Is Cauchy's formula apt for evaluating this integral 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?
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$\ds{{1 \over 2\ic}\,\pp\int_{-\infty}^{\infty}
     {s\sin\pars{sr} \over \pars{s - k}\pars{s + k}}\,\dd s:\ {\large ?}}$.

\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}

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