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 $x0$

Question

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
$$begin{align}
0&=oint_C frac{e^{iz}}{z},dz\
&=int_epsilon^R frac{e^{ix}}{x},dx +int_0^{pi/2}frac{e^{iRe^{iphi}}}{Re^{iphi}},iRe^{iphi},dphi+int_R^epsilon frac{e^{-x}}{ix},i,dx+int_{pi/2}^0 frac{e^{iepsilon e^{iphi}}}{epsilon e^{iphi}},iepsilon e^{iphi},dphi\
&=int_epsilon^R frac{e^{ix}}{x},dx -int_epsilon^R frac{e^{-x}}{x},dx-ifracpi2+O(epsilon)+Oleft(frac1Rright)tag1
end{align}$$
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
$$begin{align}-int_x^R frac{cos(x')}{x'},dx'&=log(x)-int_epsilon^R e^{-x}log(x),dx+int_epsilon^xfrac{cos(x')-1}{x'},dx'\
&-log(epsilon) -e^{-R}log(R)+e^{-epsilon}log(epsilon)+Oleft(frac1Rright)+O(epsilon)tag2
end{align}$$
Letting $Rtoinfty$ and $epsilonto 0$ 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
$$int_0^infty frac{cos(x)-e^{-x}}{x},dx=int_0^infty left(frac{x}{x^2+1}-frac1{x+1}right)=0tag3$$
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?

Danny Pak-Keung Chan · Answer

Note that for each $x>0$, the improper integral $int_{x}^{infty}frac{cos(t)}{t}dt$
converges conditionally (in the sense that $lim_{Arightarrowinfty}int_{x}^{A}frac{cos(t)}{t}dt$
exists but $lim_{Arightarrowinfty}int_{x}^{A}left|frac{cos(t)}{t}right|dt=infty$).
Define $F:(0,infty)rightarrowmathbb{R}$ 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_{hrightarrow0}frac{int_{x+h}^{infty}frac{cos(t)}{t}dt-int_{x}^{infty}frac{cos(t)}{t}dt}{h}\
 & = & -lim_{hrightarrow0}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 $tneq0$, 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 $phi:mathbb{R}rightarrowmathbb{R}$ 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 $G:(0,infty)rightarrowmathbb{R}$ by 
$$
G(x)=int_{x}^{infty}frac{cos(t)}{t}dt-left[ln(x)+int_{0}^{x}frac{cos(t)-1}{t}dtright].
$$
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.

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 $x>0$

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