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Let $x\in\mathbb{R}$, let $\alpha\in\mathbb{R}_{>0}$, define $\alpha^+=\lim_{t\downarrow0}(\alpha+it)$, and consider the integral \begin{equation} I(x)=\int_{-\infty}^\infty dy\frac{\exp(ixy)}{y^2-\alpha^+}. \end{equation} For any $t\in\mathbb{R}_{>0}$, the integral \begin{equation} \int_{-\infty}^\infty dy\frac{\exp(ixy)}{y^2-(\alpha+it)} \end{equation} is readily done using complex analysis. However, I'm a bit confused as to how the limit behaves with respect to the integral. I want to say that \begin{equation} \int_{-\infty}^\infty dy\frac{\exp(ixy)}{y^2-\alpha^+}=\int_{-\infty}^\infty dy\lim_{t\downarrow0}\frac{\exp(ixy)}{y^2-(\alpha+it)}, \end{equation} and consequently that \begin{equation} \int_{-\infty}^\infty dy\lim_{t\downarrow0}\frac{\exp(ixy)}{y^2-(\alpha+it)}=\lim_{t\downarrow0}\int_{-\infty}^\infty dy\frac{\exp(ixy)}{y^2-(\alpha+it)}, \end{equation} but I'm not sure if this is how it works and right now I can't think of any theorems or restrictions which play a role here. I think that the first equality holds because the integrand is continuous for all $y\neq\pm\sqrt{\alpha}$ and $\{\pm\sqrt{\alpha}\}$ is a set of measure zero, but I'm not sure. Any thoughts?

B. Pasternak
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  • If you take $x = 0$, doesn't the integral (without the term $it$) diverge? It tends like $c/x$ in a neighbhorhood of $\pm \sqrt{\alpha}$. – JMK Sep 14 '16 at 21:36
  • Hmm, it seems so, but since the limit is there you are never really without the term $it$, right? – B. Pasternak Sep 14 '16 at 21:50
  • well that's on the right hand side of your final equation, on the left hand side, you have

    $$\lim_{t\downarrow0}\frac{\exp(ixy)}{y^2-(\alpha+it)} = \frac{\exp(ixy)}{y^2 - \alpha^+} = \frac{\exp(ixy)}{y^2 - \alpha}$$

    I don't know why you introduced the $\alpha^+$ notation, as $\alpha^+ = \alpha$ because that limit exists. Anyway, when you plug in $x = 0$, the concern about divergence is real, so if you have an expression of the integral with the $it$ term, then it should diverge to $\infty$ as $t \to 0$ and $x = 0$.

     – JMK Sep 14 '16 at 22:00
  • Well yes this is also the source of my confusion, kind of. The expression which we are asked to derive does not diverge at $x=0$ and $t\to0$. The limit is there precisely to avoid this divergence behavior I think, since shifting the pole off of the $x$-axis allows you to calculate the integral neatly. But then it seems that one (or both) of the equalities I wrote, do not holds. – B. Pasternak Sep 14 '16 at 22:05
  • Even worse, the integral defined as such diverges for any x, simply because the poles are on the real axis. Right? – B. Pasternak Sep 15 '16 at 06:16
  • Yes, you can repeat the argument for any such $x$. So at this point I'm really not sure. Have you tried a numerical evaluation using Mathematica or something? – JMK Sep 16 '16 at 00:18

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