# Solving Inequality with Integrals: $0<\int _{ 100 }^{ 200 }{ \frac { \sin(\pi x) }{ x } dx<\frac { 1 }{ 100\pi } }$

can someone explain me how to show this in a quick way

$$0<\int _{ 100 }^{ 200 }{ \frac { \sin(\pi x) }{ x } dx<\frac { 1 }{ 100\pi } }$$

$0<\int _{ 2n }^{ 2n+2 }{ \frac { \sin(\pi x) }{ x } dx<\frac { 1 }{ \pi } } *(\frac { 1 }{ n } -\frac { 1 }{ n+1 } )$ I showed this before, by splitting the integral and estimate it against 1/n and 1/n+1 however i used the fact that i can leave out te 1/x if i give it the maximus/minimum value it can have. But I can't do this with this one because I don't know about Maximum/Minimum

• Can you show your effort? – Jaideep Khare May 29 '17 at 8:31
• You mean $\Pi = \pi$ ? – Zubzub May 29 '17 at 8:40

$$I = \int_{0}^{+\infty}\sin(\pi x)\left(\frac{1}{x+100}-\frac{1}{x+200}\right)\,dx \stackrel{\mathcal{L}}{=} \int_{0}^{+\infty}\frac{\pi}{\pi^2+s^2}\left(e^{-100s}-e^{-200 s}\right)\,ds$$ by a useful property of the Laplace transform. It follows that $I>0$ and $$I < \int_{0}^{+\infty}\frac{e^{-100s}}{\pi}\,ds = \frac{1}{100\pi}$$ as wanted. We have indeed $$0 < I < \int_{0}^{+\infty}\frac{1}{\pi}\left(e^{-100s}-e^{-200s}\right)\,ds = \frac{1}{200\pi}.$$
Notice that this function you want to study has $0$ at every integer point. This suggests one to use Euler-Maclaurin formula with $$f(x)=\frac{\sin{\pi x}}{x}$$ and move the integral to one side and the rest to the other: $$\int_{100}^{200} f(x)\,dx \approx\sum_{n= 100}^{200} f(n) - \frac{f(200) + f(100)}{2} - \sum_{k=1}^\infty \,\frac{B_{2k}}{(2k)!} (f^{(2k - 1)}(200) - f^{(2k - 1)}(100))$$ Most terms go to zero! So we get $$\int_{100}^{200} f(x)\,dx \approx\sum_{k=1}^\infty \,\frac{B_{2k}}{(2k)!} (f^{(2k - 1)}(100) - f^{(2k - 1)}(200))$$ $$=\frac{\pi }{2400}+\frac{1}{720} \left(\frac{\pi ^3}{200}-\frac{21 \pi }{4000000}\right)+ \cdots= 0.001309+0.000215298$$ As always when evaluating asymptotic sums, the error is on the same order of magnitude as the last term we omit, so we only take the first term to get $$\int_{100}^{200} f(x)\,dx \approx \frac{\pi }{2400}$$ with error less than $0.0003$. This is enough to show your bound.