solving the integral $ \int_0^\infty \dfrac{\sin xt}{x(x^2+1)} dx $ How to solve the following integral by differential equations techniques?
$$ \int_0^\infty \dfrac{\sin xt}{x(x^2+1)} dx $$

Recalling the Laplace transform

$$ F(s)=\int_{0}^{\infty} f(t) e^{-st} dt. $$

We can use it to solve the problem. Taking Laplace transform with respect to $t$ gives

$$ f(t)=\int_0^\infty \dfrac{\sin xt}{x(x^2+1)} dx \implies F(s)=\int_{0}^{\infty}\frac{dx}{(1+x^2)(s^2+x^2)}  . $$

The last integral can be evaluated using partial fraction techniques (see below) to give

$$ \implies F(s) = \frac{1}{2}\,{\frac {\pi }{s \left( s+1 \right) }}. $$

Now, we can find the inverse Laplace transform using convolution technique as

$$ f(t)=\frac{\pi}{2}\int_{0}^{t} e^{-\tau}d\tau = \frac{\pi}{2}(1-e^{-t}). $$

Alternatively, one can use partial fraction technique 

$$  \frac{\pi}{2}\frac {1}{s(s+1)}=\frac{\pi}{2}\left(\frac{1}{s}-\frac{1}{s+1}\right). $$

to find the inverse Laplace transform
Notes:

1) The Laplace transform of $\sin(az)$ with respect to $z$ is
  $$ {\frac {a}{{s}^{2}+{a}^{2}}}. $$
2) $$\frac{1}{(1+x^2)(s^2+x^2)}={\frac {1}{ \left( {s}^{2}+{x}^{2} \right)  \left(1- {s}^{2} \right)}}-{\frac {1}{ \left( {x}^{2}+1 \right)\left(1- {s}^{2} \right) }}.$$
3) Inverse Laplace transform of $\frac{1}{s}$ and $\frac{1}{1+s}$ are given by
  $$  1,\quad e^{-t} $$
  respectively.  

