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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

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

$$I = \int_{0}^{+\infty}\sin(\pi x)\left(\frac{1}{x+100}-\frac{1}{x+200}\right)\,dx \stackrel{\mathcal{L}}{=} \int_{0}^{+\infty}\frac{\pi}{\pi^2+s^2}\left(e^{-100s}-e^{-200 s}\right)\,ds $$ 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 $$ 0 < I < \int_{0}^{+\infty}\frac{1}{\pi}\left(e^{-100s}-e^{-200s}\right)\,ds = \frac{1}{200\pi}. $$

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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.

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Since i havent studied the Euler Maclaurine Formula, I can hardly understand ur Results but nevertheless i have found another Possibility to solve it. I posted the Integral with Index n at the bounderys, All one has to do is splitting the Integral in two Parts for n=50 so u calculate the integral from 100 to 150 and evaluted it to 1/100Pi and than doing the same thing again from 150 to 200.

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