Probability proof of inversion formula for Laplace transform Let $f:[0, \infty[\longrightarrow \mathbb{R}$ be  bounded and continuous and define $L(\lambda)=\int_0^\infty e^{-\lambda x}f(x)dx$.
Let $X_n$ be a sequence of independent random variables with exponential distribution of rate $\lambda$.
Using the fact that the sum $S_n=X_1+...+X_n$ has Gamma distribution we can see that
$$  (-1)^{n-1}\frac{\lambda^nL^{(n-1)}(\lambda)}{(n-1)!} =Ef(S_n)$$
where $L^{(n-1)}$ is the $n-1$ derivative of $L$.
How can we use this to 
prove $$ f(y)=\lim_{n}(-1)^{n-1}\frac{(\frac{n}{y})^nL^{(n-1)}(\frac{n}{y})}{(n-1)!} $$
We know that by the strong law of large numbers $S_n/n$ converges to $E(X_1)$ almost everywhere, so I could consider 
the parameter $\lambda=f(y)$ but that doesn't seem useful..

Old question, but I was working on the same problem recently. Here's an idea:
The functions $g_n$ defined by $g(x) = f(x/n)$ are bounded and continuous. Let $L_h$ denote the Laplace transform of a function $h$, and let the independent exponential random variables $X_1,X_2,\dots$ have rate $\frac{1}{y}$ (using $y\neq 0$). By the first part of the problem
$$
\mathbb{E}\left(f\left(\frac{S_n}{n}\right)\right) = \mathbb{E}(g_n(S_n)) = (-1)^{n-1}\frac{\frac{1}{y^n}L^{(n-1)}_{g_n}\left(\frac{1}{y}\right)}{(n-1)!}
$$
Yet the strong law from Chapter 7 gives
$$\lim_{n\to\infty}\frac{S_n}{n} = \mathbb{E}(X_1) = y \quad \text{almost surely}$$
and since $f$ is bounded the expected values converge:
$$
\lim_{n\to\infty} (-1)^{n-1}\frac{\frac{1}{y^n}L^{(n-1)}_{g_n}\left(\frac{1}{y}\right)}{(n-1)!} = \lim_{n\to\infty} \mathbb{E}\left(f\left(\frac{S_n}{n}\right)\right) = f(y)
$$
The rest is just a matter of rewriting the terms $L_{g_n}^{(n-1)}$ in terms of $L_f^{(n-1)}$. For any bounded continuous (say) function $h$ one can compute for each $\lambda > 0$ that
$$
\quad L_h^{(n-1)}(\lambda) = (-1)^{n-1}\int_0^\infty e^{-\lambda x} x^{n-1} h(x)\,dx
$$
In our case, then,
$$
L_{g_n}^{(n-1)}(\lambda) = (-1)^{n-1}\int_0^\infty e^{-\lambda x} x^{n-1} f(x/n) \, \mathrel{\underset{u:=x/n}{=}} (-1)^{n-1}\int_0^\infty e^{-n\lambda u} n^{n-1}u f(u) n \, du = n^n L_f(n\lambda)
$$
Setting $\lambda = \frac{1}{y}$ yields
$$
f(y) = \lim_{n\to\infty} (-1)^{n-1}\frac{\frac{1}{y^n}L^{(n-1)}_{g_n}\left(\frac{1}{y}\right)}{(n-1)!} = \lim_{n\to\infty} (-1)^{n-1}\frac{\frac{n^n}{y^n}L^{(n-1)}_{f}\left(\frac{n}{y}\right)}{(n-1)!}
$$
