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1.4: Integrales inpropias convergentes absolutamente uniformemente
https://espanol.libretexts.org/Matematicas/Analisis/Libro%3A_Funciones_Definidas_por_Integrales_Inpropias_(Trench)/01%3A_Cap%C3%ADtulos/1.04%3A_Integrales_inpropias_convergentes_absolutamente_uniformemente
\[\begin{aligned} \int_{r}^{\infty}e^{-px} |g(x,y)|\,dx &=& \int_{r}^{\infty} e^{-(p-p_{0})x}e^{-p_{0}x}|g(x,y)|\,dx\\ &\le& K\int_{r}^{\infty} e^{-(p-p_{0})x}\,dx= \frac{K e^{-(p-p_{0})r}}{p-p_{0}},\... $\begin{aligned} \int_{r}^{\infty}e^{-px} |g(x,y)|\,dx &=& \int_{r}^{\infty} e^{-(p-p_{0})x}e^{-p_{0}x}|g(x,y)|\,dx\\ &\le& K\int_{r}^{\infty} e^{-(p-p_{0})x}\,dx= \frac{K e^{-(p-p_{0})r}}{p-p_{0}},\end{aligned}$ para $x$ suficientemente grande si $p_{0}>0$ , Teorema [teorema: 4] implica eso $\int_{0}^{\infty}e^{-px}x^{\alpha}\sin xy\,dx$ y $\int_{0}^{\infty}e^{-px}x^{\alpha}\cos xy\,dx$ convergen absolutamente uniformemente sobre $(-\infty,\infty)$ si $p>0$ y $\alpha~\ge~0$ .

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