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.

  • $\begingroup$ @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

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 $x\in [0,\frac {\pi} 2]$


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: $$\small \frac{1}{48} \left(4 \pi -\pi ^2+40 \log\left(\frac{2+\pi }{2}\right)\right)\leq I\\ I \leq \frac{1}{2\pi^3}\left(\pi(\pi-2)(4-\pi)+2 \left(8-4 \pi +\pi ^3\right) \log\left(\frac{2+\pi }{2}\right)\right).$$


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