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Reading a paper I've come across the following bound for an integral: $$ \int_{0}^{1}\frac{\sin^{2}\left(\,2 \pi n x\,\right)} {x \left\vert\,\log\left(\,{x/2}\,\right)\,\right\vert}\,\mathrm{d}x > C\log\left(\,\log\left(\,n + 2\,\right)\,\right) $$ for a constant $C > 0$ independent of $n$. I would like to show this.

Substituting $x \to \frac{x}{2\pi n}$ we get $$ \int_{0}^{1}\frac{\sin^{2}\left(\,2\pi nx\,\right)} {x\left\vert\,\log\left(\,x/2\,\right)\,\right\vert}\,\mathrm{d}x = \int_{0}^{2\pi n}\frac{\sin^{2}\left(\,x\,\right)} {\left\vert\,\log\left(\,x/\left[\,4\pi n\,\right]\,\right)\,\right\vert}\, \frac{\mathrm{d}x}{x}. $$ This looks like the logarithmic integral $\,\mathrm{li}\left(\,x\,\right) = \int_{0}^{x}\frac{\mathrm{d}t}{\log\left(\,t\,\right)}$, which satisfies $\,\mathrm{li}\left(\,x\,\right) > \log\left(\,\log\left(\,x\,\right)\,\right)$. However, I don't see how to get around the $\frac{\sin^{2}\left(\,x\,\right)}{x}$ in the integrand.

How can I prove this lower bound ?. Thanks !.

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  • $\begingroup$ Could we not say that since $\frac{\sin^2(x)}{x}$ is a function f such that$f:\mathbb{R}\rightarrow (0,1)$ it has to be less than the logarithmic integral, implying something similar to the desiredinequality? $\endgroup$ – Shinaolord Jun 8 '17 at 1:13
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Break the integration range as $\left[0,\frac{1}{4n}\right]\cup\left[\frac{1}{4n},1\right]$. The integral on the first part is positive and convergent to zero as $n\to +\infty$, hence we may simply focus on the second part. The function $\sin^2(z)$ has mean value $\frac{1}{2}$, hence it is enough to apply integration by parts / exploit a Fourier cosine transform and notice that $$ \int_{\frac{1}{4n}}^{1}\frac{\frac{1}{2}}{z\left|\log\frac{z}{2}\right|}\,dz = \frac{1}{2}\log\left(\frac{\log(8n)}{\log 2}\right)=\frac{1}{2}\log\left(3+\log_2(n)\right)$$ to prove the given claim.

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  • $\begingroup$ I'm not quite sure if I understand the way that you sketch. Why can I all of a sudden put $\frac{1}{2}$ instead of $\sin^2(x)$? And what do you exactly mean by "exploit a Fourier cosine transform"? $\endgroup$ – Steven Jun 6 '17 at 18:39
  • $\begingroup$ @Steven: how do you usually evaluate integrals involving a rapidly oscillating function? Do you know the Riemann-Lebesgue lemma? $\endgroup$ – Jack D'Aurizio Jun 6 '17 at 18:40
  • $\begingroup$ Probably I'm missing something essential here, but doesn't the Riemann-Lebesgue Lemma just say that the Fourier-transform vanishes as $x \to \pm \infty$? How can I use this? $\endgroup$ – Steven Jun 6 '17 at 18:42
  • $\begingroup$ @Steven: long story short: $\sin^2(2\pi n x)$ equals $\frac{1}{2}$ plus a periodic function with mean zero. Exploit that and profit. $\endgroup$ – Jack D'Aurizio Jun 6 '17 at 19:16
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    $\begingroup$ @FelixMarin: I am assuming that the $n$ appearing in OP's first integral is a positive integer. $\endgroup$ – Jack D'Aurizio Jun 8 '17 at 1:39

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