Chapter 9.4, Problem 60E

a.

To determine

Tofind: the values of $f\left(x\right)$ as $x\to \infty$ .

Answer to Problem 60E

The values of function isx as $x\to \infty$ .

Explanation of Solution

Given information:

The function is $f\left(x\right)={\displaystyle {\int }_0^x\frac{\sin t}{t}dt}$ .

Calculation: to find the values of function as $x\to \infty$ , substitute the value of $x$ and the value of corresponding to it,

  $\begin{array}{l}f\left(x\right)={\displaystyle {\int }_0^x\frac{\sin t}{t}dt}\\ f\left(x\right)=\underset{x\to \infty }{\lim }{\displaystyle {\int }_0^x\frac{\sin t}{t}dt}\text{ [substitute }x=\infty ]\\ f\left(x\right)={\displaystyle {\int }_0^{x}0dt}\text{ [}\underset{x\to \infty }{\lim }\frac{\sin x}{x}dt=0]\\ f\left(x\right)=x\end{array}$

So the values of function is depends upon the values of x .

b.

To determine

To state: that the integral ${\displaystyle {\int }_0^{\infty }\frac{\sin x}{x}dx}$ is converge.

Answer to Problem 60E

The integral is converges because limit of function exist.

Explanation of Solution

Given information:

The integral is ${\displaystyle {\int }_0^{\infty }\frac{\sin x}{x}dx}$ .

the integral of ${\displaystyle {\int }_0^{\infty }\frac{\sin x}{x}dx}$ is converges because the limit of function is exist,

And $\underset{x\to \infty }{\lim }\frac{\sin x}{x}dt=0$ .