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Prove $$\int_0^{\pi/2} x\left({\sin nx\over \sin x}\right)^4\mathrm{d}x<{n^2\pi^2\over 8}.$$

My attempt: \begin{align} \int_0^{\pi/2} x\left({\sin nx\over \sin x}\right)^4\mathrm{d}x & =\sum_{k=1}^n \int_{{k-1\over 2n}\pi}^{{k\over 2n}\pi}x\left({\sin nx\over \sin x}\right)^4\mathrm{d}x\\ & \leq\sum_{k=1}^n \left({\pi\over 2}\right)^4 \int_{{k-1\over 2n}\pi}^{{k\over 2n}\pi}\left({\sin^4nx\over x^3}\right)\mathrm{d}x \quad (\text{use } \sin x\geq {2\over \pi}x ) \tag{1}\label{1}\\ &= \left({\pi\over 2}\right)^4 n^2\sum_{k=1}^n \int_{{k-1\over 2}\pi}^{{k\over 2}\pi}\left({\sin^4x\over x^3}\right)\mathrm{d}x \quad (\text{use } x\to {x\over n}).\\ \end{align} Is my direction right? If right, how can I prove the following $$\sum_{k=1}^n \int_{{k-1\over 2}\pi}^{{k\over 2}\pi}\left({\sin^4x\over x^3}\right)\mathrm{d}x\leq\int_0^{+\infty}\left({\sin^4x\over x^3}\right)\mathrm{d}x \leq {2\over \pi^2}. $$ I use Mathematica to calculate the integral $\int_0^{+\infty}\left({\sin^4x\over x^3}\right)\mathrm{d}x\simeq 0.7>{2\over\pi^2}$, hence my process (\ref{1}) seems to be wrong.

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  • $\begingroup$ I supposed $n$ positive integer. Is it right? $\endgroup$ – Raffaele Oct 3 '17 at 12:54
  • $\begingroup$ @Raffaele Yes it is. $\endgroup$ – yahoo Oct 3 '17 at 13:01
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The term $\left(\frac{\sin nx}{\sin x}\right)^4$ is associated with the Jackson kernel.
Your inequality is indeed just a minor variation on Lemma 0.5 in the linked notes, and it can be proved through the same technique: expand $|x|$ as a Fourier cosine series over $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$, do the same for $\left(\frac{\sin nx}{\sin x}\right)^4$, then apply orthogonality/Bessel's inequality.

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  • $\begingroup$ Great answer! The question is the first exercise of a chapter about integral, I thought it would be easy for me. By reading the material you give, I should estimate the $\int_0^{\pi\over 2n}$ part by using $\sin x\geq {2\over\pi}x$ and the rest using $x\geq {k\pi\over 2n}$. The first part it self still larger than the right hand side. So I need to give a more concise estimate rather than $\sin x\geq {2\over\pi} x$. By the way, if I accept the answer, how can I ask more people to find if there is a more simpler answer? $\endgroup$ – yahoo Oct 3 '17 at 14:38

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