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 the function $g$ is of class $\mathcal { C } ^ { 1 }$ on $] - \pi ; \pi [$ and that for all $\theta \in ] - \pi ; \pi [$, $$g ^ { \prime } ( \theta ) = \mathrm { i } e ^ { \mathrm { i } x \theta } \int _ { 0 } ^ { + \infty } h ^ { \prime } ( t ) \mathrm { d } t$$ where $h$ is the function defined by $$\begin{aligned}
h : ] 0 ; + \infty [ & \longrightarrow \mathbf { C } \\
t & \longmapsto \frac { t ^ { x } } { 1 + t e ^ { \mathrm { i } \theta } } .
\end{aligned}$$ Calculate $h ( 0 )$ and $$\lim _ { t \rightarrow + \infty } h ( t ) .$$ Deduce that the function $g$ is constant on $] - \pi ; \pi [$.
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 the function $g$ is of class $\mathcal { C } ^ { 1 }$ on $] - \pi ; \pi [$ and that for all $\theta \in ] - \pi ; \pi [$,
$$g ^ { \prime } ( \theta ) = \mathrm { i } e ^ { \mathrm { i } x \theta } \int _ { 0 } ^ { + \infty } h ^ { \prime } ( t ) \mathrm { d } t$$
where $h$ is the function defined by
$$\begin{aligned}
h : ] 0 ; + \infty [ & \longrightarrow \mathbf { C } \\
t & \longmapsto \frac { t ^ { x } } { 1 + t e ^ { \mathrm { i } \theta } } .
\end{aligned}$$
Calculate $h ( 0 )$ and
$$\lim _ { t \rightarrow + \infty } h ( t ) .$$
Deduce that the function $g$ is constant on $] - \pi ; \pi [$.