Transforming this serie to a definite integral.

Question

Basically i need to simplify the following summation: $$\sum_{n=0}^\infty \frac{cos(nx)}{n^2}$$ As far as i know this summation is equal to $$\frac{x^2}{2}-\frac{\pi x}{4}+\frac{\pi ^2}{6}$$ when $[0\le x \le 2\pi]$. Now for the project I'm trying to calculate this for the value of $x$ is never included in such interval. So, given that $cos(x) = cos(x +2\pi k)$ i can actually solve this equation by changing the variable to $y=x-2\pi k$ and simplifying with this. But the resulting equation is an equation with 2 variables, $x$ and $k$, that ain't the result i was looking for (i know as well that $k$ is technically not a variable since you can actually find her, but to do so you need to use modulo which has no math equation and therefore is not the thing i was looking for).

So the alternative i have is to convert such summation in a definite integral, i guess. I've spent few hours looking for an actual method to do so without any result (since I'm a computer engineer, not a mathematician, and I've never had to study deeply calculus). I was wondering if you guys can actually point me to the right direction on this. To actually give you more infos about that my summation is in the form $$2\sum_{n=0}^\infty \frac{cos(nxm)}{n^2 m^2}$$ where $m$ is a generic multiplication factor. Ultimately I apologize for my English but understand it's not my first language.

Answer 1

Addressing the content of your first paragraph...

Let $x = \hat{x}+2\pi k$ for $k$ an integer. In your sum, $n$ is also an integer. Then \begin{align*} \cos(nx) &= \cos(n(\hat{x}+2\pi k)) \\ &= \cos(n\hat{x}+2\pi k n) \\ &= \cos(n\hat{x}) \text{,} \end{align*} because $kn$, the product of two integers, is also an integer. Consequently, the value of $k$ has no effect on the sum of your series.

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \sum_{n = \color{red}{1}}^{\infty}{\cos\pars{nx} \over n^{2}} & = \Re\sum_{n = \color{red}{1}}^{\infty}{\pars{\expo{\ic x}}^{n} \over n^{2}} = \Re\mrm{Li}_{2}\pars{\expo{\ic x}} \\[5mm] & = {1 \over 2}\pars{\mrm{Li}_{2}\pars{\expo{2\pi\ic\braces{x/\bracks{2\pi}}}} + \mrm{Li}_{2}\pars{\expo{-2\pi\ic\braces{x/\bracks{2\pi}}}}} \\[5mm] & = {1 \over 2}\bracks{-\,{\pars{2\pi\ic}^{2} \over 2!} \,\mrm{B}_{2}\pars{x \over 2\pi}}\,,\qquad\qquad \left\{\begin{array}{rcl} \ds{{x \over 2\pi} \in \left[0,1\right)} & \mbox{if} & \ds{\Im\pars{x} \geq 0} \\[2mm] \ds{{x \over 2\pi} \in \left(0,1\right]} & \mbox{if} & \ds{\Im\pars{x} < 0} \end{array}\right. \end{align}

See this link. $\ds{\mrm{B}_{n}}$ is a Bernoulli Polynomial. Note that $\ds{\mrm{B}_{2}\pars{z} = z^{2} - z + {1 \over 6}}$.

Then, \begin{align} \sum_{n = \color{red}{1}}^{\infty}{\cos\pars{nx} \over n^{2}} & = \pi^{2}\bracks{\pars{x \over 2\pi}^{2} - {x \over 2\pi} + {1 \over 6}} = \bbx{{1 \over 4}\,x^{2} - {\pi \over 2}\,x + {\pi^{2} \over 6}} \end{align}