Evaluating the Integral: $\int_0^\infty \frac{( \frac{1}{2} - \cos x )}{x} dx$ Evaluating the Integral:
$\int_0^\infty\left[\frac{1}{2} - \cos\left(x\right)\right]\,{\rm dx \over x}$ 
I came upon this limit: $\lim_{x\rightarrow\infty} -Ci(x) + Ci(1/x) +\ln(x)$, is it $\gamma$ ?
Here $ Ci(x) = \gamma + \ln x + \int_0^x \frac{\cos t -1}{t} dt $ is the cosine integral and $\gamma$ is the Euler constant. The Limit and the Integral appear to be equal.

$$\int_0^{+\infty}\left(\frac 1 2 - \cos x\right)\frac {dx} x = \gamma$$
\begin{align*}\int_{1/n}^n \left(\frac 1 2 - \cos x\right)\frac {dx} x &=
\int_{1/n}^1 \left(1 - \cos x\right)\frac {dx} x
-\int_1^n \cos x\frac {dx} x+
\frac 1 2 \left(\int_1^n \frac {dx} x -\int_{1/n}^1 \frac {dx} x\right) \\
&=\int_{1/n}^1 (1-\cos x) \frac {dx} x - \int_1^n \cos x \frac {dx} x
\end{align*}
$$\bbox[0.2ex,border:0.5pt solid black]{\int_0^1 \frac {1-\cos(y)}y\,dy
-\int_1^{+\infty} \frac {\cos(y)} y\,dy=\gamma}$$
Here is the proof of Gronwall, 1918. For $n$ a positive integer
$$\begin{align}
A(n) 
&= \int_0^{n\pi} \frac{1-\cos (y)}{y} dy - \ln ( n\pi ) \\
&= \int_0^1 \frac{1-\cos (y)}{y} dy + \int_1^{n\pi} \frac{1-\cos (y)}{y} dy - \ln ( n\pi ) \\
&= \int_0^1 \frac{1-\cos(y)}{t} dy - \int_1^{n\pi} \frac{\cos(y)}{y} dy
\end{align}$$
With a change of variable $x = 2 \pi ny$:
$$\begin{align}
\int_0^{n\pi} \frac{1-\cos(x)}{x} xy 
&= \int_0^{\frac{1}{2}} \frac{1 - \cos (2\pi ny)}{y} dy \\
&= \pi \int_0^{\frac{1}{2}} \frac{1 - \cos (2 \pi ny)}{\sin(\pi y)} dy + \int_0^{\frac{1}{2}} g(y) dy - \int_0^{\frac{1}{2}} g(y) \cos ( 2 \pi ny) dy
\end{align}$$
$g(y) = \frac{1}{y} - \frac{\pi}{\sin (\pi y)}$ which is continuous on $[0,\frac{1}{2}]$ with $g(0) = 0$.
With Riemann-Lebesgue theorem
$$\int_0^\frac{1}{2} g(y) \cos ( 2 \pi n y ) dy \rightarrow 0 \text{ for } n \rightarrow + \infty$$
$$\begin{align}
\int_0^\frac12 g(y) dy 
&= \lim_{s \to 0^+} \int_s^\frac12 \left( \frac 1y - \frac \pi{\sin(\pi y)} \right) dy \\
&= \lim_{s \to 0^+} \left[ \ln \left( \frac{y}{\tan ( \pi/2\;y )} \right) \right]_s^{\frac{1}{2}} = \ln ( \pi ) - 2 \ln (2)
\end{align}$$
On the other hand
$$\begin{align}
\pi \int_{0}^{\frac{1}{2}} \frac{1 - \cos ( 2 \pi n y)}{\sin ( \pi y )} dy
&= 2\pi \int_{0}^{\frac{1}{2}} \sum_{k=1}^{n} \sin ( (2k-1) \pi y ) dy \\
&= \sum_{k=1}^{n} \frac{2}{2k-1} 
={\ln (n) + 2 \ln (2) + \gamma + o(1)}
\end{align}$$
Finally
$$A(n) = \gamma + o(1)$$
Thanks very much to Sladjan Stankovik for providing insight into this problem for me.
