How can we prove that $$\int_{\pi/4}^{\pi/2}\frac{\sin x }{x}\leq \sqrt{2}/2.$$
I tried all (easy) direct estimators but all of them gave values bigger than this estimator.
Thanks
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We have that $\sin\left(x\right) $ is a positive monotone increasing function on $\left[\frac{\pi}{4},\frac{\pi}{2}\right] $ hence $$\int_{\pi/4}^{\pi/2}\frac{\sin\left(x\right)}{x}dx\leq\sin\left(\frac{\pi}{2}\right)\int_{\pi/4}^{\pi/2}\frac{1}{x}dx=\log\left(2\right)<\frac{\sqrt{2}}{2}.$$ Another way is to observe that $\sin\left(x\right)/x $ is a monotone decreasing function on $\left[\frac{\pi}{4},\frac{\pi}{2}\right] $ hence $$\int_{\pi/4}^{\pi/2}\frac{\sin\left(x\right)}{x}dx\leq\frac{4\sin\left(\pi/4\right)}{\pi}\int_{\pi/4}^{\pi/2}1dx=\sin\left(\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}. $$
Marco Cantarini
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@Mathmath It's only a numerical computation. $\log(2)\approx 0.693$ and $\frac{\sqrt{2}}{2} \approx 0.707$. – Marco Cantarini Feb 29 '16 at 08:45
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@Mathmath The inequality $\log(x)<1/ \sqrt{x}$ holds if $0<x<e^{2W(1/2)}$ where $W(x)$ is the Lambert function. But I don't think you saw this function. Why do want to you prove it without calculator? It's just numbers. However with some numerically estimation technique you can find the result. – Marco Cantarini Feb 29 '16 at 08:51
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@Mathmath See for example here http://math.stackexchange.com/questions/75074/an-alternative-way-to-calculate-logx – Marco Cantarini Feb 29 '16 at 08:59
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@Mathmath You can also use the series representation $$\log\left(2\right)=\sum_{n\geq1}\frac{\left(-1\right)^{n+1}}{n}$$ and the error that can you obtain from an alternating series https://en.wikipedia.org/wiki/Alternating_series_test . – Marco Cantarini Feb 29 '16 at 09:07
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Thank you, I think I have a direct way $\sin x/x \leq \sin x$ and that solves the problem directly. – Mathmath Feb 29 '16 at 10:54
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On your interval, $\sin(x)$ is concave function, so $\sin(x) \le \sin(\pi/4) + \cos(\pi/4)*(x-\pi/4)=\sqrt{2}/2*(1+x-\pi/4)$. If you integrate that divided by $x$ as in your integrand, you get an upper limit $\sqrt{2}/2 \times K$, whereby $K$ comes out smaller, but close to $1$ ($K=\pi/4+(1-\pi/4)*\ln{2}\approx 0.93$. Surely, perhaps you can prove how $K<1$ to make this proof easier. Hope I am not too far from your expected proof...
Chip
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We have to estimate: $$ I = \int_{0}^{\pi/4}\frac{\cos x}{\frac{\pi}{2}-x}\,dx\stackrel{CS}{\leq}\sqrt{\int_{0}^{\pi/4}\cos^2(x)\,dx\int_{0}^{\pi/4}\frac{dx}{(\pi/2-x)^2}}=\sqrt{\frac{2+\pi}{4\pi}} $$ ($CS$ stands for Cauchy-Schwarz) and $\sqrt{\frac{2+\pi}{4\pi}}<\sqrt{\frac{1}{2}}$ since $2+\pi<2\pi$.
As an alternative, since $\frac{\sin x}{x}=\prod_{n\geq 1}\left(1-\frac{x^2}{n^2\pi^2}\right)$ due to the Weierstrass product, we have: $$ I \leq \int_{\pi/4}^{\pi/2}\left(1-\frac{x^2}{\pi^2}\right)\,dx = \frac{41\pi}{192}<\frac{41\sqrt{10}}{192}<\frac{1}{\sqrt{2}}.$$
As a third way, since $f(x)=\frac{\sin(x)}{x}$ is a concave function on the interval $\left[\frac{\pi}{4},\frac{\pi}{2}\right]$, by the Hermite-Hadamard inequality we have: $$ I = \int_{\pi/4}^{\pi/2}f(x)\,dx \leq \frac{\pi}{4}\cdot f\left(\frac{3\pi}{8}\right)=\frac{2}{3}\cdot\cos\left(\frac{\pi}{8}\right)=\frac{1}{3}\sqrt{2+\sqrt{2}}. $$
By interpolating $\frac{\cos(x)}{\frac{\pi}{2}-x}+\frac{\cos\left(\frac{\pi}{4}-x\right)}{\frac{\pi}{4}+x}$ with respect to the endpoints and the midpoint of $\left[0,\frac{\pi}{4}\right]$, we also get the improved inequality:
$$ \color{red}{\int_{\pi/4}^{\pi/2}\frac{\sin x}{x}\,dx} \leq \frac{1}{12}(1+\sqrt{2})+\frac{2}{9}\sqrt{2+\sqrt{2}}=\color{red}{0.6117}97589\ldots $$
where the difference between the two terms is smaller than $2\cdot 10^{-5}$.
Jack D'Aurizio
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