Let $g$ be the function defined by $$\begin{aligned}
g : ] - \pi ; \pi [ & \longrightarrow \mathbf { C } \\
\theta & \longmapsto e ^ { \mathrm { i } x \theta } \int _ { 0 } ^ { + \infty } \frac { t ^ { x - 1 } } { 1 + t e ^ { \mathrm { i } \theta } } \mathrm {~d} t
\end{aligned}$$ where $x$ is a fixed element of $]0;1[$. Show, using the dominated convergence theorem, that: $$\lim _ { \theta \rightarrow \pi ^ { - } } g ( \theta ) \sin ( x \theta ) = \int _ { - \infty } ^ { + \infty } \frac { \mathrm { d } u } { 1 + u ^ { 2 } }$$
Let $g$ be the function defined by
$$\begin{aligned}
g : ] - \pi ; \pi [ & \longrightarrow \mathbf { C } \\
\theta & \longmapsto e ^ { \mathrm { i } x \theta } \int _ { 0 } ^ { + \infty } \frac { t ^ { x - 1 } } { 1 + t e ^ { \mathrm { i } \theta } } \mathrm {~d} t
\end{aligned}$$
where $x$ is a fixed element of $]0;1[$. Show, using the dominated convergence theorem, that:
$$\lim _ { \theta \rightarrow \pi ^ { - } } g ( \theta ) \sin ( x \theta ) = \int _ { - \infty } ^ { + \infty } \frac { \mathrm { d } u } { 1 + u ^ { 2 } }$$