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) $
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 $f(1)$.