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Is this limit $$\lim_{\varepsilon\to 0}\,\,\varepsilon\int_{\mathbb{R}^3}\frac{e^{-\varepsilon|x|}}{|x|^2(1+|x|^2)^s}$$ with $s>\frac{1}{2}$, zero?.

The limit of a product is the product of limit, so I evaluate $$\lim_{\varepsilon\to 0}\,\,\int_{\mathbb{R}^3}\frac{e^{-\varepsilon|x|}}{|x|^2(1+|x|^2)^s}$$. With the theorem of dominated convergence the last limit equals $$\int_{\mathbb{R}^3}\frac{1}{|x|^2(1+|x|^2)^s}=4\pi\int_{0}^{+\infty}\frac{1}{(1+r^2)^s}=C<\infty$$ (I have used the fact that $s>\frac{1}{2}$)

Using the product rule I have the result. Have I made some mistake?

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1 Answer 1

What you did is correct. The only thing you have to take care is that in general, dominated convergence theorem applies for sequences. Here there is no problem since the convergence is monotonic.

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Do you really need monotonicity to justify this? I mean, if $I(\epsilon)$ denotes the integral, then $\lim_{\epsilon\rightarrow 0^+}I(\epsilon)=I$ if and only if $I(\epsilon_n)$ tends to $I$ for every sequence $\epsilon_n$ tending to $0^+$. –  1015 Apr 28 '13 at 20:38
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I agree with @Julien. –  Matt Apr 28 '13 at 21:13
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@julien I agree monotonicity is not needed (anyway I think the argument of going back to sequences should be mentioned). –  Davide Giraudo Apr 28 '13 at 21:20
    
Sure. I also always go back to sequences in such a situation. –  1015 Apr 28 '13 at 21:24
    
Going back to sequences is a standard procedure in cases like this; for this reason I didn't mention it. But you have reason, maybe if I mentioned iit my reasoning was clearer. –  Matt Apr 28 '13 at 21:24

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