Calculate $\int_0^\infty {\sin x ~dx\over x(x^2+1)}$

Question

Calculate $$I:=\int_0^\infty {\sin x ~dx\over x(x^2+1)}$$

I have show that $$J:={1\over 2}\int_{-\infty}^\infty {e^{ix}-1\over x(x^2+1)}dx={\pi(e-1)\over2e}$$ but I don't know how to show $I=J$.

I do notice that $I=\Im\int_0^\infty{e^{ix}-1\over x(x^2+1)}dx$ but the integrand $F$ is not an even function, so why does $I={1\over2}\Im \int_{-\infty}^\infty F(x)dx$?

Answer 1

$I=\frac 1 2\int_{-\infty}^{\infty} \frac {\sin\, x} {x(1+x^{2})}dx$ because the integrand is even.