Chapter 5.4, Problem 77E

a.

To determine

To show: $\text{Si}\left(x\right)$ is an odd function of x .

Answer to Problem 77E

It is shown that $\text{Si}\left(x\right)$ is an odd function of x .

Explanation of Solution

Given information:

The sine integral function $\text{Si}(x)={\displaystyle \underset{0}{\overset{x}{\int }}\frac{\sin t}{t}dt}$ and

  $f\left(t\right)=\left\{\begin{array}{c}\frac{\sin t}{t},\\ 1,\end{array}\begin{array}{c}t\ne 0\\ t=0\end{array}\right.$

the continuous extension of $\frac{\sin t}{t}$ to the origin.

Concept used:

  $\text{Even}\text{Function}:\text{A}\text{function}\text{is}\text{even}\text{if}f\left(-x\right)=f\left(x\right)\text{for}\text{all}x\in ℝ$ .

  $\text{Odd}\text{Function}:\text{A}\text{function}\text{is}\text{odd}\text{if}f\left(-x\right)=-f\left(x\right)\text{for}\text{all}x\in ℝ$

Since

  $f\left(-t\right)=\left\{\begin{array}{c}\frac{\sin t}{t},\\ 1,\end{array}\begin{array}{c}t\ne 0\\ t=0\end{array}\right.$

Thus, $f\left(t\right)$ is even.

So, ${\displaystyle {\int }_0^{x}f}(t)dt={\displaystyle {\int }_{-x}^{0}f}(t)dt.$

Thus,

  $\begin{array}{c}-{\displaystyle {\int }_0^{x}f}(t)dt=-{\displaystyle {\int }_{-x}^{0}f}(t)dt\\ ={\displaystyle {\int }_0^{-x}f}(t)dt\end{array}$

Hence, $\text{Si}\left(x\right)$ is an odd function of x .

b.

To determine

To find: The value of $\text{Si}\left(0\right).$

Answer to Problem 77E

The value of $\text{Si}\left(0\right)=0$ .

Explanation of Solution

Given information:

The sine integral function $\text{Si}(x)={\displaystyle \underset{0}{\overset{x}{\int }}\frac{\sin t}{t}dt}$ and

  $f\left(t\right)=\left\{\begin{array}{c}\frac{\sin t}{t},\\ 1,\end{array}\begin{array}{c}t\ne 0\\ t=0\end{array}\right.$

the continuous extension of $\frac{\sin t}{t}$ to the origin.

Calculation:

  $\begin{array}{c}\text{Si}(0)={\displaystyle \underset{0}{\overset{0}{\int }}\frac{\sin t}{t}dt}\\ =0\end{array}$

c.

To determine

To find: The values of x at which $\text{Si}\left(x\right)$ has a local extreme value.

Answer to Problem 77E

The required value of $x=k\pi ,\text{where }k=\pm 1,\pm 2,\pm 3,\dots$

Explanation of Solution

Given information:

The sine integral function $\text{Si}(x)={\displaystyle \underset{0}{\overset{x}{\int }}\frac{\sin t}{t}dt}$ and

  $f\left(t\right)=\left\{\begin{array}{c}\frac{\sin t}{t},\\ 1,\end{array}\begin{array}{c}t\ne 0\\ t=0\end{array}\right.$

the continuous extension of $\frac{\sin t}{t}$ to the origin.

Calculation:

Put $f\left(x\right)=0$ to get the extreme values of $\text{Si}\left(x\right).$

  $\begin{array}{l}\frac{\sin x}{x}=0\\ x=k\pi ,\text{where }k=\pm 1,\pm 2,\pm 3,\dots \end{array}$

d.

To determine

To graph: $\text{Si}\left(x\right)$ by using NINT.

Answer to Problem 77E

The graph of $\text{Si}\left(x\right)$ is shown.

  

Explanation of Solution

Given information:

The sine integral function $\text{Si}(x)={\displaystyle \underset{0}{\overset{x}{\int }}\frac{\sin t}{t}dt}$ and

  $f\left(t\right)=\left\{\begin{array}{c}\frac{\sin t}{t},\\ 1,\end{array}\begin{array}{c}t\ne 0\\ t=0\end{array}\right.$

the continuous extension of $\frac{\sin t}{t}$ to the origin.

Calculation:

The graph of $\text{NINT}\left(\text{Si}\left(x\right),x,0,x\right)$ is shown.