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Please help to improve/correct the following arguments.

I want to show that the integral $$I(x,y,z)=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}{z\over 2\pi [(\alpha-x)^2+(\beta-y)^2+z^2]^{3\over2}}f(\alpha,\beta)\;d\alpha\; d\beta$$

where $f$ is an arbitrary continuous function and $I$ is the solution of Laplace's equation $\nabla^2 I=0$, has the following properties:

1) Tends to $0$ as $x^2+y^2\to\infty$. Argument: As $x^2+y^2\to\infty$, the integrand tends to $0$, hence the integral tends to $0$. Concern: Does the arbitrariness of $f$ screw things up?

2) That $I(x,y,0)=f(x,y)$. Argument: For $\alpha,\beta\neq x,y$ respectively, the integrand is $0$ when $z=0$, but when $\alpha,\beta= x,y$ respectively, we have a singularity. This resembles the sampling property of the delta function, but how would I know that I can use that argument here, since we don't really have a delta function. Please help!

Thank you.

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    We need other assumptions over $f$ in order to make the integral convergent, for example bounded is enough. It's not clear you can switch limit and integral, but it's possible thanks to the dominated convergence theorem. I don't understand question 2, if we substitute $z$ by $0$, the integrand is $0$.2011-12-24
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    If we take $f=1$, the first result seem wrong, since after the substitutions $t_1=\alpha-x$ and $t_2=\alpha-y$, the integral is constant. Maybe we have to ask $f(x,y)\to 0$ as $x^2+y^2\to\infty$. For the second question, we have to show that $\lim_{z\to 0}I(x,y,z)=f(x,y)$, a factor $2\pi$ is missing, but it works.2011-12-24
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    See [this question](http://math.stackexchange.com/a/92959). Walter, is this a coincidence?2011-12-24
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    @Davide: About your first point: On the assumption that this question is meant to have the same conditions as that one, we only need to show that $I(x,y,z)$ goes to $0$ as $x^2+y^2\to\infty$ if $f(x,y)$ does. Irrespective of that other question, the question has to be meant that way since it acknowledges that $I(x,y,0)$ is $f(x,y)$ (under suitable missing conditions), and thus $I(x,y,z)$ can only go to $0$ if $f(x,y)$ does.2011-12-24
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    @joriki: That question is indeed very similar! In fact more or less the same context. The difference is I have to show that the boundary conditions are satisfied instead of being given them in the beginning. (And indeed, the integral is supposed to be the solution to Laplace's equation.) Re your comment to Davide, how would I know that $f\to 0$ as $x^2+y^2\to \infty$? And how would I know that in the limit of $z\to 0$ we get the delta function if I don't assume the BC's from the start?2011-12-24
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    @DavideGiraudo: You are right, a factor of $2\pi $is missing, thanks. Edited. The 2 statements are supposed to be right (because I am told to show that they are true...) How do you show that $\lim_{z\to 0}I(x,y,z)=f(x,y)$?2011-12-24
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    Take $\varepsilon>0$, make the substitutions $t_1=\alpha-x$, $t_2=\beta-y$: we have to show that $$\lim_{z\to 0}\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}\frac z{(t_1^2+t_2^2+z^2)^{3/2}}|f(t_1+x,t_2+y)-f(x,y)|=0.$$ Take $\delta$ in the definition of uniform continuity on $\overline{B(0,1)}$. Cut the integral in two parts.2011-12-24
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    @DavideGiraudo: Thanks again, but I still don't quite understand... Would you mind saying a little bit more about the limit-taking?2011-12-24
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    For the integral over the set $||(t_1,t_2)||\leq \delta$, we have $f(t_1+x,t_2+y)\leq \varepsilon$, so this integral is $\leq \varepsilon$. For the other integral, we have $\frac{|z|}{(t_1^2+t_2^2+z^2)^{3/2}}\leq \frac{|z|}{(t_1^2+t_2^2+\delta^2)^{3/2}}$, then integrate and take the limit $z\to 0$. Can you add in the question what are exactly the assumptions on $f$ and $I$?2011-12-24
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    @DavideGiraudo: I am only given that $f$ is continuous. As for $I$, it is the solution to Laplace's equation s.t. $\nabla^2 I=0$.2011-12-24

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