Calculate $\int_{-\infty}^\infty\frac{\sin t}{t(1+t^2)}\,\mathrm{d}t$ 
Evaluate $\displaystyle\int_{-\infty}^\infty\frac{\sin(t)}{t(1+t^2)}\,\mathrm{d}t.$

$$\int_{-\infty}^\infty\frac{\sin t}{t(1+t^2)}\,\mathrm{d}t= \mathfrak{Im}\left\{\int_{-\infty}^\infty\frac{\mathrm{e}^{\mathrm{i}t}}{t(1+t^2)} \,\mathrm{d}t\right\}$$
To get the roots of the denominator, we consider $t(1+t^2)=0$. Then, the roots are $\{0,\mathrm{i},-\mathrm{i}\}$.
I am going to calculate the integral by determining residues associated with $z=0$ and $z=\mathrm{i}$.
\begin{eqnarray}
\mathfrak{Im}\left\{\int_{-\infty}^\infty\frac{\mathrm{e}^{\mathrm{i}t}}{t(1+t^2)} \,\mathrm{d}t\right\}
&=&\mathfrak{Im}\left\{\pi\mathrm{i}\lim_{t\to0}\frac{\mathrm{e}^{\mathrm{i}t}}{1+t^2}+2 \pi\mathrm{i}\lim_{t\to\mathrm{i}}\frac{\mathrm{e}^{\mathrm{i}t}}{t(t+\mathrm{i})}\right\}\\
&=&\mathfrak{Im}\left\{\pi\mathrm{i}+ 2\pi\mathrm{i}\frac{\mathrm{e}^{-1}}{-2}\right\}\\
&=&\pi(1-\mathrm{e}^{-1}) 
\end{eqnarray}
Is it correct?
How can I apply Plancherel's theorem to calculate the same integral?
Thanks!
 A: Since $\mathcal{L}(\sin t)=\frac{1}{1+s^2}$ and $\mathcal{L}^{-1}\left(\frac{1}{t(t^2+1)}\right)=1-\cos(s)$, the value of the integral is
$$ 2\int_{0}^{+\infty}\frac{1-\cos(s)}{1+s^2}\,ds = \pi-2\int_{0}^{+\infty}\frac{\cos(s)}{1+s^2} =\color{red}{\pi\left(1-\frac{1}{e}\right).}$$
A: It seems no one has yet adressed the question of how to do this using Plancherel's theorem. Here it goes:
The Fourier transform of $f(x) := \chi_{[-1,1]}(x)$ is easily computed to be $$\hat f (t) = \sqrt{\frac{2}{\pi}} \frac{\sin(t)}{t},$$ and the Fourier transform of $g(x) := \exp(-|x|)$ is $$\hat g(t) =\frac{1}{\sqrt{2\pi}} \frac{2}{1+t^2}$$ (See this question for a proof of the latter). Using Plancherel's theorem, we obtain
$$\int_{-\infty}^{\infty} \frac{\sin(t)}{t(1+t^2)}\mathrm{d}t = \frac{\pi}{2} \cdot (\hat f, \hat g)_{L^2} = \frac{\pi}{2}\cdot (f,g)_{L^2} = \pi \int_{0}^{1} e^{-x} ~\mathrm{d}x = \pi(1-e^{-1})  $$
A: I verified your result using
\begin{align}
\int_{-\infty}^{\infty} \frac{\sin(t)}{t(1+t^2)} dt &= 2\int^\infty_0 \frac{t\sin(t)}{t^2(1+t^2)}\,dt\\
& = 2\int^\infty_0 \frac{\sin(t)}{t}\,dt - 2\int^\infty_0 \frac{t\sin(t)}{1+t^2} \,dt\\&= \pi -\frac{\pi}{e}
\end{align}
Both integrals are standards in contour integration. 
A: Let us consider $$I\left(x\right)=2\int_{0}^{\infty}\frac{\sin\left(xt\right)}{t\left(1+t^{2}\right)}dt$$ then $$I'\left(x\right)=2\int_{0}^{\infty}\frac{\cos\left(xt\right)}{1+t^{2}}dt$$ and following this proof we get $$I'\left(x\right)=\pi e^{-x}
 $$ then $$I\left(x\right)=-\pi e^{-x}+C
 $$ and since $I\left(0\right)=0
 $ we have $$C=\pi
 $$ so $$I\left(1\right)=2\int_{0}^{\infty}\frac{\sin\left(t\right)}{t\left(1+t^{2}\right)}dt=\color{blue}{\pi-\frac{\pi}{e}}.$$
