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Let $a\in (0,1)$ and define $$J(j):=\int_{0}^{\infty} e^{- 2^{j} s} \frac{s^{a}}{1+s^{2a}} ds,\quad j\in \mathbb{Z}.$$ Note that rescaling $2^{j} s\mapsto s$ shows that

$$J(j)\leq 2^{-j(1+a)}\int_{0}^{\infty} e^{- s} \frac{s^{a}}{1+2^{-2a j}s^{2a}} ds\sim 2^{-j(1+a)},\qquad \tag{1}$$ for all $j\in \mathbb{Z}$.

I am concerned with the dyadic sum $$\sum_{j\in \mathbb{Z}} 2^{ (\frac{d}{p}+1-a)j}J(j),$$ with $d,p\geq 1$, and given $a>{d}/{p}$. I also know that the factor $2^{ (\frac{d}{p}+1-a)j}$ is sharp. Using (1), we have

$$\sum_{j\in \mathbb{Z}^{+}} 2^{ (\frac{d}{p}+1-a)j}J(j)\lesssim \sum_{j\in \mathbb{Z}^{+}} 2^{ (\frac{d}{p}-2 a)j}\approx 1.$$

This reduces the problem to the sum over $\mathbb{Z}^{-}$. This is the part I need help with.

Notice that $s\mapsto\frac{d}{ds}\frac{s^{a}}{1+s^{2a}}$ is an $L^{1}([0,\infty))$ function. So, if one integrates by parts then uses the trivial estimate $$\int_{0}^{\infty} e^{- 2^{j} s} \left|\frac{d}{d s}\frac{s^{a}}{1+s^{2a}}\right| ds\lesssim 1$$ we get that $$J(j)\lesssim 2^{-j}.\quad (2)$$ But if we use (2) we end up with the divergent sum $\sum_{j\in \mathbb{Z}^{-}} 2^{ (\frac{d}{p}-a)j}$. This sum diverges since $a>\frac{d}{p}$.

The question is: Given $1>a>d/p>0$, does the following sum converge $$\sum_{j\in \mathbb{Z}^{-}} 2^{ (\frac{d}{p}+1-a)j}J(j).$$

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  • $\begingroup$ Unless I'm missing something, $d$, $p$ only occur in the combination $d/p$, so it would be easier to replace them by this single parameter. $\endgroup$
    – Christian Remling
    Commented Sep 29, 2023 at 16:16
  • $\begingroup$ Agreed. But, Iosif Pinelis has already answered it as it is. $\endgroup$
    – Medo
    Commented Sep 29, 2023 at 18:16
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    $\begingroup$ Another approach here is to compare the sum $\sum_{j \in {\bf Z}} 2^{(\frac{d}{p}+1-a) j} J(j)$ against the integral $\int_{\bf R} 2^{(\frac{d}{p}+1-a)u} J(u)\ du$ and then apply the Fubini-Tonelli theorem. After some changes of variable, the double integral will factor into two single integrals, both of which are easily seen to be convergent. $\endgroup$
    – Terry Tao
    Commented Sep 29, 2023 at 19:54

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$\newcommand\ep\epsilon\newcommand\Ga\Gamma\newcommand\Z{\mathbb Z}$The answer is yes.

Indeed, let $b:=d/p$, so that $1>a>b>0$. For real $\ep>0$, let $$K(\ep):=\int_0^\infty e^{-\ep s}\frac{s^a}{1+s^{2a}}\,ds \le\int_0^\infty e^{-\ep s}s^{-a}\,ds=C\ep^{a-1},$$ where $C:=\Ga(1-a)$. So, $$J(j)=K(2^j)\le C2^{(a-1)j}$$ and hence $$\sum_{j\in\Z^-} 2^{ (b+1-a)j}J(j) \le C\sum_{j\in\Z^-} 2^{bj}<\infty.\quad\Box$$

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    $\begingroup$ A useful rule of thumb is that exponential weights such as $e^{-\epsilon s}$ are morally restricting $s$ to the region $s=O(1/\epsilon)$, in which $\frac{s^a}{1+s^{2a}}$ is typically of size $\epsilon^a$, giving the $\epsilon^{a-1}$ predicted size for $K(\epsilon)$. These sorts of heuristics are good for quick back of the envelope calculations. $\endgroup$
    – Terry Tao
    Commented Sep 29, 2023 at 18:38
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    $\begingroup$ @TerryTao : Yes, this rule of thumb is certainly useful, and I did use something like this initially. Then I noticed that a simple inequality actually provides an asymptotically exact bound. $\endgroup$
    – Iosif Pinelis
    Commented Sep 29, 2023 at 19:26

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