Show that $\int_1^\infty \frac{\log(x-1)}{x^3}\, dx = -\frac 1 2$ by considering $f(z)=\frac{[\log(1-z)]^2}{z^3}$ 
Show that $\int_1^\infty \frac{\log(x-1)}{x^3}\, dx = -\frac 1 2$ by considering $f(z)=\frac{[\log(1-z)]^2}{z^3}$.

Attempt at a solution:
Perform the line integral of a small circle centered at $x=1$, lines along the $x$ axis from 1 to $\infty$ and $\infty$ to 1, as well as a circle of infinite radius joining the two lines along the $x$ axis together. The integrals along both circles are $0$.
There is a simple pole at $z=1$ with residue $1$, so by CRT the above integral is $2\pi$.
So I have $$2\pi = \int_1^\infty \frac 1 {x^3} [\log(1-x)]^2 \, dx + \int_\infty^1 \frac 1 {(xe^{2\pi i})^3} [\log(1-xe^{2\pi i})]^2 \, dx \, e^{2\pi i}$$
And I'm stuck here I can't see how to get to $\int_1^\infty  \frac{\log(x-1)}{x^3} \, dx = -\frac 1 2$. 
I can get to $\log(1-x)$ by bringing a factor of $i \pi$ outside of the $\log$ but can't see how to get rid of the square of the exponential.
 A: I recommend shifting your integral over a bit:
$$I=\int_0^{+\infty}\frac{\ln(z)}{(z+1)^3}~\mathrm dz$$
Now consider $f(z)=\frac{\ln^2(z)}{(z+1)^3}$.  Taking the same approach as you have,
$$\begin{align}\oint f(z)~\mathrm dz&=\int_0^{+\infty}f(z)~\mathrm dz+\int_{+\infty}^0f(ze^{2\pi i})~\mathrm dz\\&=\int_0^{+\infty}\frac{\ln^2(z)}{(z+1)^3}-\frac{\left[\ln(z)+2\pi i\right]^2}{(z+1)^3}~\mathrm dz\\&=\int_0^{+\infty}\frac{\require{cancel}\cancel{\ln^2(z)-\ln^2(z)}-4\pi i\ln(z)+4\pi^2}{(z+1)^3}~\mathrm dz\end{align}$$
Thus,
$$\int_0^{+\infty}\frac{\ln(z)}{(z+1)^3}~\mathrm dz=-\frac1{4\pi}\Im\left[\oint f(z)~\mathrm dz\right]$$

Also, check your residue calculations.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \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)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
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I'll comsider the following branch-cut, along $\ds{\left(-\infty,0\right]}$, for the $\ds{\ln}$-function:

$$
\ln\pars{w} = \ln\pars{\verts{w}} + \,\mrm{arg}\pars{w}\ic\,,\qquad -\pi < \mrm{arg}\pars{w} < \pi\,,\quad w \not= 0
$$
With this definition, it turns out that $\ds{\ln\pars{1 - z}}$ has a brunch-cut along $\ds{\left[0,\infty\right)}$ such that I'll perform an integration along a
key-hole' contour $\ds{\,\mc{C}}$ which 'takes care' of the $\ds{\ln\pars{1 - z}}$-branch cut.

\begin{equation}
\mbox{Note that }\ \underline{when\,\,\, x > 1}:\ 
\left\{\substack{%
\ds{\left.\vphantom{\large A}\mrm{arg}\pars{1 - z}
\right\vert_{\large\ \Im\pars{z}\ =\ 0^{+}}\ =\ -\pi}
\\[3mm]
\ds{\left.\vphantom{\large A}\mrm{arg}\pars{1 - z}
\right\vert_{\large\ \Im\pars{z}\ =\ 0^{-}}\ =\ \phantom{-A}\pi}}\right.
\label{1}\tag{1}
\end{equation}

The integrand has a single pole, inside the contour, at $\ds{z = 0}$. Namely,
\begin{align}
\oint_{C}{\ln^{2}\pars{1 - z} \over z^{3}}\,\dd z & =
2\pi\ic\,\lim_{z \to 0}\bracks{z\,{\ln^{2}\pars{1 - z} \over z^{3}}} = 2\pi\ic
\label{2}\tag{2}
\end{align}

\begin{align}
\mbox{Moreover, ( see \eqref{1} )}&
\\[2mm]
\oint_{C}{\ln^{2}\pars{1 - z} \over z^{3}}\,\dd z & =
\int_{1}^{\infty}{\bracks{\ln\pars{x - 1} - \pi\ic}^{\,2} \over x^{3}}\,\dd x +
\int_{\infty}^{1}{\bracks{\ln\pars{x - 1} + \pi\ic}^{\,2} \over x^{3}}\,\dd x
\\[5mm] & =
-4\pi\ic\int_{1}^{\infty}{\ln\pars{x - 1} \over x^{3}}\,\dd x\label{3}\tag{3}
\end{align}
By simplicity, I omitted the integration 'around' $\ds{z = 1}$ indented path which trivially vanishes out.


\eqref{2} and \eqref{3}
  $\ds{\implies \int_{1}^{\infty}{\ln\pars{x - 1} \over x^{3}}\,\dd x =
{2\pi\ic \over -4\pi\ic} = \bbx{-\,{1 \over 2}}}$.

A: Here is a real approach
Preliminary Results
$$
\int_1^\infty\frac{\log(x)}{(1+x)^3}\,\mathrm{d}x
=-\int_0^1\frac{x\log(x)}{(1+x)^3}\,\mathrm{d}x\tag{a}
$$
$$
\frac{1-x}{(1+x)^3}=\sum_{k=0}^\infty(-1)^k(k+1)^2x^k\tag{b}
$$
$$
\begin{align}
\int_0^ax^k\log(x)\,\mathrm{d}x
&=\frac1{k+1}\int_0^a\log(x)\,\mathrm{d}x^{k+1}\\
&=\frac1{k+1}\left[a^{k+1}\log(a)-\int_0^ax^k\,\mathrm{d}x\right]\\
&=\frac1{k+1}\left[a^{k+1}\log(a)-\frac1{k+1}a^{k+1}\right]\tag{c}
\end{align}
$$

Main Result
$$
\begin{align}
\int_1^\infty\frac{\log(x-1)}{x^3}\,\mathrm{d}x
&=\int_0^\infty\frac{\log(x)}{(1+x)^3}\,\mathrm{d}x\tag{1}\\
&=\int_0^1\frac{(1-x)\log(x)}{(1+x)^3}\,\mathrm{d}x\tag{2}\\
&=\lim_{a\to1^-}\int_0^a\frac{(1-x)\log(x)}{(1+x)^3}\,\mathrm{d}x\tag{3}\\
&=\lim_{a\to1^-}\sum_{k=0}^\infty(-1)^k(k+1)^2\int_0^ax^k\log(x)\,\mathrm{d}x\tag{4}\\
&=\lim_{a\to1^-}\sum_{k=0}^\infty(-1)^ka^{k+1}\left[(k+1)\log(a)-1\right]\tag{5}\\
&=\lim_{a\to1^-}\left[\frac{a\log(a)}{(1+a)^2}-\frac{a}{1+a}\right]\tag{6}\\[3pt]
&=-\frac12\tag{7}
\end{align}
$$
Explanation:
$(1)$: substitute $x\mapsto x+1$
$(2)$: apply $\text{(a)}$
$(3)$: introduce a limit for later
$(4)$: apply $\text{(b)}$
$(5)$: apply $\text{(c)}$
$(6)$: evaluate series
$(7)$: evaluate limits
