Is this an invalid way to compute $\int_{2\pi}^\infty \frac{\sin(x)}{x} \, dx$? So by using tabular integration:
$$\int_{2\pi}^\infty \frac{\sin(x)}{x} \, dx$$ $$= \left. -\frac{1}{x}\cos(x) - \frac{1}{x^2}\sin(x) + \frac{2}{x^3}\cos(x) + \frac{2\cdot3}{x^4}\sin(x) - \frac{2\cdot3\cdot4}{x^5}\cos(x) - \frac{2\cdot3\cdot4\cdot5}{x^6}\sin(x) + \cdots \right|_{2\pi}^\infty$$
adding all the cosine terms gives:
$$\left.-\frac{1}{x}\cos(x)+ \frac{2}{x^3}\cos(x)- \frac{2\cdot3\cdot4}{x^5} \cos(x) + \cdots = \sum_{n=1}^\infty \frac{(-1)^n (2n-2)!}{x^{2n-1}} \cos(x) \right|_{2\pi}^\infty$$
and similarly for the sine terms:
$$\left.-\frac{1}{x^2}\sin(x)+ \frac{2\cdot3}{x^4}\sin(x)- \frac{2\cdot3\cdot4\cdot5}{x^6}\sin(x)+\cdots =\sum_{n=1}^\infty \frac{(-1)^n \cdot (2n-1)!}{x^{2n}} 
\sin(x) \right|_{2\pi}^\infty$$
So putting everything back together we have:
$$\left.\int_{2\pi}^\infty \frac{\sin(x)}{x} \, dx = \sum_{n=1}^\infty \frac{(-1)^n (2n-2)!}{x^{2n-1}} \cos(x)+ \sum_{n=1}^\infty \frac{(-1)^n (2n-1)!}{x^{2n}} \sin(x) \right|_{2\pi}^\infty$$
I think that as $x$ approaches $\infty$ the fractions in both sums approach $0$, so the top limit does not yield anything from the sums. Likewise, $\sin(2\pi) = 0$ and $\cos(2\pi) = 1$, so all that we are left with is:
$$\int_{2\pi}^\infty \frac{\sin(x)}{x} \, dx = -\sum_{n=1}^\infty \frac{(-1)^n (2n-2)!}{(2\pi)^{2n-1}}$$
However, putting this sum into wolfram I get that the sum is divergent, which doesn't seem right to me since $ f(x) = \frac{\sin(x)}{x}$ decreases to $0$ quite fast. 
Is my logic inconsistent here? Have I done the integration incorrectly, or used a technique improperly/ inappropriately?
 A: $\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
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 \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)}
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\begin{align}
\color{#f00}{\int_{2\pi}^{\infty}{\sin\pars{x} \over x}\,\dd x}\ &\ =\
\overbrace{\int_{0}^{\infty}{\sin\pars{x} \over x}\,\dd x}^{\ds{\pi \over 2}}\ -\
\overbrace{\int_{0}^{2\pi}{\sin\pars{x} \over x}\,\dd x}^{\ds{\mrm{Si}\pars{2\pi}}}\ =\
\color{#f00}{{\pi \over 2} - \mrm{Si}\pars{2\pi}}
\end{align}

$\ds{\mrm{Si}}$ is the
  Sine Integral Function.

The $\,\mrm{Si}\pars{z}$ series expansion is given by:
$$
\mrm{Si}\pars{z} = \sum_{n = 0}^{\infty}\pars{-1}^{n}\,
{z^{2n + 1} \over \pars{2n + 1}!\pars{2n + 1}}
$$
