Show that $\int_{0}^{+\infty} \frac{\sin(x)}{x(x^2+1)} dx = \frac{\pi}{2}\left(1-\frac{1}{e}\right) $

Since it is even function, we evaluate $$ \int_{-\infty }^\infty \frac{\sin x}{x(x^2+1)}dx $$ We consider the analytic function $$ F(z)=\frac{e^{iz}}{z(z^2+1)} $$ on the contour of $C=[-R,-r]\cup[r, R]\cup C_R\cup \gamma_r$, where $C_R$ is the upper half circle with radius of $R$ and $\gamma_r$ is the small upper half circle with radius of $r<1$ bypassing $0$. By Cauchy's residue theorem, we have $$ \int_{-R}^{-r}F(x)dx+\int_{r}^RF(x)dx+\int_{C_R}F(z)dz+\int_{\gamma_r}F(z)dz=2\pi iRes(F,i) $$ Note that only pole in the $C_R$ is $z=i$. Since $$ Res(F,i)=\lim_{z\to i}(z-i)F(z)=\lim_{z\to i}\frac{e^{iz}}{z(z+i)}=-\frac{e^{-1}}{2} $$ We have $$ \int_{-R}^{-r}F(x)dx+\int_{r}^RF(x)dx+\int_{C_R}F(z)dz+\int_{\gamma_r}F(z)dz=-\frac{\pi i}{e}\tag1 $$ By Jordan lemma $$ \int_{C_R}|e^{iz}||dz|=2R\int_0^{\pi/2}e^{-R\sin t}\:dt<\pi $$ Hence $$ \left|\int_{C_R}F(z)dz\right|\leqslant \frac{R}{R^2-1}\int_{C_R}|e^{iz}||dz|<\frac{\pi R}{R^2-1}\to0 $$ as $R\to\infty$. Moreover $$ \int_{\gamma_r}F(z)dz=\int_{\gamma_r}\frac{e^{iz}}{z(z^2+1)}dz=\int_{\gamma_r}\left(\frac1{z}+g(z)\right)dz\tag2 $$ where $$ g(z)=\sum_{n=1}^{\infty}\frac{i^nz^{n-1}}{n!}\sum_{n=0}^{\infty}(-1)^nz^{2n} $$ and is analytic for $|z|<1$. Since $|g(z)|<M$ for $|z|<1$ $$ \left|\int_{\gamma_r}g(z)dz\right|<\pi Mr\to0 $$ as $r\to0$. And $$ \int_{\gamma_r}\frac1{z}dz=\int_{\pi}^{0}\frac{ire^{i\theta}}{re^{i\theta}}d\theta=-\pi i $$ So by $(2)$ $$ \lim_{r\to0}\int_{\gamma_r}F(z)dz=\lim_{r\to0}\int_{\gamma_r}\frac1{z}dz+\lim_{r\to0}\int_{\gamma_r}g(z)dz=-\pi i $$ Hence from $(1)$ $$ \int_{-\infty}^{\infty}F(x)dx=\pi i(1-\frac1{e}) $$ And $$ \int_{-\infty }^\infty \frac{\sin x}{x(x^2+1)}dx=\Im{\int_{-\infty}^{\infty}F(x)dx}=\pi (1-\frac1{e}) $$ So $$ \int_{0}^\infty \frac{\sin x}{x(x^2+1)}dx=\frac{\pi}{2} (1-\frac1{e}) $$

Let $$\begin{align} I(\alpha) &=\int_0^{\infty}\frac{\sin \alpha x}{x(1+x^2)}\,\mathrm{d}x \tag{1}\\ \mathcal{L}\left[ I(\alpha)\right] &= \int_0^{\infty}\frac{1}{x(1+x^2)}\cdot \frac{x}{s^2+x^2}\,\mathrm{d}x \tag{2}\\ &= \int_0^{\infty}\frac{1}{(1+x^2)(s^2+x^2)}\,\mathrm{d}x \tag{3}\\ &= \frac{1}{1-s^2}\int_0^{\infty}\left(\frac{1}{x^2+s^2}-\frac{1}{x^2+1}\right)\,\mathrm{d}x \tag{4}\\ \mathcal{L}\left\lvert I(\alpha)\right\rvert &= \frac{\pi}{2(s+s^2)} \tag{5}\\ I(\alpha) &= \frac{\pi}{2}\left(1-\frac1{e^\alpha}\right) \tag{6} \end{align}$$

and for $\alpha = 1 $ we get

$$ \int_0^{\infty}\frac{\sin x}{x(1+x^2)}\,\mathrm{d}x = \frac{\pi}{2}\left(1-\frac1{e}\right)$$