Is this a tempered distribution?

Question

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?

Answer 1

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.