Chapter 6.4, Problem 79E

To determine

To calculate: The approximate value of ${\displaystyle \underset{0}{\overset{x}{\int }}\frac{\sin t}{t}}dt=1$

Answer to Problem 79E

It is clear that integral has only solution at $x=1.0648397$

Explanation of Solution

Given information:

The convincing argument that the ${\displaystyle \underset{0}{\overset{x}{\int }}\frac{\sin t}{t}}dt=1$ has exactly one solution.

Formula used:

  $\frac{dy}{dx}=\frac{d}{dx}\left({\displaystyle \underset{0}{\overset{x}{\int }}\frac{\sin t}{t}dt}\right)$

Calculation:

Solve by your graphical calculator

  $\text{NINT}\left(\frac{\sin t}{t},t,0,x\right)$

Then

  $x=1.0648397$

Assume,

  $y={\displaystyle \underset{0}{\overset{x}{\int }}\frac{\sin t}{t}}dt$

Take derivative both sides

  $\begin{array}{l}\frac{dy}{dx}=\frac{d}{dx}\left({\displaystyle \underset{0}{\overset{x}{\int }}\frac{\sin t}{t}dt}\right)\\ \frac{dy}{dx}=\frac{\sin x}{x}\end{array}$

For $\left[0,\pi \right],\frac{dy}{dx}$ is positive, so y is increasing in this interval, hence it has only solution in this interval which is equal to $x=1.0648397$

For $\left[\pi ,2\pi \right],\frac{dy}{dx}$ is negative, so y is decreasing in this interval, and also

  $\begin{array}{l}y\left(2\pi \right)={\displaystyle \underset{0}{\overset{2\pi }{\int }}\frac{\sin t}{t}}dt\\ =\text{NINT}\left(\frac{\sin t}{t},t,0,2\pi \right)\\ y\left(2\pi \right)=1.42\end{array}$

Hence, $y\left(2\pi \right)$ is greater than 1 and y is decreasing in this interval. Hence y can never reach to 1 in this interval.

Furthermore, each arch of $\frac{\sin x}{x}$ is smaller in height than the previous one for the interval

  $\left[\left(2k+1\right)\pi ,\left(2k+2\right)\pi \right]$

So,

  ${\displaystyle \underset{2k\pi }{\overset{\left(2k+1\right)\pi }{\int }}\left|\frac{\sin x}{x}\right|}dx>{\displaystyle \underset{\left(2k+1\right)\pi }{\overset{\left(2k+2\right)\pi }{\int }}\left|\frac{\sin x}{x}\right|}dx$

This means that

  ${\displaystyle \underset{0}{\overset{\left(2k+2\right)\pi }{\int }}\frac{\sin x}{x}}dx-{\displaystyle \underset{0}{\overset{\left(2k\right)\pi }{\int }}\frac{\sin x}{x}dx}={\displaystyle \underset{\left(2k\right)\pi }{\overset{\left(2k+2\right)\pi }{\int }}\frac{\sin x}{x}}dx>0$

Where k is a positive integer

Thus, each successive minimum value is greater than previous one. Hence, for the interval $\left[\left(2k+1\right)\pi ,\left(2k+2\right)\pi \right]$ , function is decreasing in an interval but minimum value of any interval after $\left[\pi ,2\pi \right]$ is greater than 1.42. So for any value of x integral cannot be 1.

In the interval $\left[0,\infty \right)$ , value of integral is negative because integral is an odd function and for an odd function if it is positive for positive value of x then it is negative for negative value of x - so in this interval also, integral is not equal to 1.

Thus, from above argument, it is clear that integral has only solution at $x=1.0648397$

Conclusion:

It is clear that integral has only solution at $x=1.0648397$