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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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    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.2012-02-04

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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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    @JoseGarcia: $I$ think you should accept the answer if it is ok for you. Thanks.2012-02-04