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I am wondering if there is a uniform bound $C$ (independent of $\lambda>10$): $$\sum_{k=-\infty}^{-1}\Big|\int_{2^k}^{2^{1+k}}\frac{\sin(\lambda t^3)}{t}dt\Big|\le C.$$

Remark: (1) An easy upper bound is $C\log\lambda$, since $$\Big|\int_{2^k}^{2^{1+k}}\frac{\sin(\lambda t^3)}{t}dt\Big|\lesssim \min\{1,\lambda 2^{3k}\},$$ by $|\sin(\lambda t^3)|\le \min\{1,\lambda t^3\}.$

(2) I can verify this uniform bound by Matlab, but I am not able to find a proof.

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You are not exploiting that the integrands are oscillatory for $\lambda t^3\gtrsim 1$, and the usual way to do this is by integration by parts. Let me change notations slightly and use $n=-k$. $$ \int_{2^{-n-1}}^{2^{-n}} e^{i\lambda t^3} \, \frac{dt}{t} =\int_{2^{-3n-3}}^{2^{-3n}} e^{i\lambda s}\, \frac{ds}{3s} = O(2^{3n}/\lambda) - i \int_{2^{-3n-3}}^{2^{-3n}} e^{i\lambda s}\frac{ds}{3\lambda s^2} = O(2^{3n}/\lambda) , $$ so summing over those $n$ for which $\lambda 2^{-3n}\ge 1$ makes an overall contribution $\lesssim 1$.

This is also true for the $n$'s with $\lambda 2^{-3n}\le 1$, by the argument you already gave, so your claim follows.

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