Cauchy's Residue Theorem for Integral $\int_{c\ -\ j\infty}^{c\ +\ j\infty} \left({\sigma \over x}\right)^s{{1-\beta^{s+1}}\over s(s+1)}\,ds$

Question

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

Answer 1

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\norm}[1]{\left\vert\left\vert\, #1\,\right\vert\right\vert} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{\int_{c\ -\ \infty\ic}^{c\ +\ \infty\ic}\pars{\sigma \over x}^{s} {1 - \beta^{s + 1} \over s\pars{s + 1}}\,\dd s:\ {\large ?}.\qquad \sigma > 0\,,\quad \beta > 0\,,\quad x \in {\mathbb R}\,,\quad \verts{x} < \beta\,;\quad c > 0}$.

---

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 $\ds{\mu}$.

1. $\ds{\large\mu\ <\ 0}$
   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.
2. $\ds{\large 0\ <\ \mu\ <\ 1}$
   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 $$\left.\fermi\pars{\mu}\right\vert_{0\ <\ \mu\ <\ 1} = 0$$.
3. $\ds{\large\mu\ =\ 1}$
   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}
4. $\ds{\large \mu > 1}$
   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}