Intereting Posts

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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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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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