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Let $\delta$ be a smooth (can be also taken analytic) function on $(0,\infty)$ such that:
(i)$\lim\limits_{s\to\infty}\delta(s)=0$,
(ii)$\lim\limits_{s\to\infty}s\delta^\prime(s)=0$.

Although I do not have an explicit expression for $\delta$, I know that $\lim\limits_{s\to\infty} s^m\delta^{(m)}(s)=0$ for any $m\in\mathbb{N}$.

Suppose further that there are an $s_0>1$ and an $M>0$ such that for any $r,s>s_0$ we have: $$\left\lvert\int_r^s \frac{\delta(\lambda)}{\lambda}\,\mathrm d\lambda\right\rvert\leq M.\qquad\tag{$*$}$$

Does this imply that the improper integral $\displaystyle\lim\limits_{s\to \infty} \int_{s_0}^s \frac{\delta(\lambda)}{\lambda}\,\mathrm d\lambda$ exists?

Under quite mild assumptions the integral exists. Indeed it suffices by Dirichlet test that there exists a non-negative, increasing function $\Gamma(\lambda)$ such that $\lim\limits_{\lambda\to\infty} \Gamma(\lambda)=\infty$ and: $$\left\lvert\int_{s}^t\frac{\Gamma(s)\delta(s)}{s}\,\mathrm ds\right\rvert\leq M.$$ To apply the criterion and get the convergence it suffices to note that $1/\Gamma$ is decreasing and infinitesimal at infinity.

Remark: for $\delta$ monotone or with fixed sign the result holds too.

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  • $\begingroup$ can u prove what u claimed is sufficient is sufficient? i'm skeptical $\endgroup$ – mathworker21 Apr 17 at 22:19
  • $\begingroup$ Can you specify you do not understand? $\endgroup$ – Diesirae92 Apr 17 at 22:27
  • $\begingroup$ I'm asking you to prove the Dirichlet test thing. I.e., if there is a such a $\Gamma$, then the answer is 'yes'. $\endgroup$ – mathworker21 Apr 17 at 22:28
  • $\begingroup$ +1 for this nice question with informative updates. $\endgroup$ – Saad Apr 18 at 6:42
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$\def\R{\mathbb{R}}\def\N{\mathbb{N}}\def\d{\mathrm{d}}\def\e{\mathrm{e}}\def\peq{\mathrel{\phantom{=}}{}}$The proposition is not necessarily true.

Lemma 1: If $f \in C^∞((0, +∞))$ and $g(x) := f(\e^x)$, then $g \in C^∞(\R)$ and the following conditions are equivalent:

  1. $\lim\limits_{s → +∞} s^m f^{(m)}(s) = 0$ for all $m \in \N$;
  2. $\lim\limits_{x → +∞} g^{(m)}(x) = 0$ for all $m \in \N$.

(In fact, $g^{(m)}(x)$ is a linear combination of $f(s), s f'(s), \cdots, s^m f^{(m)}(s)$ with $s = \e^x$.)

Lemma 2: If $h \in C^∞((0, +∞))$, then for each $m \in \N_+$, there exist polynomials $P_m, Q_m \in \R[h_1, \cdots, h_m]$ such that\begin{align*} (\sin(h(x)))^{(m)} &= P_m(h'(x), \cdots, h^{(m)}(x)) \sin(h(x))\\ &\peq + Q_m(h'(x), \cdots, h^{(m)}(x)) \cos(h(x)).\quad \forall x > 0 \end{align*}

(It can be easily proved by induction on $m$.)

Now define $G(x) = \sin(h(x))$, where $h(x) = x^a$ and $a \in (0, 1)$ is a constant, and take $δ(s) = G'(\ln s)$. Since $\lim\limits_{x → +∞} h^{(m)}(x) = 0$ for any $m \in \N_+$, then Lemma 2 yields that $\lim\limits_{x → +∞} G^{(m)}(x) = 0$ for all $m \in \N_+$, and combining Lemma 1 shows that $\lim\limits_{s → +∞} s^m δ^{(m)}(s) = 0$ for all $m \in \N$.

For $s > r > 1$, making the substitution $u = \e^x$ yields\begin{gather*} \left| \int_r^s \frac{δ(u)}{u} \,\d u \right| = \left| \int_{\ln r}^{\ln s} δ(\e^x) \,\d x \right| = \left| \int_{\ln r}^{\ln s} G'(x) \,\d x \right|\\ = |G(\ln s) - G(\ln r)| \leqslant |G(\ln s)| + |G(\ln r)| \leqslant 2. \end{gather*}

For $B > A > 1$ and any $s > 1$, because\begin{align*} &\peq |G(\ln(Bs)) - G(\ln(As)))| = |\sin(h(\ln(Bs))) - \sin(h(\ln(Bs)))|\\ &= 2 \left| \sin\left( \frac{1}{2} (h(\ln(Bs)) - h(\ln(As))) \right) \right| · \left| \cos\left( \frac{1}{2} (h(\ln(As)) + h(\ln(Bs))) \right) \right|\\ &\leqslant 2 \left| \frac{1}{2} (h(\ln(Bs)) - h(\ln(As))) \right| = |(\ln s + \ln B)^a - (\ln s + \ln A)^a| \end{align*} and $\lim\limits_{s → +∞} ((\ln s + \ln B)^a - (\ln s + \ln A)^a) = 0$, so$$ \lim_{s → +∞} \int_{As}^{Bs} \frac{δ(u)}{u} \,\d u = \lim_{s → +∞} (G(\ln(Bs)) - G(\ln(As))) = 0. $$

However, for any $s > s_0 > 1$, since$$ \int_{s_0}^s \frac{δ(u)}{u} \,\d u = G(\ln s) - G(\ln s_0) $$ and $\lim\limits_{x → +∞} G(x)$ does not exist, then $\displaystyle \int_{s_0}^s \frac{δ(u)}{u} \,\d u$ does not exist.

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  • $\begingroup$ Very good. Seems correct. Let me check the details and if everything works the bounty is yours. $\endgroup$ – Diesirae92 Apr 18 at 9:03
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    $\begingroup$ @Diesirae92 Thanks. Note that this counterexample works even under the additional condition deleted in your newest edits. $\endgroup$ – Saad Apr 18 at 9:05
  • $\begingroup$ very very good indeed. Beautifully done. My guess was that something like $\sin(\log(\lambda))/\log(\log(\lambda))$ would provide a counterexample. The bounty will be yours as soon as I can assign it. $\endgroup$ – Diesirae92 Apr 18 at 9:13
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I don't think so.

Let $\delta(\lambda) = \epsilon(\lambda)\frac{1}{\log\lambda}$ where $\epsilon(\lambda)$ is always $+1$ or $-1$. Take $s_0 = 10$ (just so $\log\lambda$ is fine) and $M=1$. Since $\int_{s_0}^\infty \frac{1}{\lambda\log\lambda} = +\infty$, there is some $s_1$ with $\int_{s_0}^{s_1} \frac{1}{\lambda\log\lambda} = +1$. Take $\epsilon(\lambda)$ to be $+1$ on $[s_0,s_1]$. Then since $\int_{s_1}^{\infty} \frac{-1}{\lambda\log\lambda} = -\infty$, there is some $s_2$ with $\int_{s_1}^{s_2} \frac{-1}{\lambda\log\lambda} = -1$. Take $\epsilon(\lambda)$ to be $-1$ on $[s_1,s_2]$. Then take $\epsilon(\lambda)$ to be $+1$ on $[s_2,s_3]$, where $\int_{s_2}^{s_3} \frac{1}{\lambda\log\lambda} = +1$. Etc. etc.

You might complain that $\delta$ is not smooth. If you don't complain, this answer is complete. If you do complain, then you can just modify $\delta$ by making $\epsilon$ smooth instead of sharply switching from $-1$ to $+1$, with the switch happening fast enough to at most have (*) hold for $M=2$, say, instead of $M=1$.

I started off this answer with "I don't think so" instead of "No", since I feel like there might be a $\Gamma$ as you described for my $\delta$, which would contradict my answer, so I'm partially confused.

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  • $\begingroup$ @Diesirae92 can you explain how? I did read properly $\endgroup$ – mathworker21 Apr 17 at 22:36
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    $\begingroup$ I've edited the question to make your counterexample not valid (sorry I was the one not reading properly). $\endgroup$ – Diesirae92 Apr 17 at 22:55
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    $\begingroup$ cmon, that's not fair. i spent time writing this whole answer $\endgroup$ – mathworker21 Apr 18 at 0:15
  • $\begingroup$ Sorry about that, I upvoted your answer since it provides a nice solution in an elementary case. Unfortunately I realized about the convergence of the derivatives only after the question was posted. $\endgroup$ – Diesirae92 Apr 18 at 7:57

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