About the proof that $\int_0^\infty\frac{dx}{x^2+6x+8} =\frac12\log2$ via residue formula In the text "Functions of one Complex Variable" by Robert E.Greene and Steven G.Krantz is my understanding of the proof to $\text{Proposition (1.1)}$ correct ?

$\text{Proposition (1.1)}$
$$\int_{0}^{ \infty} \frac{dx}{x^{2} + 6x + 8} = \frac{1}{2} \log(2) \, \, $$

$\text{Proof}$
For the sake and using Complex-Analytic techniques the author considers the following integral.
$$\oint_{\eta_{R}} \frac{\log(z)}{z^{2} + 6z + 8}dz$$
As an exercise, it was left to us by the author that $\log(r)$ is a well defined holomorphic function. To address a trivial proof, one can define $\log(z)$ on $U \equiv \mathbb{C} 
\setminus 
\{x : x \geq 0 
\}$ by $\{ \log(re^{i \theta}) = (\log(r)) + i \theta$  when $0 < \theta < 2 \pi, r > 0 \}$. 
Before proceeding any further, take note that 
$$u(r, \theta)=\log(r) \ \ \ \text{ and } \ \ \ v(r, \theta) =\theta.$$
Now it's easy to note that
$$
 \big(  \partial_{r}u \big) =\frac{1}{r}= \frac{1}{r} \cdot 1 = \frac{1}{r} \cdot \left(  \partial_{\theta} v\right)\ \ \ \ \ \text{and } \ \ \ \  \big(  \partial_{r}u \big) = 0 = \frac{-1}{r}\cdot 0 = \frac{-1}{r} \cdot \big( \partial_{\theta} u \big)
$$
So indeed, $log(z)$ is analytic.
But before proceeding further he defines $\eta_{R}$ such that, 
$$\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \eta_{R}^{1}(t)  =  t + i/\sqrt{2R},  \, \, \, \,   1/\sqrt{2R} \leq t \leq R,$$
$$\eta_{R}^{2}(t)= Re^{it}, \, \, \, \,  \theta_{0} \leq t \leq 2 \pi - \theta_{0},$$
where $\theta_{0} = \theta_{0}(R) = \sin^{-1}(1/(R \sqrt{2R}))$
$$\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \eta_{R}^{3}(t)  =  R -t -i/\sqrt{2R},  \, \, \, \, 0 \leq t \leq R-1/\sqrt{2R},$$
$$\eta_{R}^{4}(t)  =  e^{it}/\sqrt{R}, \, \, \, \, \,  \, \, \, \, \, \, \, \, \, \, \, \, \, \pi/4 \leq t \leq 7 \pi /4.$$
$\text{Remark}$
For those who don't have the book on hand a picture of the Contour employed can be found in $\text{Figure (1.1)}$
$\text{Figure (1.1)}$
$\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, $
The author now says that:
$(*)$
$$ \bigg| \lim_{R \rightarrow \infty}\oint_{\eta_{R}^{4}} \frac{\log(z)}{z^{2} + 6z + 8}dz\bigg| \rightarrow 0$$
, and that 
$(**)$
$$ \bigg| \lim_{R \rightarrow \infty}\oint_{\eta_{R}^{2}} \frac{\log(z)}{z^{2} + 6z + 8}dz\bigg| \rightarrow 0.$$
A particular device that the author cites to justify convergence over $\eta_{R}^{2}$ and $\eta_{R}^{4}$ consider on faith
$$\bigg(\log \bigg( \frac{x + i \sqrt{2R}}{(x-i/\sqrt{2R}} \bigg) \bigg)\rightarrow -2 \pi i.$$
We will come back to this after dealing with the integrals over $\eta_{R}^{2}$ and $\eta_{R}^{4}$.
One should note that
$$ \sum_{\psi}^{4} \bigg(\oint_{\eta_{R}^{\psi}} \frac{\log(z)}{z^{2} + 6z + 8}dz \bigg).$$
Now over $\eta_{R}^{2}$ we have,  
\begin{align*}
\bigg| \oint_{\eta_{R}^{2}}\frac{\log(z)}{z^{2} + 6z + 8}dz\bigg|& = \bigg| \int_{-R}^{+Ri} \frac{\log(Re^{it})}{(Re^{it})^{2} + 6(Re^{it}) + 8} iRe^{i \theta} d \theta\bigg|\\&=  \int_{-R}^{+Ri} \bigg|\frac{\log(Re^{it})}{(Re^{it})^{2} + 6(Re^{it}) + 8} \bigg| \big| iRe^{i \theta} d \theta \big|\\&= \int_{-R}^{+Ri} \frac{\bigg|\log(Re^{it}) \bigg|}{\bigg|(Re^{it})^{2} + 6(Re^{it}) + 8 \bigg|}  \bigg|iRe^{i \theta} \bigg| \bigg|d \theta  \bigg| \\& = \int_{\theta_{0}}^{2 \pi - \theta_{0}} \frac{\bigg|\log(Re^{it}) \bigg|}{\bigg|(Re^{it})^{2} + 6(Re^{it}) + 8 \bigg|}  \bigg|iRe^{i \theta} \bigg| \bigg|d \theta  \bigg|
\end{align*}
Now we can establish a precise estimate over $\eta_{R}^{2}$, 
$$\bigg| \oint_{\eta_{R}^{2}} \frac{\log(z)}{z^{2} + 6z + 8}dz\bigg| \leq  \frac{\ln(R) + \pi }{R^{2} - 13} \pi r \, \, \text{as} \, \, \, R \rightarrow \infty $$
There by proving $(*)$.
A similar process can be done for $\eta_{R}^{4}$, hence:
\begin{align*}
\bigg| \oint_{\eta_{R}^{4}} \frac{\log(e^{it}/\sqrt{R})}{(e^{it}/ \sqrt{R})^{2} + (e^{it} / \sqrt{R})(6) +8}  dz\bigg|& =  \oint_{\eta_{R}^{4}} \bigg| \frac{\log(e^{it}/\sqrt{R})}{(e^{it}/ \sqrt{R})^{2} + (e^{it} / \sqrt{R})(6) +8}  iRe^{i \theta} d \theta\bigg|\\&= \oint_{\eta_{R}^{4}}  \frac{\bigg|\log(e^{it}/\sqrt{R}) \bigg|}{\bigg|(e^{it}/ \sqrt{R})^{2} + (e^{it} / \sqrt{R})(6) +8 \bigg|}  iRe^{i \theta} d \theta \\&= \oint_{\eta_{R}^{4}}  \frac{\bigg| \log(e^{it})- \frac{1}{2}\log(R^{}) \bigg|}{ \bigg|\frac{e^{2it}}{\sqrt{2R}} + (e^{it} / \sqrt{R})(6) +8 \bigg|} \bigg|  iRe^{i \theta} d \theta \bigg|\\& =\oint_{\frac{\pi}{4}}^{\frac{7 \pi}{4}}  \frac{\bigg| it\log(e^{})- \frac{1}{2}\log(R^{}) \bigg|}{ \bigg|\frac{e^{2it}}{\sqrt{2R}} + (e^{it} / \sqrt{R})(6) +8 \bigg|} \bigg|  iRe^{i \theta}\bigg| d \theta \bigg|.  \end{align*}
Now finally a precise estimate for $\eta_{R}^{4}$
$$\bigg| \oint_{\eta_{R}^{4}} \frac{\log(e^{it}/\sqrt{R})}{(e^{it}/ \sqrt{R})^{2} + (e^{it} / \sqrt{R})(6) +8}  dz\bigg|  \leq  \text{length}(\eta_{R}^{4})  \cdot \sup_{\eta_{R}^{4}}(g) \leq \pi R \frac{O(\log(R))}{\sqrt{R}} \, \text{as} \, R \rightarrow \infty  $$
Thus proving $(**)$
After achieving our preliminary results now we have that, 
$(***)$
\begin{align*}
 \bigg( \oint_{\eta_{R}^{1}} g(z) dz + \oint_{\eta_{R}^{3}} g(z) dz \bigg)& = \lim_{R \rightarrow \infty }  \bigg( \oint_{\mu_{R}^{1} } \frac{\log(x+  \sqrt{2R})}{(\log(x+  \sqrt{2R}))^{2} + 6(\log(x+  \sqrt{2R})) + 8} - \oint_{\mu_{R}^{3} } \frac{\log(x - i/ \sqrt{2R})}{(\log(x -i /\sqrt{2R}))^{2} + 6(\log(x - i  /\sqrt{2R})) + 8}   \bigg) \\&= -2 \pi i \lim_{R \rightarrow \infty}\int_{0}^{R} \frac{dt}{t^{2} + 6t + 8} \\&
 \end{align*}
Using the Residue Theorem it's easy to observe that:
$(****)$
$$ \oint_{\eta_{R}} g(z) dz  = 2 \pi i (\operatorname{Res_{g}}(-2) \cdot + Res_{g}(-4) \cdot 1) = - \pi i \log(2)$$
Finally putting $(****)$, $(***)$, $(**)$ and $(*)$ together yields that the, 
$$\lim_{R \rightarrow \infty}\int_{0}^{R} \frac{dt}{t^{2} + 6t + 8} = \frac{1}{2}\log(2).$$
 A: First of all, in your proposition, since $x$ is a dummy variable, it makes no sense to say '$\text{for all }x\in\mathbb R$'.
Besides, it is not quite clear how you obtained $(***)$.
Here, I provide a lemma, which can be applied to derive $(***)$, as well as explaining the motivation to introduce $\log(z)$ at the first place.

Lemma
$$\lim_{\Delta\to0^+}\left(\int_{\gamma_1}+\int_{\gamma_2}\right)f(z)\ln(z-s)dz=-2\pi i\int_{pe^{i\theta}}^{qe^{i\theta}}f(t)dt$$
Proof:
Let $\hat{k}=i\frac{s}{|s|}$
\begin{align*}
&~~~~\lim_{\Delta\to0^+}\left(\int_{\gamma_1}+\int_{\gamma_2}\right)f(z)\ln(z-s)dz \\
&=\lim_{\Delta\to0^+}\left(\int_{\gamma_1}+\int_{\gamma_2}\right)f(z)\ln|z-s|dz
+i\lim_{\Delta\to0^+}\left(\int_{\gamma_1}+\int_{\gamma_2}\right)f(z)\arg(z-s)dz \\
&=\left(\int_{pe^{i\theta}}^{qe^{i\theta}}+\int^{pe^{i\theta}}_{qe^{i\theta}}\right)f(z)\ln|z-s|dz \\
&~~~~+i\lim_{\Delta\to0^+}\int^{qe^{i\theta}+\Delta\hat{k}}_{pe^{i\theta}+\Delta\hat{k}} f(z)\arg(z-s)dz 
+i\lim_{\Delta\to0^+}\int^{qe^{i\theta}-\Delta\hat{k}}_{pe^{i\theta}-\Delta\hat{k}} f(z)\arg(z-s)dz\\
\end{align*}
Obviously the first term is zero. 
For the second term, by the substitution $z=ue^{i\theta}+\Delta\hat{k}$
\begin{align*}
&~~~~ i\lim_{\Delta\to0^+}\int^{qe^{i\theta}+\Delta\hat{k}}_{pe^{i\theta}+\Delta\hat{k}} f(z)\arg(z-s)dz \\
&=i\lim_{\Delta\to0^+}\int^q_p f(ue^{i\theta}+\Delta\hat{k})\arg(ue^{i\theta}+\Delta\hat{k}-s)e^{i\theta}du \\
&=i\int^q_p f(ue^{i\theta})\theta e^{i\theta}du \\
&=i\theta\int_{pe^{i\theta}}^{qe^{i\theta}}f(t)dt
\end{align*}
From the second line to the third line, dominated convergence theorem is applied to exchange limit and integral, and $\lim_{\Delta\to 0^+}\arg(ue^{i\theta}+\Delta\hat{k}-s)=\theta$ is used.
For the third term, by the substitution $z=ue^{i\theta}-\Delta\hat{k}$
\begin{align*}
&~~~~ i\lim_{\Delta\to0^+}\int_{qe^{i\theta}-\Delta\hat{k}}^{pe^{i\theta}-\Delta\hat{k}} f(z)\arg(z-s)dz \\
&=i\lim_{\Delta\to0^+}\int_q^p f(ue^{i\theta}-\Delta\hat{k})\arg(ue^{i\theta}-\Delta\hat{k}-s)e^{i\theta}du \\
&=-i\int^q_p f(ue^{i\theta})(2\pi+\theta) e^{i\theta}du \\
&=-i(2\pi+\theta)\int_{pe^{i\theta}}^{qe^{i\theta}}f(t)dt
\end{align*}
Similarly, $\lim_{\Delta\to 0^+}\arg(ue^{i\theta}-\Delta\hat{k}-s)=2\pi+\theta$ is used.
As a result,
\begin{align*}
&~~~~\lim_{\Delta\to0^+}\left(\int_{\gamma_1}+\int_{\gamma_2}\right)f(z)\ln(z-s)dz \\
&=0+i\theta\int_{pe^{i\theta}}^{qe^{i\theta}}f(t)dt-i(2\pi+\theta)\int_{pe^{i\theta}}^{qe^{i\theta}}f(t)dt\\
&=-2\pi i\int_{pe^{i\theta}}^{qe^{i\theta}}f(t)dt
\end{align*}
Q.E.D.
A: This post is to address, a complete derivation of $(***)$ which where my understanding of the proof breaks down.
$\text{Proof}$
Now recall the Lemma given by Szeto,

$\text{Szeto's Lemma}$
$(1)$
  $$ \lim_{\Delta\to0^+}\left(\int_{\gamma_1}+\int_{\gamma_2}\right)f(z)\ln(z-s)dz=-2\pi i\int_{pe^{i\theta}}^{qe^{i\theta}}f(t)dt.$$

Now what we are aiming to derive, is that
\begin{align*}
 \bigg( \oint_{\eta_{R}^{1}} g(z) dz + \oint_{\eta_{R}^{3}} g(z) dz \bigg)& \rightarrow - 2 \pi i \int_{0}^{\infty} \frac{dt}{t^{2} + 6t + 8}. \tag{1.1}\\&
 \end{align*}
Applying $(1)$ to $\text{(1.1)}$ we note that
$$ \bigg( \lim_{R \rightarrow \infty}\oint_{\mu_{R}^{1} } \frac{\log(x+  \sqrt{2R})}{(\log(x+  \sqrt{2R}))^{2} + 6(\log(x+  \sqrt{2R})) + 8}\ln(z-s) \, dz  \bigg) + \bigg( \lim_{R \rightarrow \infty}\oint_{\mu_{R}^{3} } \frac{\log(x - i/ \sqrt{2R})}{(\log(x -i /\sqrt{2R}))^{2} + 6(\log(x - i  /\sqrt{2R})) + 8} \ln(z-s) dz  \bigg).$$
Further analysis of $\eta_{R}^{1}$ revels that, 
\begin{align*}
 \lim_{R \rightarrow \infty} \bigg( \int_{1 / \sqrt{2R}}^{R} \frac{\log(x + \sqrt{2R})}{(\log(x + \sqrt{2R}^{2} + 6(\log(x + \sqrt{2R}) + 8}\ln(z-s)dz \bigg)&=&
\\& \lim_{ R \rightarrow \infty} \bigg( \int_{0}^{R} \frac{\log(x + \sqrt{2R})}{(\log(x + \sqrt{2R}^{2} + 6(\log(x + \sqrt{2R}) + 8} \ln(z-s)dz\bigg) + \lim_{ R \rightarrow \infty} \bigg(  \int_{0}^{\frac{1}{\sqrt{2R}}}\frac{\log(x+  \sqrt{2R})}{(\log(x - i/\sqrt{2R}))^{2} + 6(\log(x - i/  \sqrt{2R})) + 8}\ln(z-s)dz\bigg).
\end{align*}
Similarly for $\eta_{R}^{3}$ we have that,
\begin{align*}
  \bigg(\lim_{R \rightarrow \infty} \int_{R - 1 / \sqrt{2R}}^{0} \frac{\log(x + \sqrt{2R})}{(\log(x + \sqrt{2R}^{2} + 6(\log(x + \sqrt{2R}) + 8}\ln(z-s)dz \bigg)&&
\\& 
\end{align*}
Putting everything together it's easy to note that, 
$$ \bigg( \oint_{\eta_{R}^{1}} g(z) dz + \oint_{\eta_{R}^{3}} g(z) dz \bigg) =2 \pi i \int_{0}^{\infty} \frac{dt}{t^{2} + 6t + 8}. $$
