Another theorem of Principal value Let $f$ holomorphic function with isolated singularities in neighborhood of $ \overline{\mathbb{H}}^+ = \{ z\in \mathbb{C} : \operatorname{Im} z \geqslant 0\}$ and suppose that f only have one sigularity on $\mathbb{R}$, a simple pole in z=0. 
If exist $R,C,\varepsilon>0$ such that $\vert f(z)\vert\leq C\vert z\vert^{-(1+\varepsilon)}$ for $z\in\bar{\mathbb{H}}^+$ and $\vert z\vert>R$ then show
$$(\int_{-\infty}^{-1}+PV\int_{-1}^{1}+\int_{1}^{\infty})f(x)dx=2\pi i\sum_{Im(z)>0}res_{z}f+\pi i(res_{0}f)$$
I can use the next theorem for principal values but I dont know how:
Let $f$ holomorphic function with isolated singularities in neighborhood of $\bar{\mathbb{H}}^+\lbrace z\in \mathbb{C}\vert 0\leq\vert z\vert\rbrace$. Suppose that singularities of $f$ on $\mathbb{R}$ they are all simple pole and is a finite subset in $(-R,R)$ for $R>0$. In addition $$\lim_{\vert z\vert\rightarrow\infty}_{z\in\bar{\mathbb{H}}^+}f(z)=0$$
Then
$$(\int_{-\infty}^{-R}+PV\int_{-R}^{R}+\int_{R}^{\infty})f(x)e^{ix}dx=2\pi i\sum_{Im(z)>0}res_{z}(f(\zeta)e^{i\zeta})+\pi i\sum_{z\in\mathbb{R}}res_{z}(f(\zeta)e^{i\zeta})$$
Let me a hand. Thanks
 A: You have to observe that the first question in not derived from the second. They look very similar, but they are different.
In order to answer the first question we can proceed as follows:
First, We can assume without loss of generality that $R>1$. From the hypothesis we observe that $$\lim_{\vert z\vert\rightarrow\infty}_{z\in\bar{\mathbb{H}}^+}f(z)=0,$$
Then we have that there exist $M>0$ such that all the possible poles of $f$ are inside of $\overline{\mathbb{H}}^+\cap M$. Let $r:=\max\{R,M+1\}$, $\rho\in(0,1)$, and consider the following curves $ \gamma_\rho,\,\gamma_r:[0,\pi]\to\mathbb{C}$ such that $\gamma_\rho(t)=\rho e^{it}$, and $\gamma_r(t)=r e^{it}$. The, by the Residue Theorem follows that
\begin{equation}
\left( \int_{\gamma_r}+\int_{[-r,-1]}+ \int_{[-1,-\rho]} - \int_{\gamma_\rho}+ \int_{[\rho,1]} +  \int_{[1,r]}  \right) f(z)\, dz = 2\pi i \sum_{Im (z)>0} res_z f 
\end{equation}
Consequently, again hypothesis we obtain:
$$\left| \int_{\gamma_r} f \right|\leq l(\gamma_r)\cdot \max_{\vert z\vert=r}_{z\in\bar{\mathbb{H}}^+} {\vert f(z)\vert}\leq (\pi r) \cdot C r^{-(1+\varepsilon)} \to 0, \text{ when } r\to\infty .$$
Then, $$ \lim_{r\to \infty}\int_{\gamma_r} f=0.$$
From this, considering the limit when $r\to \infty$, the above equation reduces to the next
\begin{equation}
\left( \int_{-\infty}^{-1} + \int_{[-1,-\rho]} - \int_{\gamma_\rho}+ \int_{[\rho,1]} +  \int_{1}^{\infty}  \right) f(z)\, dz = 2\pi i \sum_{Im (z)>0} res_z f .
\end{equation}
Moreover, because $0$ if a simple pole of $f$, we have that there exist an holomorphic function $g$  in a small open neighborhood of $0$ such that
$$f(z)=\frac{res_0 f}{z}+g(z),$$
then, if we have that $D_{\rho}\subset U$, it follows that
$$\int_{\gamma_\rho} f= \int_{\gamma_\rho}\frac{res_0 f}{z} +\int_{\gamma_\rho}g, $$
but, a simple calculation show us that $$ \int_{\gamma_\rho}\frac{res_0 f}{z}= i \pi \, res_0f, $$
and 
$$ \int_{\gamma_\rho}g = \rho \cdot \int_0^{\pi}g(\rho e^{it}) i e^{it} dt,$$
Hence, when $\rho$ goes to $0$, we obtain
$$\lim_{\rho \to 0} \int_{\gamma_\rho} f = i\pi\,  res_0 f. $$
From this and the noted before, we finally get
 \begin{equation}
\left( \int_{-\infty}^{-1} + PV\int_{-1}^1 +  \int_{1}^{\infty}  \right) f(x)\, dx = 2\pi i \sum_{Im(z)>0} res_z f + i\pi\,  res_0 f.
\end{equation}
