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how could i prove that the sucession

$ \frac{k^{k+1}}{k!}t^{k}e^{-kt} $ tends in the limit $ k \to \infty$ to the delta function $ \delta (t-1) $

this is used inside the post 's inversion formula for the Laplace transform.

should i use the Saddle point method perhaps ?? :)

also.. what is the compact support of the Heaviside function $ H(x) $ or of the dirac delta function $ \delta (sin \pi x) $

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I don't understand your questions: the succession is for $t$ fixed, right? And what do you mean by "compact support of" a function? –  Davide Giraudo Feb 4 '12 at 11:00
    
Jose, I assume you are asking about compact support because you are working with distributions. If this is the case, then you are confusing definitions. Test functions need compact support, functions which define distributions do not. The Heaviside function has support on the entire positive line (it has to, or else the boundary term from integration by parts would not vanish for some test functions and it would not be the derivative of the delta function). In the case of your latter question, $\delta (\cdot)$ is not really a function and so only has support in the sense of distributions. –  Chris Janjigian Feb 4 '12 at 15:19

1 Answer 1

up vote 3 down vote accepted

You should use Stirling formula as

$$k!\approx \sqrt{2\pi k}k^k e^{-k}$$

and you will get

$$S_k\approx \frac{k^\frac{1}{2}}{\sqrt{2\pi}}t^ke^{-k(t-1)}$$

that can be rewritten as

$$S_k\approx \frac{1}{\sqrt{2\pi\frac{1}{k}}}e^{-\frac{(t-1)}{\frac{1}{k}}+k\ln t}$$

and put $\epsilon=\frac{1}{k}$. Now consider a compact support function $f(t)$ (this means that this function goes fastly enough to zero when its argument $t$ goes to $\pm\infty$). Then

$$\lim_{\epsilon\rightarrow 0}\int_{-\infty}^{+\infty}\frac{1}{\sqrt{2\pi\epsilon}}e^{-\frac{t-1}{\epsilon}+\frac{1}{\epsilon}\ln t}f(t)dt.$$

Now, applying saddle point method we take the derivative of the argument of the exponential. You will get an extremum for $t=1$. Expanding $-(t-1)+\ln t$ till second order and integrating you will get the proof of your assertion, after extracting $f(1)$.

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That is not what compact support means! That is "vanishing at infinity". Interesting answer nevertheless. Compact support of the function $f$ means that the closure of the set where $f$ is nonzero is compact. On the Euclidean space this means that it is bounded and closed. So your function is actually $0$ way before infinity :-). –  Jonas Teuwen Feb 4 '12 at 12:52
    
@JonasTeuwen: Thank you a lot Jonas. You are right of course but I am a physicist and this is just my limitation. –  Jon Feb 4 '12 at 13:03
    
Ok, thanks everybody :) for your answers ;) –  Jose Garcia Feb 4 '12 at 22:22
    
@JoseGarcia: I think you should accept the answer if it is ok for you. Thanks. –  Jon Feb 4 '12 at 22:49

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