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 that for all $\theta \in ] 0 ; \pi [$, $$g ( \theta ) \sin ( x \theta ) = \frac { 1 } { 2 \mathrm { i } } \left( g ( - \theta ) e ^ { \mathrm { i } x \theta } - g ( \theta ) e ^ { - \mathrm { i } x \theta } \right) = \sin ( \theta ) \int _ { 0 } ^ { + \infty } \frac { t ^ { x } } { t ^ { 2 } + 2 t \cos ( \theta ) + 1 } \mathrm {~d} t$$
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 that for all $\theta \in ] 0 ; \pi [$,
$$g ( \theta ) \sin ( x \theta ) = \frac { 1 } { 2 \mathrm { i } } \left( g ( - \theta ) e ^ { \mathrm { i } x \theta } - g ( \theta ) e ^ { - \mathrm { i } x \theta } \right) = \sin ( \theta ) \int _ { 0 } ^ { + \infty } \frac { t ^ { x } } { t ^ { 2 } + 2 t \cos ( \theta ) + 1 } \mathrm {~d} t$$