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It is well known that the Poisson kernel

$$\eta_y(x)=\frac{1}{\pi}\frac{y}{x^2+y^2}$$

Provides a limit representation of the Dirac delta function $\lim_{y\to0}\eta_y(x)=\delta(x)$. In an attempt to verify this fact in Mathematica, I take:

Limit[y/(x^2+y^2) , y->0]

0

which gives the proper result for all x except for x=0. In fact, I would have preferred the Limit function to actually return Pi DiracDelta[x], similarly to what happens if we evaluate FourierTransform[1,x,w]. Nevertheless, we can also plot, e.g.:

Plot[y/(x^2 + y^2) /. y -> 10^-1, {x, -1, 1}, PlotRange -> All]

enter image description here

and convince ourselves that for ever-smaller y the function indeed approaches a delta function.

Now, consider changing the sign of x^2 in the above expression:

Plot[y/(-x^2 + y^2) /. y -> 5^-1, {x, -1, 1}, PlotRange -> All]

enter image description here

In this case, apart from the non-zero region around x=0 there also appear negative and positive infinities symmetrically on both sides. Considering that the normalization changes a bit:

Integrate[y/(-x^2 + y^2), {x, -Infinity, Infinity}]

enter image description here

one might think that

$$\lim_{y^2\to 0+i0}\eta_y(x)=\lim_{y^2\to 0+i0}\frac{i}{\pi}\frac{y}{-x^2+y^2}$$

is also a limit representation of the delta function, if we take y to zero along a path such that y^2 is slightly complex. However, so far this is just an assumption, and the Limit routine cannot be used to verify this as we have seen above. Is it possible to use Mathematica to verify this claim?

Kagaratsch
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    Just for completeness, I think your second graph is not for $y=0.1$ – mikuszefski Jun 01 '17 at 06:06
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    Also if you have y= I w you just get I w/(-x^2-w^2)=-I w/(x^2+w^2), i.e. basically the same as before. – mikuszefski Jun 01 '17 at 06:50
  • @mikuszefski You are right, it was actually y=1/5, I fixed it. Also, nice observation! Even though I'd like to keep y mostly real (due to its interpretation in my specific case) this shows that analytic continuation to complex y ensures that it is still the delta function. You should post this as an answer and I will upvote and accept! – Kagaratsch Jun 01 '17 at 12:56
  • (1) Definitely a math rather than Mathematica question. (2) Usually delta-type functionals are defined in real space. Might or might not make sense to try to extend to C in this case. – Daniel Lichtblau Jun 01 '17 at 15:58
  • I'm voting to close this question as off-topic because the issue it raises is not really a Mathematica issue but a matter of the OP not having grasped the mathematics involved. – m_goldberg Jun 02 '17 at 01:21
  • @m_goldberg You are right. I will delete this question in a couple hours, since it is off topic here, if no one comments to suggest otherwise until then. – Kagaratsch Jun 02 '17 at 03:22
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    @Kagaratsch the main question remains: *Is it possible to use Mathematica to verify this claim?* Would you consider this a Mathematica question? – mikuszefski Jun 02 '17 at 05:52
  • @mikuszefski I would give this the benefit of the doubt, even if only to answer "no" (but more knowledgeable folks could say a lot about the symbolic capabilities of MMA and such). (Voting to leave open) – LLlAMnYP Jun 02 '17 at 10:12
  • @mikuszefski Yes, that is what I thought too, when asking the question. OK, since the opinions are split, I will not delete the question. – Kagaratsch Jun 02 '17 at 12:38
  • BTW I'll not post my comment from above as answer, as it only deals with a mathematical but not the Mathematica aspect. – mikuszefski Jun 02 '17 at 13:20
  • There is a math question one could extract and post here at math.stackexchange.com, but that is not this question. This question is "How do I compute limits of distributions in mathematica? It keeps computing the limit as functions instead! –  Jun 02 '17 at 14:19
  • @Hurkyl Indeed, help, how do I compute a limit of distributions? Seems to me that the question is pretty evident from the above and I would love to know the answer. – Kagaratsch Jun 02 '17 at 14:23

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