Is this a tempered distribution? I want to check if

$$\phi \mapsto \lim_{\epsilon \to 0} \int_\mathbb {R} \frac{\phi(x)}{x^2-(1+i\epsilon)^2} dx, \ \phi \in \mathcal{S}(\mathbb{R})$$

is a Tempered Distribution. This problem arises in solvin a PDE. Any ideas or hints?
 A: I am skipping some details but I hope that the idea is correct. Write
\begin{align*}
&  \int_{\mathbb{R}}\frac{\phi(x)}{x^{2}-(1+i\varepsilon)^{2}}dx=\int%
_{1/2}^{3/2}\frac{\phi(x)-\phi(1)}{x^{2}-(1+i\varepsilon)^{2}}dx\\
&  +\int_{-3/2}^{-1/2}\frac{\phi(x)-\phi(-1)}{x^{2}-(1+i\varepsilon)^{2}
}dx+\int_{\mathbb{R}\setminus\lbrack-3/2,-1/2]\cup\lbrack1/2,3/2]}\frac
{\phi(x)}{x^{2}-(1+i\varepsilon)^{2}}dx\\
&  +\phi(1)\int_{1/2}^{3/2}\frac{1}{x^{2}-(1+i\varepsilon)^{2}}dx+\phi
(-1)\int_{-3/2}^{-1/2}\frac{1}{x^{2}-(1+i\varepsilon)^{2}}dx\\
&  =I+II+III+IV+V.
\end{align*}
By the mean value theorem $|\phi(x)-\phi(1)|\leq\Vert\phi^{\prime}%
\Vert_{\infty}|x-1|$ and so
\begin{align*}
|I|  & \leq\Vert\phi^{\prime}\Vert_{\infty}\int_{1/2}^{3/2}\frac
{|x-1|}{|x-1-i\varepsilon|\,|x+1+i\varepsilon|}dx\\
& \leq\Vert\phi^{\prime}\Vert_{\infty}\int_{1/2}^{3/2}\frac{1}{x+1}dx
\end{align*}
since $|x-1-i\varepsilon|=\sqrt{(x-1)^{2}+\varepsilon^{2}}\geq|x-1|$. The term
$II$ can be treated similarly.
\begin{align*}
|III|  & \leq\Vert\phi\Vert_{\infty}\int_{\mathbb{R}\setminus\lbrack
-3/2,-1/2]\cup\lbrack1/2,3/2]}\frac{1}{|x^{2}-(1+i\varepsilon)^{2}|}dx\\
& \leq\Vert\phi\Vert_{\infty}\int_{3/2}^{\infty}\frac{1}{x^{2}-5/4}
dx+\Vert\phi\Vert_{\infty}\int_{-\infty}^{-3/2}\frac{1}{x^{2}-5/4}dx+\Vert
\phi\Vert_{\infty}\int_{-1/2}^{1/2}\frac{1}{3/4-x^{2}}dx
\end{align*}
where $|x^{2}-(1+i\varepsilon)^{2}|=|x^{2}-1-\varepsilon^{2}+2i\varepsilon
|\geq|x^{2}-1-\varepsilon^{2}|$. Since $$\frac{2(1+i\varepsilon)}
{x^{2}-(1+i\varepsilon)^{2}}=\frac{1}{x+1+i\varepsilon}-\frac{1}
{x-(1+i\varepsilon)}$$ and
\begin{align*}
\int_{1/2}^{3/2}\frac{1}{x-1-i\varepsilon}dx  & =\int_{-1/2}^{1/2}\frac
{1}{s-i\varepsilon}ds=Log\left(  \frac{1}{2}-\varepsilon i\right)  -Log\left(
-\frac{1}{2}-\varepsilon i\right)  \\
& =\log\sqrt{\frac{1}{4}+\varepsilon^{2}}+i\arg\left(  \frac{1}{2}-\varepsilon
i\right)  -\log\sqrt{\frac{1}{4}+\varepsilon^{2}}-i\arg\left(  -\frac{1}
{2}-\varepsilon i\right)  \\
& =i\arg\left(  \frac{1}{2}-\varepsilon i\right)  -i\arg\left(  -\frac{1}
{2}-\varepsilon i\right)  ,
\end{align*}
the terms $IV$ and $V$ can be bound by $c\Vert\phi\Vert_{\infty}$. In
conclusion $|T(\phi)|\leq c(\Vert\phi\Vert_{\infty}+\Vert\phi^{\prime}
\Vert_{\infty})$ and so $T$ is a tempered distribution.
