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.

Thanks in advance.

Please read my question looking for mistakes. I hope an answer can help me in understanding the basics of these techniques. This is my first post about this topic. Thanks again!

  • $\begingroup$ Thanks for your edit giobrach. I hope that my computations were mathematical meaning, and learn a few about this theory in next months. $\endgroup$ – user243301 May 5 '16 at 15:58
  • $\begingroup$ Also I know the table of the most usual Laplace transforms. $\endgroup$ – user243301 May 6 '16 at 5:43
  • $\begingroup$ I would like learn some easy facts and computations from your answer about computations of this kind, of the Question: compute previous integral transform, or an approximation of it and its abscissa of convergence. Many thanks. $\endgroup$ – user243301 May 11 '16 at 9:45
  • $\begingroup$ Was a typo in the name of the theorem, the right is Ikehara when I've written "...first reading of a theorem due to Ikeara"". $\endgroup$ – user243301 May 13 '16 at 12:33

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