# On the Laplace transform $\int_0^\infty e^{-sx}d \left( \ \int_2^{e^{1+x}}\frac{dt}{\log t}\right)$

I've read the basics about Laplace transform, and I know that since for $\Re s>1$, $\frac{e^x}{1+x}$ has exponential order, then $$F(s)=\int_0^\infty e^{-sx}\frac{e^{1+x}}{1+x}dx$$ is well defined, when $x>0$.

Let $\text{Li}(x)$ be the logarithmic integral $$\int_2^x \frac{dt}{\log t}$$

First for $X>0$ my computations were for the integral $\int_0^X$ $$e^{-sX}(\text{Li}(e^{X+1})-\text{Li}(e))+s\int_0^X e^{-sx}(\text{Li}(e^{x+1})-\text{Li}(e))dx.$$

Also by integration I can write $F(s)$ as $$F(s)=e\sum_{k=0}^\infty (-1)^k\int_0^1e^{(1-s)x}x^kdx+e\int_0^1e^{(1-s)}\sum_{k=0}^\infty\frac{1}{t^{k+2}}dt.$$

Secondly $e^{-\rho R}\text{Li}(e^{X+1})$ is a bounded function of $X$ for each $\rho >h$, but on what assumptions? I don't know what is this abscissa $h$. Also I know the basic $$\int_0^\infty e^{(s-h)x}dx=\frac{1}{s-h}.$$

My context was a first reading of a theorem due to Ikeara, but my purpose with this current post is understanding and learning, if it is possible.

Question. Justify and compute, if possible, or otherwise give an approximation of the Laplace transform $$\int_0^\infty e^{-sx}d \left( \ \int_2^{e^{1+x}}\frac{dt}{\log t}\right)$$ stating for which complex abscissa $h$ (if such computation is feasible) it is defined.