Chapter 9.4, Problem 61E

a.

To determine

Toshow: that improper integral gives the same value if

  ${\displaystyle {\int }_{-\infty }^{\infty }\frac{1}{1+x^2}dx=}{\displaystyle {\int }_{-\infty }^1\frac{1}{1+x^2}dx+{\displaystyle {\int }_1^{\infty }\frac{dx}{1+x^2}}}$

Explanation of Solution

Given information:

  ${\displaystyle {\int }_{-\infty }^{\infty }\frac{1}{1+x^2}dx=}{\displaystyle {\int }_{-\infty }^1\frac{1}{1+x^2}dx+{\displaystyle {\int }_1^{\infty }\frac{dx}{1+x^2}}}$ .

Proof: to prove the same value, find individual value of the integrals,

  $\begin{array}{l}{\displaystyle {\int }_{-\infty }^1\frac{dx}{1+x^2}}={\left[{\tan }^{-1}x\right]}_{-\infty }^1\text{ [}{\displaystyle \int \frac{1}{1+x^2}dx={\tan }^{-1}x]}\\ {\displaystyle {\int }_{-\infty }^1\frac{dx}{1+x^2}}={\tan }^{-1}1-{\tan }^{-1}\left(-\infty \right)\\ {\displaystyle {\int }_{-\infty }^1\frac{dx}{1+x^2}}=\frac{\pi }{4}-\left(-\frac{\pi }{2}\right)\text{ [simplify]}\\ {\displaystyle {\int }_{-\infty }^1\frac{dx}{1+x^2}}=\frac{3\pi }{4}\text{ [add]}\end{array}$

  $\begin{array}{l}{\displaystyle {\int }_1^{\infty }\frac{dx}{1+x^2}}={\left[{\tan }^{-1}x\right]}_1^{\infty }\text{ [}{\displaystyle \int \frac{1}{1+x^2}dx={\tan }^{-1}x]}\\ {\displaystyle {\int }_1^{\infty }\frac{dx}{1+x^2}}={\tan }^{-1}\infty -{\tan }^{-1}\left(1\right)\\ {\displaystyle {\int }_1^{\infty }\frac{dx}{1+x^2}}=\frac{\pi }{2}-\left(\frac{\pi }{4}\right)\text{ [simplify]}\\ {\displaystyle {\int }_1^{\infty }\frac{dx}{1+x^2}}=\frac{\pi }{4}\text{ [add]}\end{array}$

  $\begin{array}{l}{\displaystyle {\int }_{-\infty }^{\infty }\frac{1}{1+x^2}dx=}{\displaystyle {\int }_{-\infty }^1\frac{1}{1+x^2}dx+{\displaystyle {\int }_1^{\infty }\frac{dx}{1+x^2}}}\\ {\displaystyle {\int }_{-\infty }^{\infty }\frac{1}{1+x^2}dx=}\frac{3\pi }{4}+\frac{\pi }{4}\text{ [substitute]}\\ {\displaystyle {\int }_{-\infty }^{\infty }\frac{1}{1+x^2}dx=}\pi \end{array}$

b.

To determine

To show: that it doesn’t matter what is the close limit .

Explanation of Solution

Given information:

  ${\displaystyle {\int }_{-\infty }^{\infty }f\left(x\right)dx=}{\displaystyle {\int }_{-\infty }^{c}f\left(x\right)}dx+{\displaystyle {\int }_c^{\infty }f\left(x\right)dx}$

Proof:

  $\begin{array}{l}{\displaystyle {\int }_{-\infty }^{c}f\left(x\right)dx=}{\displaystyle {\int }_{-\infty }^{0}f\left(x\right)}dx+{\displaystyle {\int }_0^{c}f\left(x\right)dx}\\ {\displaystyle {\int }_c^{\infty }f\left(x\right)dx=}{\displaystyle {\int }_c^{0}f\left(x\right)}dx+{\displaystyle {\int }_0^{\infty }f\left(x\right)dx}\end{array}$

Thus,

  $\begin{array}{l}{\displaystyle {\int }_{-\infty }^{c}f\left(x\right)dx}+{\displaystyle {\int }_c^{\infty }f\left(x\right)dx}={\displaystyle {\int }_{-\infty }^{0}f\left(x\right)}dx+{\displaystyle {\int }_0^{c}f\left(x\right)dx}+{\displaystyle {\int }_c^{0}f\left(x\right)}dx+{\displaystyle {\int }_0^{\infty }f\left(x\right)dx}\\ {\displaystyle {\int }_{-\infty }^{c}f\left(x\right)dx}+{\displaystyle {\int }_c^{\infty }f\left(x\right)dx}={\displaystyle {\int }_{-\infty }^{0}f\left(x\right)}dx+{\displaystyle {\int }_0^{\infty }f\left(x\right)dx}\end{array}$

Because ,

  ${\displaystyle {\int }_0^{c}f\left(x\right)dx}+{\displaystyle {\int }_c^{0}f\left(x\right)}dx$