Application of the mean-value theorem for integrals

Let $\varphi$ be a function with second derivative $\varphi''$ continuous and nonzero on an interval $[a,b]$ . Furthermore, let $m > 0$ be a constant such that
$\varphi'(t) \geq m \qquad \text{for all } t \in [a,b].$

Use the second mean-value theorem for integrals (Theorem 5.5 in Apostol) to prove the inequality

$\left| \int_a^b \sin \varphi(t) \, dt \right| \leq \frac{4}{m}.$
If $a > 0$ prove that
$\left| \int_a^x \sin (t^2) \, dt \right| \leq \frac{2}{a} \qquad \text{for } x > a.$

Proof. Since we have the assumption that $\varphi'(t) \geq m > 0$ for all $t \in [a,b]$ we may divide by $\varphi'(t)$ , to obtain
$\left| \int_a^b \sin \varphi(t) \, dt \right| = \left| \int_a^b \frac{\sin \varphi(t)}{\varphi'(t)} \cdot \varphi'(t) \, dt \right|.$

Then, to apply the second mean-value theorem for integrals (Theorem 5.5 of Apostol) we define functions

$f(t) = \frac{1}{\varphi'(t)} \qquad \text{and} \qquad g(t) = \varphi'(t) \sin \varphi(t).$

The function $g$ is continuous since $\sin \varphi(t)$ is a composition of continuous functions (we know $\varphi(t)$ is continuous since it is differentiable) and $\varphi'(t)$ is continuous (again, it is differentiable since $\varphi''(t)$ exists and is continuous by assumption). Then the product of continuous function is also continuous, which establishes that $g$ is continuous. We also know that $f$ meets the conditions of the theorem since it has derivative given by

$f'(t) = - \frac{\varphi''(t)}{(\varphi'(t))^2}.$

This derivative is continuous since $\varphi''(t)$ and $\varphi'(t)$ are continuous and $\varphi'(t)$ is nonzero. Furthermore, this derivative does not change since on $[a,b]$ since $\varphi''(t)$ is nonzero on $[a,b]$ (and by Bolzano’s theorem we know that a continuous function that changes sign must have a zero). Therefore, we can apply the second mean-value theorem:

$\begin{align*} \left| \int_a^b \frac{\varphi'(t) \sin \varphi(t)}{\varphi'(t)} \, dt \right| &= \left| \frac{1}{\varphi'(a)} \int_a^c \varphi'(t) \sin \varphi(t) \, dt + \frac{1}{\varphi'(b)} \int_c^b \varphi'(t) \sin \varphi(t) \, dt \right| \\ &\leq \left| \frac{1}{m} \int_a^c \varphi'(t) \sin \varphi(t) \, dt + \frac{1}{m} \int_c^b \varphi'(t) \sin \varphi(t) \, dt \right| \\ &\leq \left| -\frac{1}{m} \cos \varphi(t) \Big \rvert_a^c + \left(-\frac{1}{m} \right) \cos \varphi(t) \Big \rvert_c^b \right| \\ & \leq \left| -\frac{1}{m} \cos \varphi(t) \Big \rvert_a^c \right| + \left| -\frac{1}{m} \cos \varphi(t) \Big \rvert_c^b \right| \\ \intertext{(by the Triangle Inequality)} &\leq \left| -\frac{2}{m} \right| + \left| -\frac{2}{m} \right| \\ & = \frac{2}{m} + \frac{2}{m} \\ &= \frac{4}{m}. \qquad \blacksquare \end{align*}$
Proof. Using part (a), we take $\varphi(t) = t^2$ , giving us
$\varphi(t) = t^2 \quad \implies \quad \varphi'(t) = 2t \quad \implies \quad \varphi''(t) = 2.$

Thus, $\varphi''(t)$ is continuous and never changes sign. Furthermore, $\varphi' \geq m = 2a$ (where $a$ is a given constant) and $2a > 0$ since $a> 0$ . Thus,

$\left| \int_a^x \sin (t^2) \, dt \right| \leq \frac{4}{2a} = \frac{2}{a} \qquad \text{for all } x > a. \qquad \blacksquare$

3 comments

April 8, 2017 at 8:23 am

P. says:

Hi, this resource was a great find!

One thing left me stumped working on this. On the second row of the solution to part one, I initially summed the two integrals, getting 2/m rather than 4/m.

I figured that I could use the triangle inequality; but I can’t understand if summing the two integrals gives an incorrect result. Could you explain?

- April 8, 2017 at 9:09 am
  
  P. says:
  
  On second thoughts… you can’t actually sum the two integrals when 1/(phi’) is actually evaluated at a and b. But the two integrals can still both be equal to 2.
  
  - July 28, 2019 at 9:54 am
    
    TRC YX says:
    
    The second line should be incorrect, but easily recoverable (by splitting the absolute value before this).
    
    A counterexample can possibly be constructed in this way: if two integrals are about to be equal in absolute value but have different signs, say -1 and 1, then the second line evaluates to something around 0, but by choosing the correct phi’, the first line can be farther from 0. Thus the first line > the second line.

Stumbling Robot

Application of the mean-value theorem for integrals

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