I'm trying to use Tauber's theorem below (Feller 1971, chapter XIII.5)
"Let U be a measure with a Laplace transform $\omega(\lambda)$ defined $\forall \lambda >0$ and $t,\tau>0$ s.t. $t\tau=1$,
then $\frac{\omega(\tau\lambda)}{\omega(\tau)} \rightarrow \frac{1}{\lambda^\rho} \ as \ \tau \rightarrow 0$ iff $\frac{U(tx)}{U(t)} \rightarrow x^\rho \ as\ t \rightarrow \infty$ "
to find a limit distribution of $B$ where $B$ has Laplace transform $\omega(\lambda)=\int e^{-\lambda x}u(x)dx$ that satisfies the implicit definition $\omega(\lambda) = e^{-\lambda-1+\omega(\lambda)}$ such that $\omega(o)=1$.
Now I figured this definition would give me $\log(\omega(\lambda))=(\omega(\lambda)-1) - \frac{1}{2}(\omega(\lambda)-1)^2 + ... \ = -\lambda-1+\omega(\lambda)$
thus $-\lambda = -\frac{1}{2}(\omega(\lambda)-1)^2$ and $\omega(\lambda)) = 1-\sqrt{2\lambda}+...$
If I define $\eta(\lambda):=\frac{1-\omega(\lambda)}{\lambda}$ then $\lim_{\tau \rightarrow 0} \frac{\eta(\lambda \tau)}{\eta(\tau)} = \lim_{\tau \rightarrow 0} \frac{1-\omega(\lambda)}{1-\omega(\tau)}\times \frac{\tau}{\tau \lambda} = \lim_{\tau \rightarrow 0} \frac{\sqrt{2\tau \lambda}\times\tau}{2\sqrt{\tau}\times\tau \lambda} = \lambda^{-\frac{1}{2}}$
So applying Tauber I would get $\lim_{t \rightarrow \infty} \frac{W(tx)}{W(t)} = x^{\frac{1}{2}}$ for $W$ a distribution that has Laplace $\eta(\lambda)$.
Calculating the inverse Laplace transform of $\eta(\lambda)$ however I get $1-U(x)$ where $U'(x)=u(x)$: $\int e^{-\lambda x}(1-U(x))dx = \int e^{-\lambda x}dx + \int{}e^{-\lambda x}U(x)dx = \frac{1}{\lambda} - \frac{\omega(\lambda)}{\lambda} = \frac{1-\omega(\lambda)}{\lambda} = \eta(\lambda)$.
Hence $\lim_{t \rightarrow \infty} \frac{1-U(tx)}{1-U(t)}=x^{\frac{1}{2}}$. Now this can't be true as $U(x)$ is an increasing cumulative distribution function and thus $1-U(x)$ is decreasing and $\frac{1-U(tx)}{1-U(t)} \leq 1$. What went wrong here?
I think plugging $u(x)$ and $U(x)$ into the Tauber theorem does not work the way I did, but I don't know how to fix this. I used another method (Lagrange inversion) to estimate $\mathbb{P}(B=x) \approx \frac{1}{\sqrt{2\pi}} x^{-\frac{3}{2}}$ so Tauber should somehow give me this $x^{-\frac{3}{2}}$.