Show that $\int_{0}^{+\infty} \frac{\sin(x)}{x(x^2+1)} dx = \frac{\pi}{2}\left(1-\frac{1}{e}\right) $ I'm trying to show that 
$$\int_{0}^{+\infty}  \frac{\sin(x)}{x(x^2+1)} dx = \frac{\pi}{2}\left(1-\frac{1}{e}\right) $$
using Jordan's lemma and contour integration. 
MY ATTEMPT: The function in the integrand is even, so I have:
$$\int_{0}^{+\infty}  \frac{\sin(x)}{x(x^2+1)} dx =\frac{1}{2}\int_{-\infty}^{+\infty}  \frac{\sin(x)}{x(x^2+1)} dx$$ 
There is a simple pole at $z=0$ and poles at $z=+i, z=-i$. 
A method in the chapter I am working on (Ablowitz & Fokas sections 4.2 & 4.3) usually considers the integral
$$\int_{-\infty}^{+\infty}  \frac{e^{ix}}{x(x^2+1)} dx=2\pi iRes\left(\frac{e^{ix}}{x(x^2+1)},z=i,-i,0\right)$$ Which when I compute results in $\dfrac{-(-1+e)^2\pi}{2e}$, which is close but not quite the answer. (Notice that factored in another way the answer is also equal to $\dfrac{(-1+e)\pi}{2e}$.
But I am not sure if this will work, instead another example builds a contour $C_r+C_e+(-R,-e)+(R,e)$ which avoids the poles and thus integrating over that yields zero and helps me get my answer. Unfortunately, this attempt does not give me the right value either. 
Do any of you integration whizzes out there have anything for me? Many thanks.

Worked this out, it's a combination of both methods I mention. You need to go around the contour $C_r+C_e+(-R,-e)+(R,e)$. By the only theorem on section 4.3 of Ablowitz & Fokas the integral around $C_e$ will go to $i\pi$. By the approach I mentioned in the first part of my attempt, the integral around $C_R$ will go to $2\pi i(-\frac{1}{2e})$ adding these two solutions up, and remembering to divide by $2$ (notice the integrand is even) yields the answer.
