I have got the Laplace transform as follows:

LX(s)=exp(−s)∞∑t=0stt!exp(−Ct2/α),L_X(s) = \exp(-s)\sum_{t = 0}^{\infty} \frac{s^t}{t!}\exp(-Ct^{2/\alpha}),

where CC and α\alpha are positive constants.

I want to prove that, using Stirling’s approximation, we have

−log(LX(s))∼Cs2/α-\log(L_X(s)) \sim Cs^{2/\alpha} as s→∞s \to \infty.

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OK….. Sounds great. Good luck with that. So where’s the question? Where’s the work? What part of it are you struggling with?

– Brevan Ellefsen

yesterday

Thanks. I tried using Stirling’s approximation for t!t! and then simplifying the expression for s→∞s \to \infty. But, I could not reach the answer I want to prove.

– ksank43

6 hours ago

alrighty, good to know. I’ll try a stab at the problem later if I can get some free time with my computer.

– Brevan Ellefsen

6 hours ago

Thanks. I faced this question when I was trying to understand the exponential Tauberian theorem and apply it in a problem. I am working with the following form of the Tauberian theorem: E(e−sX)∼ersα\mathbb{E}(e^{-sX}) \sim e^{rs^{\alpha}} for s→∞s \to \infty is equivalent to P(X≤ϵ)∼et/ϵβ\mathbb{P}(X \leq \epsilon) \sim e^{t/\epsilon^{\beta}} for ϵ→0\epsilon \to 0 (for 1/α=1/β+11/\alpha = 1/\beta +1).

– ksank43

6 hours ago

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