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I'm reading a paper on PDEs in preparation for some research.

In it, integrals like this appear repreatedly: $$ I(x,t) = \int_{|y|>1} e^{-i\alpha xy} \frac{e^{i\beta y^2}}{(1+y^2)^m} dy. $$ Here $y \in \mathbb{R}$, $m \in (1/4,1/2]$. The values of $\alpha = \frac{1}{2t}$ and $\beta = \frac{1}{4t}-1$ depend on $t$, but are nonzero.

The paper says to integrate by parts to obtain $$ I(x,t) = \frac{C}{\beta} \int_{|y| > 1} \frac{1}{y} \left( \frac{d}{dy} \left( \frac{e^{-i\alpha xy}}{(1+y^2)^m} \right) \right) e^{i\beta y^2} dy + \frac{1}{\beta} F(x), $$ where $F$ is a bounded continuous function. The function $F$ would be the boundary terms, though I'm not sure why the $t$ dependence disappeared (unless they're just suppressing it).

I apologize if this should be obvious, but I can't see how to get this answer from integrating by parts. I get $$\frac{1}{\beta} \int_{|y| > 1} \frac{1}{y} \left( \frac{d}{dy} \left( \frac{e^{-i\alpha xy}}{(1+y^2)^m} \right) \right) e^{i\beta y^2} dy $$ $$ = \int_{|y|>1} e^{-i\alpha xy} \frac{2ie^{i\beta y^2} - \frac{e^{i\beta y^2}}{\beta y^2}}{(1+y^2)^m} dy + \text{ boundary terms}.$$

It doesn't make sense to me to assume that the authors included the $$\int_{|y|>1} e^{-i\alpha xy} \frac{\frac{e^{i\beta y^2}}{\beta y^2}}{(1+y^2)^m} dy $$ term in the function $F$, since they assert $F$ is continuous and the whole point of this discussion in the paper is to carefully show that such integrals are continuous...

Any insight would be appreciated. I don't have much experience with PDEs, and this is proving a bit of a snag for me. Thanks.

[This instance is on p. 798 of Bona and Saut's paper Dispersive Blow-Up II. Schrodinger-Type Equations, Optical and Oceanic Rogue Waves.]

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