Cauchy Principal Value of $\int_{-\infty}^{\infty} \dfrac{\sin x}{x(x^2+1)}\mathrm dx$ How to evaluate this integral
$$
\int_{-\infty}^\infty\frac{\sin x}{x(x^2 + 1)}dx
$$
I am having a problem to solve this because of two poles when I solve it by integration first from $-R$ to $R$ and then along a semicircle in the upper half plane.
 A: Let us avoid contour-hunting. The integrand function is even, hence it is enough to compute
$$ F(a)=\int_{0}^{+\infty}\frac{\sin(ax)}{x(x^2+1)}\,dx $$
(defined over $\text{Re}(a)>0$) at $a=1$. We may notice that
$$ (\mathcal{L} F)(s) = \int_{0}^{+\infty}\frac{dx}{(s^2+x^2)(1+x^2)}=\frac{\pi}{2s(s+1)} $$
hence
$$F(a) = \frac{\pi}{2}\mathcal{L}^{-1}\left(\frac{1}{s}-\frac{1}{s+1}\right)(a)=\frac{\pi}{2}(1-e^{-a}) $$
and
$$ \int_{-\infty}^{+\infty}\frac{\sin x}{x(x^2+1)}\,dx = \color{blue}{\pi(1-e^{-1})}.$$
There is no need to introduce a principal value: the $\text{sinc}$ function is continuous and bounded over $\mathbb{R}$.
A: Here is another approach that like @Jack D'Aurizio approach avoids contour integration altogether.
Let 
$$F(a) = \int_{-\infty}^\infty \frac{\sin (ax)}{x(x^2 + 1)} \, dx = 2 \int_0^\infty \frac{\sin (ax)}{x(x^2 + 1)} \, dx, \qquad a > 0.$$
We are required to find $F(1)$.
Using Feynman's trick of differentiating under the integral sign we have
$$F'(a) = 2 \int_0^\infty \frac{\cos (ax)}{x^2 + 1} \, dx,$$
and
$$F''(a) = - 2 \int_0^\infty \frac{x \sin (ax)}{x^2 + 1} \, dx.$$
Observe that 
$$F''(a) - F(a) = -2 \int_0^\infty \frac{\sin (ax)}{x} \, dx = -\pi.$$
Here the well-known result of
$$\int_0^\infty \frac{\sin (ax)}{x} \, dx = \frac{\pi}{2}, \qquad a > 0,$$
has been used. 
One solving the differential equaltion $F''(a) - F(a) = -\pi$ we have
$$F(a) = C_1 e^a + C_2 e^{-a} + \pi.$$
The two unknown constants $C_1$ and $C_2$ can be found by noting that $F(0) = 0$ and $F'(0) = \pi$. Doing so one finds
$$F(a) = \pi(1 - e^{-a}).$$
Finally, setting $a = 1$ one has
$$\int_{-\infty}^\infty \frac{\sin x}{x(x^2 + 1)} \, dx = \pi(1 - e^{-1}),$$
as required.
A: According to your choice of the contour: $C$ is the upper half-plane and $\Gamma$ is the semicircular arc of radius $R$ (say).
$$\int_C \frac{\sin z}{z(z^2+1)}\mathrm dz=\text{P.V.} \int_{-\infty}^{+\infty} \frac{\sin x}{x(x^2+1)}\mathrm dx+\lim_{R \to \infty}\int_{\Gamma}\frac{\sin z}{z(z^2+1)}\mathrm dz$$
Now, use the fact that if $f(z)=\frac{g(z)}{h(z)}$ where $f$ and $g$ are analytic near $z_0$ and $h$ has a simple zero at $z_0$, then $\text{Res}(f(z), z_0) = \frac{g(z_0)}{h'(z_0)}$. Note that $g$ is $\sin z$ and $h$ is $z(z^2+1)$. For the evaluation of the integral over the arc $\Gamma$, use the ML-inequality and you should be good to go.
