Cauchy's Residue Theorem for Integral $\int_{c\ -\ j\infty}^{c\ +\ j\infty} \left({\sigma \over x}\right)^s{{1-\beta^{s+1}}\over s(s+1)}\,ds$ This is a similar problem to the one I posted here. I am at this point of integration where:
$$\int_{c\ -\ j\infty}^{c\ +\ j\infty} \left({\sigma \over x}\right)^s{{1-\beta^{s+1}}\over s(s+1)}\,ds$$
whereby $\beta > 0$, $\sigma > 0$, $c>0$ and $|x|<\beta$. $\beta$, $\sigma$, $c$ and $x$ are all real numbers 
 A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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$\ds{\int_{c\ -\ \infty\ic}^{c\ +\ \infty\ic}\pars{\sigma \over x}^{s}
    {1 - \beta^{s + 1} \over s\pars{s + 1}}\,\dd s:\ {\large ?}.\qquad
    \sigma > 0\,,\quad  \beta > 0\,,\quad x \in {\mathbb R}\,,\quad
    \verts{x} < \beta\,;\quad c > 0}$.


First, we split the integral in two similar integrals:

\begin{align}&\color{#66f}{\large%
\int_{c\ -\ \infty\ic}^{c\ +\ \infty\ic}\pars{\sigma \over x}^{s}
{1 - \beta^{s + 1} \over s\pars{s + 1}}\,\dd s}
\\[5mm]&=\int_{c\ -\ \infty\ic}^{c\ +\ \infty\ic}\pars{\sigma \over x}^{s}
{1 - \beta^{s + 1} \over s}\,\dd s
-\int_{c\ -\ \infty\ic}^{c\ +\ \infty\ic}\pars{\sigma \over x}^{s}
{1 - \beta^{s + 1} \over s + 1}\,\dd s
\\[5mm]&=\int_{c\ -\ \infty\ic}^{c\ +\ \infty\ic}
\pars{\sigma \over x}^{s}{1 - \beta^{s + 1} \over s}\,\dd s
-{x \over \sigma}\int_{c\ +\ 1\ -\ \infty\ic}^{c\ +\ 1\ +\ \infty\ic}
\pars{\sigma \over x}^{s}{1 - \beta^{s} \over s}\,\dd s\tag{1}
\end{align}

Because the singularity is at $\ds{s = 0}$, the second second integral can be evaluated from $\ds{c - \infty\ic}$ to $\ds{c + \infty\ic}$ such that
  $\pars{1}$ is reduced to:

\begin{align}&\color{#66f}{\large%
\int_{c\ -\ \infty\ic}^{c\ +\ \infty\ic}\pars{\sigma \over x}^{s}
{1 - \beta^{s + 1} \over s\pars{s + 1}}\,\dd s}
\\[5mm]&=\int_{c\ -\ \infty\ic}^{c\ +\ \infty\ic}
\pars{\sigma \over x}^{s}{1 - \beta^{s + 1} \over s}\,\dd s
-{x \over \sigma}\int_{c\ -\ \infty\ic}^{c\ +\ \infty\ic}
\pars{\sigma \over x}^{s}{1 - \beta^{s} \over s}\,\dd s
\\[5mm]&=\pars{1 - {x \over \sigma}}\int_{c\ -\ \infty\ic}^{c\ +\ \infty\ic}
\pars{\sigma \over x}^{s}{1 \over s}\,\dd s
-\pars{\beta - {x \over \sigma}}
\int_{c\ -\ \infty\ic}^{c\ +\ \infty\ic}
\pars{\sigma\beta \over x}^{s}{1 \over s}\,\dd s
\\[5mm]&=\pars{1 - {x \over \sigma}}\fermi\pars{\sigma \over x}
-\pars{\beta - {x \over \sigma}}\fermi\pars{\sigma\beta \over x}\tag{2}
\end{align}

where
  $\ds{\quad\fermi\pars{\mu}\equiv
     \int_{c\ -\ \infty\ic}^{c\ +\ \infty\ic}{\mu^{s} \over s}\,\dd s\quad}$ with
  $\ds{\quad\mu \not= 0.\quad}$ There a few cases we have to consider given the sign and magnitude of $\ds{\mu}$.



*

*
$\ds{\large\mu\ <\ 0}$

We don't really know how the OP handles this case: Some definition should be established for $\ds{\mu^{s}}$. However, lets
$\ds{\mu = \verts{\mu}\exp\pars{\bracks{2n + 1}\ic\pi}}$ with
$\ds{n \in {\mathbb Z}}$. $\ds{\mu^{s}}$ becomes
$\ds{\pars{~\mbox{with}~s = c + \ic y\,,\ y\ \in {\mathbb R}}}$
$$
\mu^{s}=\verts{\mu}^{c + \ic y}\exp\pars{\bracks{2n + 1}\ic\pi c}
\exp\pars{-\bracks{2n + 1}\pi y}
$$
such that
$$
\left.\fermi\pars{\mu}\right\vert_{\mu\ <\ 0}
=\ic\exp\pars{\bracks{2n + 1}\ic\pi c}\verts{\mu}^{c}
\int_{-\infty}^{\infty}
{\verts{\mu}^{\ic y} \exp\pars{-\bracks{2n + 1}\pi y} \over c + \ic y}\,\dd y
$$
We can see that, for any value of $\ds{n \in {\mathbb Z}}$, the integral diverges.


*
$\ds{\large 0\ <\ \mu\ <\ 1}$

In this case we can write $\ds{\mu^{s} = \expo{-ts}}$ with $\ds{t > 0}$. By 'closing' the contour to the 'right' of the complex plane we conclude that
$$\left.\fermi\pars{\mu}\right\vert_{0\ <\ \mu\ <\ 1} = 0$$. 


*
$\ds{\large\mu\ =\ 1}$

The integral diverges because
$$
\fermi\pars{1}=\int_{c - \infty\ic}^{c + \infty\ic}{\dd s \over s}
=\int_{-\infty}^{\infty}{\ic\,\dd y \over c + \ic y}
$$
Note that
\begin{align}
\left.\int_{-a}^{b}{\ic\,\dd y \over c + \ic y}\right\vert_{a\ >\ 0\,,\ b\ >\ 0}&=
\int_{-a}^{b}{y + \ic c \over y^{2} + c^{2}}\,\dd y
\\[5mm]&=\half\,\ln\pars{b^{2} + c^{2} \over a^{2} + c^{2}}
+\bracks{\arctan\pars{b \over c} + \arctan\pars{a \over c}}\ic
\end{align}


*
$\ds{\large \mu > 1}$

In this case we can write $\ds{\mu^{s} = \expo{ts}}$ with $\ds{t > 0}$. By 'closing' the contour to the 'left' of the complex plane we conclude that
$$\left.\fermi\pars{\mu}\right\vert_{\mu\ >\ 1} = 2\pi\ic$$. 




In conclussion,

$$
\mbox{with}\quad \mu \in \pars{0,\infty} \verb*\* \braces{1}\,,\quad
\fermi\pars{\mu}=\left\{\begin{array}{lcrcccl}
0 & \mbox{if} & 0 & < & \mu & < & 1
\\[2mm]
2\pi\ic & \mbox{if} &&& \mu & > & 1
\end{array}\right.
$$

Finally $\ds{\pars{~\mbox{see expression}\ \pars{2}~}}$,

\begin{align}&\color{#66f}{\large%
\int_{c\ -\ \infty\ic}^{c\ +\ \infty\ic}\pars{\sigma \over x}^{s}
{1 - \beta^{s + 1} \over s\pars{s + 1}}\,\dd s}
\\[5mm]&=\left\{\begin{array}{lcrcl}
0 & \mbox{if}\quad & 0 < {\sigma \over x} < 1\ & \mbox{and} &
0 < {\sigma\beta \over x} < 1 
\\[3mm]
2\pi\ic\pars{{x \over \sigma} - \beta}
& \mbox{if}\quad & 0 < {\sigma \over x} < 1 & \mbox{and} &
{\sigma\beta \over x} > 1 
\\[3mm]
2\pi\ic\pars{1 - {x \over \sigma}}
& \mbox{if}\quad & {\sigma \over x} > 1 & \mbox{and} &
0 < {\sigma\beta \over x} < 1 
\\[3mm]
2\pi\ic\pars{1 - \beta}
& \mbox{if}\quad & {\sigma \over x} > 1 & \mbox{and} &
{\sigma\beta \over x} > 1 
\end{array}\right.
\end{align}
