Is Cauchy's formula apt for evaluating this integral

Question

I'm trying to evaluate the following. $$\frac{1}{2i}\int_{-\infty}^\infty \frac{s \sin{(sr)}}{(s-k)(s+k)}\mathrm{d}s,$$ with $k$ and $r$ being real constants.

The integral could be written as $$\int_{-\infty}^\infty \frac{s e^{isr}}{(s-k)(s+k)}\mathrm{d}s-\int_{-\infty}^\infty \frac{s e^{-isr}}{(s-k)(s+k)}\mathrm{d}s,$$ which makes it nicer, as it looks appropriate to use Cauchy's integral formula. But the problem I have is that the poles lie right on the real interval, so is it possible to exploit Cauchy's formula in such a case?

Answer 1

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{{1 \over 2\ic}\,\pp\int_{-\infty}^{\infty} {s\sin\pars{sr} \over \pars{s - k}\pars{s + k}}\,\dd s:\ {\large ?}}$.

\begin{align}&\color{#c00000}{{1 \over 2\ic}\,\pp\int_{-\infty}^{\infty} {s\sin\pars{sr} \over \pars{s - k}\pars{s + k}}\,\dd s} ={\sgn\pars{r} \over 2\ic}\,\pp\int_{-\infty}^{\infty} {s\sin\pars{\verts{r}s} \over \pars{s - k}\pars{s + k}}\,\dd s \\[3mm]&={\sgn\pars{r} \over 4\ic}\,\pp\bracks{% \int_{-\infty}^{\infty}{\sin\pars{\verts{r}s} \over s + k}\,\dd s +\int_{-\infty}^{\infty}{\sin\pars{\verts{r}s} \over s - k}\,\dd s} \\[3mm]&={\sgn\pars{r} \over 4\ic}\,\pp\bracks{% \int_{-\infty}^{\infty}{\sin\pars{\verts{r}s}\cos\pars{\verts{r}k} \over s}\,\dd s +\int_{-\infty}^{\infty}{\sin\pars{\verts{r}s}\cos\pars{\verts{r}k} \over s} \,\dd s} \\[3mm]&={\sgn\pars{r} \over 2\ic}\,\cos\pars{\verts{r}k}\int_{-\infty}^{\infty}{\sin\pars{s} \over s}\,\dd s ={\sgn\pars{r} \over 2\ic}\,\cos\pars{\verts{r}k}\,\pi \end{align}

$$ \color{#66f}{\large{1 \over 2\ic}\,\pp\int_{-\infty}^{\infty} {s\sin\pars{sr} \over \pars{s - k}\pars{s + k}}\,\dd s ={\pi \over 2\ic}\,\sgn\pars{r}\cos\pars{r\,k}} $$