# Improper integrals and Dirichlet criterion

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.

• can u prove what u claimed is sufficient is sufficient? i'm skeptical – mathworker21 Apr 17 at 22:19
• Can you specify you do not understand? – Diesirae92 Apr 17 at 22:27
• I'm asking you to prove the Dirichlet test thing. I.e., if there is a such a $\Gamma$, then the answer is 'yes'. – mathworker21 Apr 17 at 22:28
• +1 for this nice question with informative updates. – Saad Apr 18 at 6:42

$$\def\R{\mathbb{R}}\def\N{\mathbb{N}}\def\d{\mathrm{d}}\def\e{\mathrm{e}}\def\peq{\mathrel{\phantom{=}}{}}$$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.

• Very good. Seems correct. Let me check the details and if everything works the bounty is yours. – Diesirae92 Apr 18 at 9:03
• @Diesirae92 Thanks. Note that this counterexample works even under the additional condition deleted in your newest edits. – Saad Apr 18 at 9:05
• 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. – Diesirae92 Apr 18 at 9:13

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.

• @Diesirae92 can you explain how? I did read properly – mathworker21 Apr 17 at 22:36
• I've edited the question to make your counterexample not valid (sorry I was the one not reading properly). – Diesirae92 Apr 17 at 22:55
• cmon, that's not fair. i spent time writing this whole answer – mathworker21 Apr 18 at 0:15
• 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. – Diesirae92 Apr 18 at 7:57