Encountered $\displaystyle{\sum_{k=1}^\infty\frac{(-1)^k\sin(kt)}{k}}$ while solving another integral

Question

I am trying to evaluate the following, $$\zeta=\sum_{k=1}^{\infty}\frac{1}{k}\int_{\pi k}^{\infty}\frac{\sin(x)}{x}\ dx$$ but I encountered the $\color{orange}{\text{series}}$, $$\color{orange}{\sum_{k=1}^\infty\frac{(-1)^k\sin(kt)}{k},\quad t\in(0,2\pi)}$$ which I do not know how to compute. By a little search I found that, $$\sum_{k=1}^\infty\frac{\sin(kt)}{k}=\frac{\pi-t}{2},\quad t\in(0,2\pi)$$ which would be useful if I was evaluating, $$\sum_{k=1}^\infty\frac{1}{k}\int^\infty_{\color{red}{2\pi k}}\frac{\sin(x)}{x}\ dx$$ but that is not the case. Any ideas on $\zeta$ or the series is welcome. My attempt is below.

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Let $x=u+\pi k$, $$\sum_{k=1}^{\infty}\frac{1}{k}\int_{\pi k}^{\infty}\frac{\sin(x)}{x}\ dx=\sum_{k=1}^\infty\frac{1}{k}\int_{0}^\infty\frac{\sin(u+\pi k)}{u+\pi k}\ du$$ note that for $k\in\mathbb{N}$, $\sin(u+\pi k)=\sin(u)\cos(\pi k)+\sin(\pi k)\cos(u)=(-1)^k\sin(u)$, $$\sum_{k=1}^\infty\frac{(-1)^k}{k}\int_{0}^\infty\frac{\sin(u)}{u+\pi k}\ du=\sum_{k=1}^\infty\frac{(-1)^k}{k}\int_0^\infty\frac{\sin(ks)}{s+\pi }\ ds$$ after $s=u/k$. Then we can exploit the periodicity of the integrand as follows, $$\int_0^\infty\frac{\sin(ks)}{s+\pi}\ ds=\sum_{n=0}^\infty\int_{2\pi n}^{2\pi(n+1)}\frac{\sin(ks)}{s+\pi}\ ds=\sum_{n=1}^\infty\int_0^{2\pi}\frac{\sin(kt)}{t+(2n-1)\pi}\ dt$$ where we made the substitution $t=s-2\pi n$ and re-indexed the series. After interchanging the series' and summing over $k$ pops up, $$\sum_{n=1}^\infty\int_0^{2\pi}\color{orange}{\sum_{k=1}^\infty\frac{(-1)^k\sin(kt)}{k\color{black}{(t+(2n-1)\pi)}}}\ dt$$

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Edit: Looking at the answers I am receiving, evaluating $\zeta$ may not be as straightforward as I initially thought; the $\color{\orange}{\text{series}}$ seems to complicate convergence a bit. I am going to add calculations for the analogous series, so maybe someone can help me find a solution to $\zeta$.

Let $u=x-2\pi k$ then $t=u/k$ (we could combine these substutions into one), $$\sum_{k=1}^\infty\frac{1}{k}\int^\infty_{\color{red}{2\pi k}}\frac{\sin(x)}{x}\ dx=\sum_{k=1}^\infty\frac{1}{k}\int_0^\infty\frac{\sin(u)}{u+2\pi k}\ du=\sum_{k=1}^\infty\frac{1}{k}\int_0^\infty\frac{\sin(kt)}{t+2\pi}\ dt$$ working similarly as above, $$\sum_{k=1}^\infty\frac{1}{k}\int_0^\infty\frac{\sin(kt)}{t+2\pi}\ dt=\sum_{k=1}^\infty\frac{1}{k}\sum_{n=0}^\infty\int_{2\pi n}^{2\pi(n+1)}\frac{\sin(kt)}{t+2\pi}\ dt=\sum_{k=1}^\infty\frac{1}{k}\sum_{n=1}^\infty\int_0^{2\pi}\frac{\sin(ks)}{s+2\pi n}\ ds$$ after $s=t-2\pi n$ and re-indexing. Switching the order of summation, $$\sum_{n=1}^\infty\sum_{k=1}^\infty\int_0^{2\pi}\frac{\sin(ks)}{k(s+2\pi n)}\ ds=\frac{1}{2}\sum_{n=1}^\infty\int_0^{2\pi}\frac{\pi-s}{s+2\pi n}\ ds$$ where I used the nicer result, $$\sum_{k=1}^\infty\frac{\sin(ks)}{k}=\frac{\pi-s}{2},\quad s\in(0,2\pi)$$ the integral is now easy considering the two numerators. Simple integration yields, $$\frac{\pi}{2}\sum_{n=1}^\infty\left((1+2n)\log\left(1+\frac{1}{n}\right)-2\right)$$ distributing the $1/2$ factor into the sum and rewriting as a product (we take limits), $$\lim_{N\to\infty}\pi\log\frac{1}{e^N}\prod_{n=1}^N\left(\frac{n+1}{n}\right)^{(n+1/2)}$$ the product telescopes, $$\lim_{N\to\infty}\pi\log\frac{\sqrt{N+1}}{e^N}\frac{(N+1)^N}{N!}$$ by Taylor's and Stirling's formula, $$\frac{\sqrt{N+1}}{e^N}\frac{(N+1)^N}{N!}\sim\frac{e}{\sqrt{2\pi}}\left(1-\frac{1}{12N}+\frac{25}{288N^2}-\cdots\right)$$ as $N\to+\infty$, hence, $$\sum_{k=1}^\infty\frac{1}{k}\int^\infty_{\color{red}{2\pi k}}\frac{\sin(x)}{x}\ dx=\pi \log\frac{e}{\sqrt{2\pi}}.$$

Answer 1

This does not answer the question.

I have been interested by your goal.

Consider the partial sum $$\zeta_n=\sum_{k=1}^n\frac{1}{k}\int_{\pi k}^{\infty}\frac{\sin(x)}{x}\ dx$$

$$\zeta_n=\frac{\pi }{2}H_n-\sum_{k=1}^n \frac{\text{Si}(k \pi )}{k}$$

Since $$\int \frac{\text{Si}(k \pi )}{k}\,dk=\frac{1}{2} \pi k \left(\, _3F_3(1,1,1;2,2,2;i k \pi )+\, _3F_3(1,1,1;2,2,2;-i k \pi )\right)$$ we can use Euler-MacLaurin summation.

The formula is quite nasty (in particular because of the hypergeometric functions).

Using it for small values of $n$

$$\left( \begin{array}{cccc} n & \frac{\pi }{2}H_n&\sum_{k=1}^n \frac{\text{Si}(k \pi )}{k} & \zeta_n \\ 10 & 4.60081 & 4.82672 & -0.225907 \\ 20 & 5.65132 & 5.87828 & -0.226963 \\ 30 & 6.27531 & 6.50248 & -0.227170 \\ 40 & 6.72072 & 6.94796 & -0.227244 \\ 50 & 7.06734 & 7.29461 & -0.227278 \\ 60 & 7.35112 & 7.57842 & -0.227297 \\ 70 & 7.59140 & 7.81871 & -0.227309 \\ 80 & 7.79976 & 8.02707 & -0.227316 \\ 90 & 7.98368 & 8.21100 & -0.227321 \\ 100 & 8.14831 & 8.37564 & -0.227325 \\ \end{array} \right)$$

slowly converging toward $-0.2273411730696824$ because

$$a_k=\frac{\pi -2 \text{Si}(k \pi )}{2 k} \quad \implies \quad \frac{a_{k+1}}{a_k}=-1+\frac{2}{k}-\frac{3}{k^2}+O\left(\frac{1}{k^3}\right)$$

Answer 2

If you put $\phi=\pi-x$ in the formula $\sum_{n=1}^{\infty}\frac{\sin n\phi}{n}=\frac{\pi-\phi}{2}$ that you saw in the book, you get $$\sum_{n=1}^{\infty}\frac{(-1)^n\sin nx}{n}=-\frac x2$$ for $-\pi<x<\pi.$ You may substract $2\pi$ from $x$ to find $$\sum_{n=1}^{\infty}\frac{(-1)^n\sin nx}{n}=-\frac x2+\pi$$ for $\pi<x<2\pi.$

Answer 3

$$\sin (x) = \Im (e^{ix}) \implies \sum_{k=1}^\infty\frac{(-1)^k\sin(kt)}{k} = \Im (\sum_{k=1}^\infty\frac{(-1)^ke^{ikt}}{k})$$

$$\sum_{k=1}^{\infty} x^k = \frac x{1-x} \implies \sum_{k=1}^{\infty} \frac{x^{k+1}}{k+1} = \int_{t=0}^x \frac t{1-t} dt = \\ =\int_{t=0}^x (-1) dt + \int_{t=0}^x \frac 1{1-t} dt = -x - \ln (1-x)$$

so $$\sum_{k=2}^{\infty} \frac{x^{k}}{k} = -x - \ln (1-x) \iff \sum_{k=1}^{\infty} \frac{x^{k}}{k} = -\ln(1-x)$$

done:

$$\Im (\sum_{k=1}^\infty\frac{(-1)^ke^{kt}}{k}) = \Im(-\ln(1+e^{it})) = -\arg(1+e^{it})$$

now $$1+e^{it} = 1 + \cos(t) + i \sin (t) \implies |1+e^{it}| = \sqrt{2+2\cos(t)} = 2 |\cos (\frac t2)|$$

so finally $$1+e^{it} = 2\cos(\frac t2)( \cos(\frac t2) + i \sin(\frac t2))$$

your answer is $$\sum_{k=1}^\infty\frac{(-1)^k\sin(kt)}{k} = -\frac t2, \text{for } t \in [0, \pi), 0 \text{ if } t=\pi$$

if $ \pi < t < 2\pi$, then $$\cos (\frac t2) <0$$ and the argument used is then $-\frac t2 + \pi$

Answer 4

Supplimentary Answer
Consider following where $m\in\mathbb{N}$: $$ \mathcal{S} _m=\sum_{n=1}^{\infty}{\frac{1}{n}\int_{mn\pi}^{\infty}{\frac{\sin \left( x \right)}{x}}\mathrm{d}x}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\mathcal{S} _{m}^{*}=\sum_{n=1}^{\infty}{\frac{\left( -1 \right) ^n}{n}\int_{mn\pi}^{\infty}{\frac{\sin \left( x \right)}{x}}\mathrm{d}x} $$ It can be shown that $$ \mathcal{S} _m=\begin{cases} \frac{\pi}{2}\ln \left( \Gamma ^2\left( \frac{m}{2} \right) \left( \frac{2e}{m} \right) ^m\frac{m}{4\pi} \right) \,\,\,\,\,\,\,\,\,\,\!\:\!\:\!\:, 2\mid m\\ \frac{\pi}{2}\ln \left( \Gamma ^2\left( \frac{m}{2}+\frac{1}{2} \right) \left( \frac{2e}{m} \right) ^m\frac{1}{2\pi} \right) , 2\nmid m\\ \end{cases} $$ $$ \mathcal{S} _{m}^{*}=\begin{cases} \frac{\pi}{2}\ln \left( \Gamma ^2\left( \frac{m}{2}+\frac{1}{2} \right) \left( \frac{2e}{m} \right) ^m\frac{1}{2\pi} \right) , 2\mid m\\ \frac{\pi}{2}\ln \left( \Gamma ^2\left( \frac{m}{2} \right) \left( \frac{2e}{m} \right) ^m\frac{m}{4\pi} \right) \,\,\,\,\,\,\,\,\,\,\!\:\!\:\!\:, 2\nmid m\\ \end{cases} $$ The reason for this "switching" phenomenon is due to $(-1)^n$ factor, two copies of it cancel each other out. If $m$ is a odd number, the sine term produces a $(-1)^n$, therefore the two cases switchs. Therefore, we can set $m$ to be a even number and be done with it. We need few lemmas before the whole proof

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lemma 1 $$ \int_0^{\infty}{\left( \frac{e^{-at}}{t}-\frac{be^{-ct}}{1-e^{-bt}} \right) \mathrm{d}t}=\ln \left( \frac{b}{a} \right) +\psi \left( \frac{c}{b} \right) $$ proof \begin{align*} \int_0^{\infty}{\left( \frac{e^{-at}}{t}-\frac{be^{-ct}}{1-e^{-bt}} \right) \mathrm{d}t}\stackrel{bt\rightsquigarrow t}{=}&\int_0^{\infty}{\left( \frac{e^{-\frac{a}{b}t}}{t}-\frac{e^{-\frac{c}{b}t}}{1-e^{-t}} \right) \mathrm{d}t} \\ =&\int_0^{\infty}{\frac{e^{-\frac{a}{b}t}-e^{-t}}{t}\mathrm{d}t}+\int_0^{\infty}{\left( \frac{e^{-t}}{t}-\frac{e^{-\frac{c}{b}t}}{1-e^{-t}} \right) \mathrm{d}t} \\ =&\ln \left( \frac{b}{a} \right) +\psi \left( \frac{c}{b} \right) \end{align*} It is required to use frullani integral and one of the integral representation of digamma function at the last step.

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lemma 2
\begin{align*} &\frac{\pi}{2z}\left( \coth \left( \frac{z}{m} \right) -\frac{m}{z} \right) =\sum_{k=1}^{\infty}{\frac{m\pi}{m^2z^2+m^2n^2\pi ^2}} \\ &\frac{\pi}{2z}\left( \mathrm{csch} \left( \frac{z}{m} \right) -\frac{m}{z} \right) =\sum_{k=1}^{\infty}{\frac{m\pi \left( -1 \right) ^n}{z^2+m^2n^2\pi}} \end{align*} proof
The results is followed by Mittag-Leffler expansion of $\mathrm{csch}$ and $\mathrm{coth}$

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lemma 3
$$ \ln \left( \Gamma \left( z \right) \right) =z\ln \left( z \right) -z-\frac{1}{2}\ln \left( z \right) +\frac{1}{2}\ln \left( 2\pi \right) +\mathcal{O} \left( \frac{1}{z} \right) $$ proof
Omited.

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Now, let's evaluate the first sum \begin{align*} \sum_{n=1}^{\infty}{\frac{1}{n}\int_{mn\pi}^{\infty}{\frac{\sin \left( x \right)}{x}}\mathrm{d}x}=&\sum_{n=1}^{\infty}{\frac{1}{n}\int_0^{\infty}{\frac{\sin \left( x \right)}{x+mn\pi}}\mathrm{d}x} \\ =&\sum_{n=1}^{\infty}{\frac{1}{n}\int_0^{\infty}{\sin \left( x \right) \int_0^{\infty}{e^{-xt}e^{-mn\pi t}\mathrm{d}t}}\mathrm{d}x} \\ =&\sum_{n=1}^{\infty}{\frac{1}{n}\int_0^{\infty}{e^{-mn\pi t}\int_0^{\infty}{\sin \left( x \right) e^{-xt}}}\mathrm{d}x\mathrm{d}t} \\ =&\sum_{n=1}^{\infty}{\frac{1}{n}\int_0^{\infty}{\frac{e^{-mn\pi t}}{1+t^2}}\mathrm{d}t} \\ =&\sum_{n=1}^{\infty}{\frac{1}{n}\int_0^{\infty}{e^{-mn\pi t}\int_0^{\infty}{\cos \left( xt \right) e^{-x}\mathrm{d}x}}\mathrm{d}t} \\ =&\sum_{n=1}^{\infty}{\frac{1}{n}\int_0^{\infty}{e^{-x}\int_0^{\infty}{\cos \left( xt \right) e^{-mn\pi t}\mathrm{d}t\mathrm{d}x}}} \\ =&\sum_{n=1}^{\infty}{\frac{1}{n}\int_0^{\infty}{e^{-x}\frac{mn\pi}{m^2n^2\pi ^2+x^2}}}\mathrm{d}x \\ =&\int_0^{\infty}{e^{-x}\sum_{n=1}^{\infty}{\frac{m\pi}{m^2n^2\pi ^2+x^2}}\mathrm{d}x} \\ =&\int_0^{\infty}{e^{-x}\frac{\pi}{2x}\left( \coth \left( \frac{x}{m} \right) -\frac{m}{x} \right) \mathrm{d}x} \\ =&\frac{\pi}{2}\int_0^{\infty}{\frac{e^{-x}}{x}\left( \frac{1+e^{-2x/m}}{1-e^{-2x/m}}-\frac{m}{x} \right) \mathrm{d}x} \end{align*} Now, by using Feynman's technique $$ \mathcal{S} \left( a \right) :=\frac{\pi}{2}\int_0^{\infty}{\frac{e^{-ax}}{x}\left( \frac{1+e^{-2x/m}}{1-e^{-2x/m}}-\frac{m}{x} \right) \mathrm{d}x} $$ Differentiate \begin{align*} \mathcal{S}' \left( a \right) =&\frac{\pi m}{4}\left( \int_0^{\infty}{\left( \frac{e^{-ax}}{x}-\frac{2}{m}\frac{e^{-ax}}{1-e^{-2x/m}} \right) \mathrm{d}x}+\int_0^{\infty}{\left( \frac{e^{-ax}}{x}-\frac{2}{m}\frac{e^{-\left( a+2/m \right) x}}{1-e^{-2x/m}} \right) \mathrm{d}x} \right) \\ =&\frac{\pi m}{4}\left( \ln \left( \frac{2}{am} \right) +\psi \left( \frac{m}{2}a \right) +\ln \left( \frac{2}{am} \right) +\psi \left( \frac{m}{2}a+1 \right) \right) \\ =&\pi \left( \frac{m}{2}\psi \left( \frac{m}{2}a \right) +\frac{1}{2a}-\frac{m}{2}\ln \left( \frac{m}{2}a \right) \right) \end{align*} Hence, \begin{align*} \sum_{n=1}^{\infty}{\frac{1}{n}\int_0^{\infty}{\frac{\sin \left( x \right)}{x+mn\pi}}\mathrm{d}x}=&\int_{\infty}^1{\mathcal{S} '\left( a \right) \mathrm{d}a}=\pi \int_{\infty}^1{\left( \frac{m}{2}\psi \left( \frac{m}{2}a \right) +\frac{1}{2a}-\frac{m}{2}\ln \left( \frac{m}{2}a \right) \right) \mathrm{d}a} \\ =&\pi \left[ \ln \left( \Gamma \left( \frac{m}{2}a \right) \right) +\frac{1}{2}\ln \left( a \right) -\frac{m}{2}a\ln \left( \frac{m}{2}a \right) +\frac{m}{2}a \right] _{R\rightarrow \infty}^{1} \\ =&\frac{\pi}{2}\ln \left( \Gamma ^2\left( \frac{m}{2} \right) \frac{2^me^mm}{4\pi m^m} \right) \end{align*} To evaluate the lower bound, stirling approximation is needed.

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One down, one to go \begin{align*} \sum_{n=1}^{\infty}{\frac{\left( -1 \right) ^n}{n}\int_{mn\pi}^{\infty}{\frac{\sin \left( x \right)}{x}}\mathrm{d}x}=&\sum_{n=1}^{\infty}{\frac{\left( -1 \right) ^n}{n}\int_0^{\infty}{\frac{\sin \left( x \right)}{x+mn\pi}}\mathrm{d}x} \\ =&\sum_{n=1}^{\infty}{\frac{\left( -1 \right) ^n}{n}\int_0^{\infty}{\sin \left( x \right) \int_0^{\infty}{e^{-xt}e^{-mn\pi t}\mathrm{d}t}}\mathrm{d}x} \\ =&\sum_{n=1}^{\infty}{\frac{\left( -1 \right) ^n}{n}\int_0^{\infty}{e^{-mn\pi t}\int_0^{\infty}{\sin \left( x \right) e^{-xt}\mathrm{d}x\mathrm{d}t}}} \\ =&\sum_{n=1}^{\infty}{\frac{\left( -1 \right) ^n}{n}\int_0^{\infty}{\frac{e^{-mn\pi t}}{1+t^2}\mathrm{d}t}} \\ =&\sum_{n=1}^{\infty}{\frac{\left( -1 \right) ^n}{n}\int_0^{\infty}{e^{-mn\pi t}\int_0^{\infty}{\cos \left( xt \right) e^{-x}\mathrm{d}x\mathrm{d}t}}} \\ =&\sum_{n=1}^{\infty}{\frac{\left( -1 \right) ^n}{n}\int_0^{\infty}{e^{-x}\int_0^{\infty}{\cos \left( xt \right) e^{-mn\pi t}\mathrm{d}t\mathrm{d}x}}} \\ =&\sum_{n=1}^{\infty}{\frac{\left( -1 \right) ^n}{n}\int_0^{\infty}{e^{-x}\frac{mn\pi}{m^2n^2\pi ^2+x^2}}}\mathrm{d}x \\ =&\int_0^{\infty}{e^{-x}\sum_{n=1}^{\infty}{\frac{m\left( -1 \right) ^n\pi}{m^2n^2\pi ^2+x^2}}}\mathrm{d}x \\ =&\int_0^{\infty}{e^{-x}\frac{\pi}{2x}\left(\mathrm{csch} \left( \frac{x}{m} \right) -\frac{m}{x} \right)}\mathrm{d}x \\ =&\frac{\pi}{2}\int_0^{\infty}{\frac{e^{-x}}{x}\left( \frac{2e^{-x/m}}{1-e^{-2x/m}}-\frac{m}{x} \right)}\mathrm{d}x \end{align*} Feynman's technique $$ \mathcal{S} _*\left( a \right) =\frac{\pi}{2}\int_0^{\infty}{\frac{e^{-ax}}{x}\left( \frac{2e^{-x/m}}{1-e^{-2x/m}}-\frac{m}{x} \right)}\mathrm{d}x $$ Then \begin{align*} \mathcal{S} _*'\left( a \right) =&\frac{\pi m}{2}\int_0^{\infty}{\left( \frac{e^{-ax}}{x}-\frac{2}{m}\frac{e^{-\left( a+1/m \right) x}}{1-e^{-2x/m}} \right)}\mathrm{d}x \\ =&\pi \left( \frac{m}{2}\psi \left( \frac{m}{2}a+\frac{1}{2} \right) -\frac{m}{2}\ln \left( \frac{m}{2}a \right) \right) \end{align*} Therefore \begin{align*} \sum_{n=1}^{\infty}{\frac{\left( -1 \right) ^n}{n}\int_0^{\infty}{\frac{\sin \left( x \right)}{x+mn\pi}}\mathrm{d}x}=\int_{\infty}^1{\mathcal{S} _*'\left( a \right) \mathrm{d}a}=&\pi \int_{\infty}^1{\left( \frac{m}{2}\psi \left( \frac{m}{2}a+\frac{1}{2} \right) -\frac{m}{2}\ln \left( \frac{m}{2}a \right) \right) \mathrm{d}a} \\ =&\pi \left[ \ln \left( \Gamma \left( \frac{m}{2}a+\frac{1}{2} \right) \right) -\frac{m}{2}a\ln \left( \frac{m}{2}a \right) +\frac{m}{2}a \right] _{R\rightarrow \infty}^{1} \\ =&\frac{\pi}{2}\ln \left( \Gamma ^2\left( \frac{m}{2}+\frac{1}{2} \right) \frac{e^m2^m}{2\pi m^m} \right) \end{align*} To evaluate the lower bound, stirling approximation is needed.

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result
With the formula proven, the following equalities become trivial. \begin{align*} &\sum_{n=1}^{\infty}{\frac{\left( -1 \right) ^n}{n}\int_{2n\pi}^{\infty}{\frac{\sin \left( x \right)}{x}}\mathrm{d}x}=\frac{\pi}{2}\ln \left( \frac{e^2}{8} \right) \\ &\sum_{n=1}^{\infty}{\frac{\left( -1 \right) ^n}{n}\int_{n\pi}^{\infty}{\frac{\sin \left( x \right)}{x}}\mathrm{d}x}=\frac{\pi}{2}\ln \left( \frac{e}{2} \right) \\ &\sum_{n=1}^{\infty}{\frac{1}{n}\int_{2n\pi}^{\infty}{\frac{\sin \left( x \right)}{x}\mathrm{d}x}}=\frac{\pi}{2}\ln \left( \frac{e^2}{2\pi} \right) \\ &\sum_{n=1}^{\infty}{\frac{1}{n}\int_{n\pi}^{\infty}{\frac{\sin \left( x \right)}{x}\mathrm{d}x}}=\frac{\pi}{2}\ln \left( \frac{e}{\pi} \right) \end{align*} Last thing I would add is that, after some investigation, I don't think the cos variant of this sum has a closed form, otherwise I would type the solution out here.