How to show $\int^{\infty}_{-\infty}\frac{\sin(ax)}{x(x^2+1)}dx=\pi(1-e^{-a})$? ($a\ge0$)
The integral depends on the value of $a$, which I assume is a real number.
Assuming that $a\ge 0$, You can use this kind of contour to integrate $f(z) = \dfrac{e^{iaz}}{z(z^2+1)}$. The point is that you have to go around the singularity at $z=0$. (An alternative is to follow Did's suggestion).
Using this contour, you obtain:
$$2\pi i Res(f,i) = \left(\int_R - \int_\epsilon\right) + \left(\int_r^R f(x) dx + \int_{-R}^{-r} f(x) dx\right),$$
$\int_R$ and $\int_\epsilon$ both represent integral of $f$ over the respective semicircles. A quick computation shows that $R\to \infty$ makes $\int_R\to 0$ and $\epsilon \to 0$ makes $\int_\epsilon\to i\pi$.
Therefore,
$$2\pi i Res(f,i) = - i\pi + \int_{-\infty}^\infty f(x) dx$$
Computing the residue and solving for the integral gives:
$$ \int \dfrac{\sin(ax)}{x(x^2+1)} dx = \pi (1-e^{-a})$$
For $a<0$, you can use a similar contour, but it has to be upside down from the one used above in order for the big $R$ contour to vanish as $R\to \infty$. The answer comes out to be $\pi (e^a -1)$.
I will assume $a > 0$. Call the integral $I(a)$. Differentiate w.r.t. $a$, $$ I'(a) = \int_{-\infty}^{\infty} \frac{\cos(a x)}{1+x^2} \ \mathrm{d}x = 2\int_{0}^{\infty}\frac{\cos(a x)}{1+x^2} \ \mathrm{d}x,$$ then note that $$\mathcal{L}(I'(a)) = 2\int_{0}^{\infty}\int_{0}^{\infty} \frac{\cos(a x)}{1+x^2}\exp(-as) \ \mathrm{d}a \ \mathrm{d}x,$$ $$\mathcal{L}(I'(a)) = 2\int_{0}^{\infty} \frac{s}{(x^2+s^2)(1+x^2)} \ \mathrm{d}x.$$ Using partial fractions and integrating, $$\mathcal{L}(I'(a)) = \frac{\pi}{s+1},$$ taking the inverse transform gives $$ I'(a) = \pi \mathrm{e}^{-a}, $$ integrate to obtain $$ I(a) = -\pi \mathrm{e}^{-a} + c, $$ and using $I(0) = 0$, we see that $c=\pi$ and $$ I(a) = \pi(1-\mathrm{e}^{-a}). $$
$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\int_{-\infty}^{\infty}{\sin\pars{ax} \over x\pars{x^{2} + 1}}\,\dd x =\pi\sgn\pars{a}\pars{1 - \expo{-\verts{a}}}:\ {\large ?}.\qquad a \in {\mathbb R}}$
\begin{align} &\bbox[#ffd,5mm]{\int_{-\infty}^{\infty}{\sin\pars{ax} \over x\pars{x^{2} + 1}}\,\dd x} = a\int_{-\infty}^{\infty}{1 \over x^{2} + 1}\, {\sin\pars{ax} \over ax}\,\dd x \\[5mm] = &\ a\int_{-\infty}^{\infty}{1 \over x^{2} + 1}\, \pars{\half\int_{-1}^{1}\expo{\ic kax}\,\dd k}\,\dd x \\[5mm] = &\ \half\,a\int_{-1}^{1}\pars{% \int_{-\infty}^{\infty}{\expo{\ic\verts{ka}x} \over x^{2} + 1}\,\dd x}\,\dd k \\[5mm] = &\ \half\,a\int_{-1}^{1}\pars{% 2\pi\ic\,{\expo{\ic\verts{ka}\ic} \over 2\ic}}\,\dd k \\[5mm] = &\ \half\,\pi a\int_{-1}^{1}\expo{-\verts{ka}}\,\dd k = \pi a\int_{0}^{1}\expo{-\verts{a}k}\,\dd k = \pi a\,{\expo{-\verts{a}} - 1 \over -\verts{a}} \\[5mm] = &\ \bbox[15px,border:1px groove navy]{\pi\sgn\pars{a} \pars{1 - \expo{-\verts{a}}}} \\ & \end{align}