Real Analysis Methodologies to Prove $-\int_x^\infty \frac{\cos(t)}{t}\,dt=\gamma+\log(x)+\int_0^x \frac{\cos(t)-1}{t}\,dt$ for $x>0$ 
MOTIVATION:

In this question, the OP asked to prove the Cosine Integral identity
$$-\int_x^\infty \frac{\cos(t)}{t}\,dt=\gamma+\log(x)+\int_0^x \frac{\cos(t)-1}{t}\,dt$$
where $\gamma $ is the Euler-Mascheroni constant.  However, the two posted answers are incomplete and seem unsatisfactory.


Proof Using Complex Anaysis

One can show for $x>0$ that $-\int_x^\infty \frac{\cos(t)}{t}\,dt=\gamma+\log(x)+\int_0^x \frac{\cos(t)-1}{t}\,dt$ using contour integration.  To wit, Cauchy's Integral Theorem guarantees that
$$\begin{align}
0&=\oint_C \frac{e^{iz}}{z}\,dz\\\\
&=\int_\epsilon^R \frac{e^{ix}}{x}\,dx +\int_0^{\pi/2}\frac{e^{iRe^{i\phi}}}{Re^{i\phi}}\,iRe^{i\phi}\,d\phi+\int_R^\epsilon \frac{e^{-x}}{ix}\,i\,dx+\int_{\pi/2}^0 \frac{e^{i\epsilon e^{i\phi}}}{\epsilon e^{i\phi}}\,i\epsilon e^{i\phi}\,d\phi\\\\
&=\int_\epsilon^R \frac{e^{ix}}{x}\,dx -\int_\epsilon^R \frac{e^{-x}}{x}\,dx-i\frac\pi2+O(\epsilon)+O\left(\frac1R\right)\tag1
\end{align}$$
whence after taking the real part of both sides of $(1$ and integrating by parts the integral $\int_\epsilon^R \frac{e^{-x}}{x}\,dx$ with $u=e^{-x}$ and $v=\log(x)$, we find
$$\begin{align}-\int_x^R \frac{\cos(x')}{x'}\,dx'&=\log(x)-\int_\epsilon^R e^{-x}\log(x)\,dx+\int_\epsilon^x\frac{\cos(x')-1}{x'}\,dx'\\\\
&-\log(\epsilon) -e^{-R}\log(R)+e^{-\epsilon}\log(\epsilon)+O\left(\frac1R\right)+O(\epsilon)\tag2
\end{align}$$
Letting $R\to\infty$ and $\epsilon\to 0$ in $(2)$ yields the sought relationship.


Proof Using Real Analysis

Alternatively, we can use either the Laplace Transform or "Feynman's Trick" to show that
$$\int_0^\infty \frac{\cos(x)-e^{-x}}{x}\,dx=\int_0^\infty \left(\frac{x}{x^2+1}-\frac1{x+1}\right)=0\tag3$$
Starting with $(3)$, is tantamount to starting with the real part of $(1)$ and we are done.


QUESTION:  So, what are other ways to prove the coveted relationship using real analysis tools only?

 A: Note that for each $x>0$, the improper integral $\int_{x}^{\infty}\frac{\cos(t)}{t}dt$
converges conditionally (in the sense that $\lim_{A\rightarrow\infty}\int_{x}^{A}\frac{\cos(t)}{t}dt$
exists but $\lim_{A\rightarrow\infty}\int_{x}^{A}\left|\frac{\cos(t)}{t}\right|dt=\infty$).
Define $F:(0,\infty)\rightarrow\mathbb{R}$ by $F(x)=\int_{x}^{\infty}\frac{\cos(t)}{t}dt$.
Note that we still have $F'(x)=-\frac{\cos(x)}{x}$. For, 
\begin{eqnarray*}
F'(x) & = & \lim_{h\rightarrow0}\frac{\int_{x+h}^{\infty}\frac{\cos(t)}{t}dt-\int_{x}^{\infty}\frac{\cos(t)}{t}dt}{h}\\
 & = & -\lim_{h\rightarrow0}\frac{\int_{x}^{x+h}\frac{\cos(t)}{t}dt}{h}\\
 & = & -\frac{\cos(x)}{x}.
\end{eqnarray*}
On the other hand, $0$ in the integral $\int_{0}^{x}\frac{\cos(t)-1}{t}dt$
is a removable singularity. For $t\neq0$, we have 
\begin{eqnarray*}
\frac{\cos(t)-1}{t} & = & \frac{(1-\frac{1}{2!}t^{2}+\frac{1}{4!}t^{4}-\cdots)-1}{t}\\
 & = & -\frac{1}{2!}t+\frac{1}{4!}t^{3}-\frac{1}{6!}t^{5}+\cdots.
\end{eqnarray*}
Define $\phi:\mathbb{R}\rightarrow\mathbb{R}$ by $\phi(t)=-\frac{1}{2!}t+\frac{1}{4!}t^{3}-\frac{1}{6!}t^{5}+\cdots$.
It can be proved that the power series converges everywhere (using
root-test), and hence $\phi$ is an analytic function. Therefore $\int_{0}^{x}\frac{\cos(t)-1}{t}dt=\int_{0}^{x}\phi(t)dt$.
Now $\frac{d}{dx}\int_{0}^{x}\frac{\cos(t)-1}{t}dt=\frac{\cos(x)-1}{x}$.
Finally, define $G:(0,\infty)\rightarrow\mathbb{R}$ by 
$$
G(x)=\int_{x}^{\infty}\frac{\cos(t)}{t}dt-\left[\ln(x)+\int_{0}^{x}\frac{\cos(t)-1}{t}dt\right].
$$
Then $G$ is differentiable and $G'(x)=-\frac{\cos(x)}{x}-\frac{1}{x}-\frac{\cos(x)-1}{x}=0$.
This shows that $G$ is a constant function.
