\textbf{Problem 1: calculation of an integral}
For $x \geqslant 0$ we define
$$f ( x ) = \int _ { 0 } ^ { \infty } \frac { e ^ { - t x } } { 1 + t ^ { 2 } } \mathrm {~d} t \quad \text { and } \quad g ( x ) = \int _ { 0 } ^ { \infty } \frac { \sin t } { t + x } \mathrm {~d} t$$
Using the expression of $g$ obtained in question 2.b., show that $g$ is continuous at 0.