Recall that $x$ is a fixed element of $]0;1[$. Show that:
$$\int _ { 0 } ^ { + \infty } \frac { t ^ { x - 1 } } { 1 + t } \mathrm {~d} t = \int _ { 0 } ^ { 1 } \left( \frac { t ^ { x - 1 } } { 1 + t } + \frac { t ^ { - x } } { 1 + t } \right) \mathrm { d } t$$