\textbf{Problem 1: calculation of an integral}
For $x \geqslant 0$ we define
$$f ( x ) = \int _ { 0 } ^ { \infty } \frac { e ^ { - t x } } { 1 + t ^ { 2 } } \mathrm {~d} t \quad \text { and } \quad g ( x ) = \int _ { 0 } ^ { \infty } \frac { \sin t } { t + x } \mathrm {~d} t$$
Show that on $] 0 , \infty [$ we have $f = g$. For this you may use a differential equation satisfied by $( f - g )$ and use the behavior of $f$ and $g$ at $+ \infty$.