Given $F(0) = \frac{1}{\pi D} \int_{0}^{\frac{\pi}{D}} \frac{\sin^2[Dy]}{\sin^2[y]} dy$. Assume D is some constant.
Using the following properties prove: $F(0) \geq \frac{4}{\pi^2} > 0.405$
Properties:
- $|\sin [x]| \leq |x| ~\forall x \in \mathrm{R}$,
- $|\sin [x]| \geq |(\frac{2}{\pi})x| ~\forall x \in [0,\frac{\pi}{2}]$
- $|\sin [x]| \geq |-(\frac{2}{\pi})x+2| ~\forall x \in [\frac{\pi}{2},\pi]$
I did not understand the following proof:
Note: This is not my homework problem. I encountered this calculation in a paper and got stuck.
You are right in your last comment. For the second bit one has $\frac{{6 - 8\log 2}}{{\pi ^2 }}$, so the whole thing is at least $\frac{8}{{\pi ^2 }}(1 - \log 2) \approx 0.248$, which is weaker than what is claimed in the paper. However, you can use the following simpler argument to obtain a better lower bound: $$ \frac{1}{{\pi D}}\int_0^{\frac{\pi }{D}} {\frac{{\sin ^2 (Dy)}}{{\sin ^2 y}}dy} = \frac{1}{{\pi D^2 }}\int_0^\pi {\frac{{\sin ^2 t}}{{\sin ^2 (t/D)}}dt} \ge \frac{1}{{\pi D^2 }}\int_0^\pi {\frac{{\sin ^2 t}}{{(t/D)^2 }}dt} = \frac{1}{\pi }\int_0^\pi {\frac{{\sin ^2 t}}{{t^2 }}dt} \\ = 0.4514116667 \ldots > \frac{4}{{\pi ^2 }} = 0.4052847345 \ldots > 0.405. $$