infty }{\lim }{-\frac{1}{2}{e}^{-x^2}|}_R^0\\ \\ \gg \gg -\frac{1}{2}+-\frac{1}{2}\gg \gg 0\text{(converges)}\end{array}$$

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(c)

(1) Definition of improper integral:

$${\int }_1^2\frac{1}{{(x-1)}^2}dx=\underset{R\to 1^{+}}{\lim }{\int }_R^2\frac{1}{{(x-1)}^2}dx$$

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(2) Antiderivative and limit:

$$\begin{array}{c}\gg \gg \underset{R\to 1^{+}}{\lim }{-\frac{1}{x-1}|}_R^2\\ \\ \gg \gg \underset{R\to 1^{+}}{\lim }-1--\frac{1}{R-1}\gg \gg +\infty \text{(diverges)}\end{array}$$