Find the limit $\lim\limits_{s\to0^+}\sum_{n=1}^\infty\frac{\sin n}{n^s}$

Question

This is a math competition problem for college students in Sichuan province, China. As the title, calculate the limit $$\lim_{s\to0^+}\sum_{n=1}^\infty\frac{\sin n}{n^s}.$$ It is clear that the Dirichlet series $\sum_{n=1}^\infty\frac{\sin n}{n^s}$ is convergent for all complex number $\Re s>0$. Here we only consider the case of real numbers.

Let $$A(x)=\sum_{n\leq x}\sin n,$$ then we have that $$A(x)=\frac{\cos\frac{1}{2}-\cos([x]+\frac{1}{2})}{2\sin\frac{1}{2}},$$ here $[x]$ is the floor function. Obviously, $A(x)$ is bounded and $|A(x)|\leq\frac{1}{\sin(1/2)}$. Using Abel's summation formula, we have that $$\sum_{n=1}^\infty\frac{\sin n}{n^s}=s\int_1^\infty\frac{A(x)}{x^{s+1}}\,dx =\frac{\cos\frac{1}{2}}{2\sin\frac{1}{2}}-s\int_1^\infty\frac{\cos([x]+\frac{1}{2})}{x^{s+1}}\,dx.$$

The integral $\int_1^\infty\frac{\cos([x]+\frac{1}{2})}{x^{s+1}}\,dx$ or $\int_1^\infty\frac{\cos([x])}{x^{s+1}}\,dx$ is also convergent for $s>-1$ (am I right? use Dirichlet's test)

My question: Is there an easy way to prove $$\lim_{s\to0^+}\int_1^\infty\frac{\cos([x])}{x^{s+1}}\,dx= \int_1^\infty\lim_{s\to0^+}\frac{\cos([x])}{x^{s+1}}\,dx= \int_1^\infty\frac{\cos([x])}{x}\,dx.$$ If the above conclusion is correct, we have that $$\lim_{s\to0^+}\sum_{n=1}^\infty\frac{\sin n}{n^s}=\frac{\cos\frac{1}{2}}{2\sin\frac{1}{2}}.$$

More generally, consider the Mellin tranform $g(s)=\int_1^\infty\frac{f(x)}{x^{s+1}}\,dx$, here $f(x)$ is continuous except integers and have left and right limit at integers.

If for $s=0$, the integral $\int_1^\infty\frac{f(x)}{x}\,dx$ is convergent, do we have that $$\lim_{s\to0^+}\int_1^\infty\frac{f(x)}{x^{s+1}}\,dx\stackrel{?}=\int_1^\infty\frac{f(x)}{x}\,dx\,$$(In Jameson's book The prime number theorem, page 124, there is a Ingham-Newman Tauberian thereom, but the conditions of the theorem there are slightly different from here.)

If the condition is strengthened to for all $s\geq-1/2$, $\int_1^\infty\frac{f(x)}{x^{s+1}}\,dx$ is convergent, is the following correct $$\lim_{s\to0^+}\int_1^\infty\frac{f(x)}{x^{s+1}}\,dx\stackrel{?}=\int_1^\infty\frac{f(x)}{x}\,dx\,$$ If this is correct, is there a simple way to prove it?

(2022/3/24/21:53) If the Dirichlet integral $\int_1^\infty\frac{f(x)}{x^s}dx$ converges at $s_0$, then it converges uniformly in $$|\arg(s-s_0)|\leq\alpha<\frac{\pi}{2}$$ for any fixed $0<\alpha<\frac{\pi}{2}$, and thus $$\lim_{s\to s_0^+}\int_1^\infty\frac{f(x)}{x^s}dx=\int_1^\infty\frac{f(x)}{x^{s_0}}dx.$$

For the uniform convergence of Dirichlet integral, see Uniform convergence about Dirichlet integral $f(s):=\int_1^\infty\frac{a(x)}{x^s}\,dx =\lim\limits_{T\to\infty}\int_1^T\frac{a(x)}{x^s}\,dx$

Answer 1

A Couple of Trigonometric Sums

First, we evaluate $$\newcommand{\Re}{\operatorname{Re}}\newcommand{\Im}{\operatorname{Im}} \begin{align} S_n &=\sum_{k=1}^n\sin(k)\tag{1a}\\ &=\Im\left(\frac{e^{i(n+1)}-1}{e^i-1}\right)\tag{1b}\\ &=\Im\left(e^{in/2}\right)\frac{\sin\left(\frac{n+1}2\right)}{\sin\left(\frac12\right)}\tag{1c}\\ &=\sin\left(\frac n2\right)\frac{\sin\left(\frac{n+1}2\right)}{\sin\left(\frac12\right)}\tag{1d}\\ &=\frac{\cos\left(\frac12\right)-\cos\left(n+\frac12\right)}{2\sin\left(\frac12\right)}\tag{1e} \end{align} $$ Explanation:
$\text{(1a)}$: definition
$\text{(1b)}$: apply Euler's Formula
$\phantom{\text{(1b):}}$ and the Formula for the Sum of a Geometric Series
$\text{(1c)}$: $\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$
$\text{(1d)}$: apply Euler's Formula
$\text{(1e)}$: $\sin(x)\sin(y)=\frac{\cos(x-y)-\cos(x+y)}2$

Similarly, $$ \begin{align} C_n &=\sum_{k=1}^n\cos\left(k+\frac12\right)\tag{2a}\\ &=\Re\left(\frac{e^{i(n+3/2)}-e^{i3/2}}{e^i-1}\right)\tag{2b}\\ &=\Re\left(e^{i(n+3)/2}\right)\frac{\sin\left(\frac{n}2\right)}{\sin\left(\frac12\right)}\tag{2c}\\ &=\cos\left(\frac{n+3}2\right)\frac{\sin\left(\frac{n}2\right)}{\sin\left(\frac12\right)}\tag{2d}\\ &=\frac{\sin\left(n+\frac32\right)-\sin\left(\frac32\right)}{2\sin\left(\frac12\right)}\tag{2e} \end{align} $$ Explanation:
$\text{(2a)}$: definition
$\text{(2b)}$: apply Euler's Formula
$\phantom{\text{(2b):}}$ and the Formula for the Sum of a Geometric Series
$\text{(2c)}$: $\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$
$\text{(2d)}$: apply Euler's Formula
$\text{(2e)}$: $\sin(x)\cos(y)=\frac{\sin(x+y)+\sin(x-y)}2$

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Estimating the Sum

Therefore, $$ \begin{align} \sum_{k=1}^\infty\frac{\sin(k)}{k^s} &=\sum_{k=1}^\infty\frac{S_k-S_{k-1}}{k^s}\tag{3a}\\ &=\sum_{k=1}^\infty S_k\left(\frac1{k^s}-\frac1{(k+1)^s}\right)\tag{3b}\\ &=\frac12\cot\left(\frac12\right)-\frac12\csc\left(\frac12\right)\sum_{k=1}^\infty\cos\left(k+\frac12\right)\left(\frac1{k^s}-\frac1{(k+1)^s}\right)\tag{3c}\\ &=\frac12\cot\left(\frac12\right)+O(s)\tag{3d} \end{align} $$ Explanation:
$\text{(3a)}$: $\sin(k)=S_k-S_{k-1}$
$\text{(3b)}$: Summation by Parts
$\text{(3c)}$: apply $(1)$ and $\sum\limits_{k=1}^\infty\left(\frac1{k^s}-\frac1{(k+1)^s}\right)=1$
$\text{(3d)}$: $\frac1{k^s}-\frac1{(k+1)^s}$ is monotonic decreasing and $1-\frac1{2^s}\le s$
$\phantom{\text{(3d):}}$ $(2)$ says that $\sup\limits_{n\ge0}\left|\sum\limits_{k=1}^n\cos\left(k+\frac12\right)\right|\le\csc\left(\frac12\right)$
$\phantom{\text{(3d):}}$ Dirichlet says $\left|\sum\limits_{k=1}^\infty\cos\left(k+\frac12\right)\left(\frac1{k^s}-\frac1{(k+1)^s}\right)\right|\le s\csc\left(\frac12\right)$

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The Requested Result

Estimate $(3)$ yields $$ \bbox[5px,border:2px solid #C0A000]{\lim_{s\to0^+}\sum_{k=1}^\infty\frac{\sin(k)}{k^s}=\frac12\cot\left(\frac12\right)}\tag4 $$

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Clarification

More than one comment has shown that the bound on the sum $$ \sum_{k=1}^\infty\cos\left(k+\frac12\right)\left(\frac1{k^s}-\frac1{(k+1)^s}\right)\tag5 $$ given in $\text{(3d)}$ requires clarification.

Using the Generalized Dirichlet Convergence Test, as presented in this answer, we will set $$ a_k=\cos\left(k+\frac12\right)\tag6 $$ and $$ b_k=\frac1{k^s}-\frac1{(k+1)^s}\tag7 $$ In $(2)$, it is shown that $$ \left|\,\sum_{k=1}^na_k\,\right|\le\csc\left(\frac12\right)\tag8 $$ Since $x^{-s}$ is convex, $$ \begin{align} &\overbrace{\left(\frac1{(k+1)^s}-\frac1{(k+2)^s}\right)}^{\large b_{k+1}}-\overbrace{\left(\frac1{k^s}-\frac1{(k+1)^s}\right)}^{\large b_k}\tag{9a}\\ &=2\left(\frac1{(k+1)^s}-\frac12\left(\frac1{k^s}+\frac1{(k+2)^s}\right)\right)\tag{9b}\\[4pt] &\le0\tag{9c} \end{align} $$ Explanation:
$\text{(9a)}$: definition of $b_k$
$\text{(9b)}$: combine terms
$\text{(9c)}$: $f\left(\frac{x+y}2\right)\le\frac{f(x)+f(y)}2$ for convex $f$

Thus, $b_k$ decreases monotonically to $0$. Therefore, the total variation of $b_k$ is $$ \begin{align} \sum_{k=1}^\infty|b_k-b_{k+1}| &=\sum_{k=1}^\infty(b_k-b_{k+1})\tag{10a}\\ &=b_1\tag{10b}\\[9pt] &=1-2^{-s}\tag{10c}\\[9pt] &=1-(1+1)^{-s}\tag{10d}\\[9pt] &\le1-(1-s)\tag{10e}\\[9pt] &=s\tag{10f} \end{align} $$ Explanation:
$\text{(10a)}$: $b_k$ is monotonically decreasing
$\text{(10b)}$: telescoping sum
$\text{(10c)}$: evaluate $b_1$
$\text{(10d)}$: $1+1=2$
$\text{(10e)}$: Bernoulli's Inequality
$\text{(10f)}$: simplify

Applying the Generalized Dirichlet Convergence Test to $(8)$ and $(10)$, we get $$ \left|\,\sum_{k=1}^\infty\cos\left(k+\frac12\right)\left(\frac1{k^s}-\frac1{(k+1)^s}\right)\,\right|\le s\csc\left(\frac12\right)\tag{11} $$

Answer 2

Your claim is correct and the tool is integration by parts. The function $f(x)=\cos(\lfloor x \rfloor + k)$ admits a primitive $F(x)$ which is uniformly bounded, and after IP, the new integral involves integrating $F(x)/(x^{s+2})$. It behaves nicely and you may take limits.

IP also provides some natural conditions for your second question. For example, it suffices that $f$ has a primitive $F$ such that $F(x)|< C(1+|x|^\alpha)$ with $\alpha<1$.

Incidentally, doing repeated IP also implies that $s\mapsto \sum_{n\geq 1} \frac{\sin(n)}{n^s}$ extends to an entire function in the complex plane.

For you last question in the generality stated I don't know.

Answer 3

I put Rugh's answer in detail here.

Let $f$ be an integrable function on $[a,b]$, $g$ a differentiable function on $[a,b]$ and let $$F(x)=\int_a^xf(t)\,dt,$$ then we have the integration by parts formula (see:Integration by Parts without Differentiation, Vicente Munoz,Mathematics Magazine , Vol. 85, No. 3 (June 2012), pp. 211-213): $$\int_a^bf(x)g(x)\,dx=g(x)\cdot F(x)\Big|_a^b-\int_a^bF(x)\,dg(x).$$ Here, all we need is that $f$ is integrable, not continuous.

Now, let $$F(x)=\int_1^x\cos([t]+a)\,dt,$$ Obviously, $|F(x)|\leq C$. or more generally, let $$F(x)=\int_1^x f(t)\,dt\quad\text{and assume that}\quad |F(x)|\leq C(1+|x|^\alpha),\quad \alpha<1.$$

Now $$\int_1^\infty\frac{f(x)}{x^{s+1}}\,dx= \frac{F(x)}{x^{s+1}}\Big|_1^\infty+(s+1)\int_1^\infty\frac{F(x)}{x^{s+2}}\,dx=(s+1)\int_1^\infty\frac{F(x)}{x^{s+2}}\,dx.$$ It is clear that $$\int_1^\infty\left|\frac{F(x)}{x^{s+2}}\right|\,dx\leq\int_1^\infty\frac{C(1+|x|^\alpha)}{x^{2}}\,dx<+\infty.$$ Hence $$\lim_{s\to0^+}\int_1^\infty\frac{f(x)}{x^{s+1}}\,dx=\lim_{s\to0^+}(s+1)\int_1^\infty\frac{F(x)}{x^{s+2}}\,dx=\int_1^\infty\frac{F(x)}{x^{2}}\,dx.$$ Since $$\int_1^\infty\frac{f(x)}{x}\,dx=\int_1^\infty\frac{dF}{x}=\int_1^\infty\frac{F(x)}{x^2}\,dx,$$ we have that $$\lim_{s\to0^+}\int_1^\infty\frac{f(x)}{x^{s+1}}\,dx=\int_1^\infty\frac{f(x)}{x}\,dx.$$

Answer 4

For $\Re(s) >1$ $$F(s)=\Gamma(s)\sum_{n\ge 1} \sin(n)n^{-s}=\sum_{n\ge 1}\int_0^\infty x^{s-1} \sin(n)e^{-nx}dx=\int_0^\infty x^{s-1} \Im(\frac{1}{e^{x-i}-1})dx $$ Next, note that $$F(s)- \Im(\frac{1}{e^{-i}-1})\Gamma(s)=\int_0^\infty x^{s-1} (\Im(\frac{1}{e^{x-i}-1})-\Im(\frac{1}{e^{-i}-1}) e^{-x})dx$$ converges and is continuous for $\Re(s) >-1$.

Whence $$\lim_{s\to 0}\sum_{n\ge 1} \sin(n)n^{-s}=\Im(\frac{1}{e^{-i}-1})=\frac{\cos(1/2)}{2\sin(1/2)}$$

Answer 5

Let $\operatorname{Li}_{s}(z)$ be the polylogarithm function of order $s$.

For $\Re(s) >0$, we have $$\sum_{n=1}^{\infty} \frac{\sin(n)}{n^{s}} = \Im \operatorname{Li}_{s}(e^{i}). $$

And since $\frac{\partial}{\partial z} \operatorname{Li}_{s+1}(z) = \frac{\operatorname{Li}_{s}(z)}{z}$, we have$$\operatorname{Li}_{0}(z) = z\frac{\mathrm d }{\mathrm d z}\operatorname{Li}_{1}(z) = - z \frac{\mathrm d}{\mathrm d z} \ln(1-z) = \frac{z}{1-z} . $$

Therefore, $$ \lim_{s \to 0^{+}}\sum_{n=1}^{\infty}\frac{\sin(n)}{n^{s}} = \Im \lim_{s \to 0^{+}} \operatorname{Li}_{s}(e^{i}) = \Im \, \frac{e^{i}}{1-e^{i}} = \Im \, \frac{i e^{i/2}}{2\sin \left(\frac{1}{2} \right)} = \frac{1}{2} \cot \left(\frac{1}{2} \right). $$

Similarly, we have$$\lim_{s \to 0^{+}}\sum_{n=1}^{\infty}\frac{\cos(n)}{n^{s}} = \Re \, \frac{e^{i}}{1-e^{i}} = - \frac{1}{2}. $$

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UPDATE:

To prove that $\lim_{s \to 0^{+}} \operatorname{Li}_{s}(e^{i}) = \operatorname{Li}_{0}(e^{i})$, we need to show that $\operatorname{Li}_{s}(e^{i})$ is continuous at $s=0$.

For $\Re(s) >0$ and all $z \in \mathbb{C}$ except $z$ real and $\ge 1$, the polylogarithm has the Mellin transform representation $$\operatorname{Li}_{s}(z) = \frac{z}{\Gamma(s)} \int_{0}^{\infty} \frac{x^{s-1}}{e^{x} -z} \, \mathrm dx. $$

If we integrate by parts, we get $$\operatorname{Li}_{s}(z) = \frac{z}{\Gamma(s+1)} \int_{0}^{\infty} \frac{x^{s}e^{x}}{\left(e^{x}-z\right)^{2}} \, \mathrm dx. \tag{1} $$

Since the Mellin transform defines an analytic function in the vertical strip where it converges converges absolutely (https://dlmf.nist.gov/1.14.iv), the above representation defines an analytic function in $s$, and thus a continuous function in $s$, in the half-plane $\Re(s) >-1$ for any fixed value of $z$ except $z$ real and $\ge 1$.

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We can check to make sure that $(1) $ gives the correct expression for $\operatorname{Li}_{0}(z)$.

$$\operatorname{Li}_{0}(z) = \frac{z}{\Gamma(1)} \int_{0}^{\infty} \frac{e^{x}}{(e^{x}-z)^{2}} = - \frac{z}{e^{x}-z} \Bigg|^{\infty}_{0} =0 + \frac{z}{1-z} = \frac{z}{1-z} $$