3
$\begingroup$

Prove the following inequality

$$\ln \frac{\pi + 2}{2} \cdot \frac{2}{\pi} < \int \limits_0^{\pi/2} \frac{\sin\ x}{x^2 + x} < \ln \frac{\pi + 2}{2}$$

I can prove that $\frac{\sin\ x}{x^2 + x} < \frac{1}{x + 1} \ \forall x \in (0, +\infty) \Rightarrow \int \limits_0^{\pi/2} \frac{\sin\ x}{x^2 + x} < \int \limits_0^{\pi/2} \frac{1}{x + 1} = \ln \frac{\pi + 2}{2}$.

Unfortunately, I don't know what happens when $x = 0$.

I can prove $\leqslant$ on LHS using Mean Value Theorem, but I have no idea how to prove that the sign is strict.

$\endgroup$
  • $\begingroup$ @SimplyBeautifulArt Why is it false? I checked Wolfram and $\frac{2}{\pi} \cdot \ln \frac{\pi + 2}{2} \approx 0.60$, $\int \limits_0^{\pi/2} \frac{\sin\ x}{x^2 + x} \approx 0.84 $, finally $\ln \frac{\pi + 2}{2} \approx 0.94$. $\endgroup$ – Welez Apr 23 '17 at 12:31
  • $\begingroup$ Whoops, I typed it into my calculator wrong. Sorry! $\endgroup$ – Simply Beautiful Art Apr 23 '17 at 12:33
2
$\begingroup$

Using $$\frac{\sin x}{x}>\frac{2}{\pi}\forall x\in \left(0,\frac{\pi}{2}\right)$$

Proving $\frac2\pi x \le \sin x \le x$ for $x\in [0,\frac {\pi} 2]$

$\endgroup$
1
$\begingroup$

The given inequality just depends on the convexity inequality $$ \forall x\in\left[0,\frac{\pi}{2}\right],\qquad \frac{2}{\pi}x\leq \sin(x)\leq x.$$ We may use another convexity inequality $$ \forall x\in\left[0,\frac{\pi}{2}\right],\qquad x-\frac{1}{6}x^3\leq \sin(x)\leq x-\left(\frac{4}{\pi^2}-\frac{8}{\pi^3}\right)x^3$$ and derive a much better inequality: $$\small \frac{1}{48} \left(4 \pi -\pi ^2+40 \log\left(\frac{2+\pi }{2}\right)\right)\leq I\\ I \leq \frac{1}{2\pi^3}\left(\pi(\pi-2)(4-\pi)+2 \left(8-4 \pi +\pi ^3\right) \log\left(\frac{2+\pi }{2}\right)\right).$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.