Searching for an asymptotic to exp[0.5]

[sheldonison]

I looked at sheldon's 2nd answer and noticed what partially was the motivation for one of my last questions :

the plot of fake ln(x) resembles integral (1 - exp(-x) ) / x !

It was "Visually Obvious" from the psuedoperiodicity and the growth rate for Re(z) << 0.

I havent considered the signs of the derivatives of integral (1 - exp(-x) / x yet.
Maybe that explains alot.

regards

tommy1729

I reread your post. So, to clarify, you are interested in comparing the fake function for \( \ln(x) = \exp(-z)\text{fake}(\exp(z)\ln(z)) \) with the integral (which I don't yet understand), integral (1 - exp(-x) ) / x.

So far, I have only answered the Op's question, and generated the "fake" function for \( \ln(z+1) \approx \exp(-z)\text{fake}(\exp(z)\ln(z+1))
\) but doing so for \( \ln(x) \) or \( \sqrt{x} \) or \( \exp^{0.5}(x) \) or \( \frac{1}{x} \) would presumably be straightforward enough using the equation I posted in my 2nd answer; copied below.

\( a_n =\lim_{r\to\infty} \int_{-\pi}^{\pi} \frac{1}{2\pi}e^{-n(r+ix)}(\exp(e^{r+ix})\ln(1+e^{r+ix}))\; \mathrm{d}x\;\; \) fake function for exp(z)ln(z+1), replace ln(1+e^{}) with other positive valued function
\( \text{fake}(z) = \exp(-z) \, \sum_{n=0}^{\infty} a_n x^n\;\; \) the fake(z) zeros are probably always near the imaginary axis \( \approx 2\pi i \) apart. Also, I posted the first 18 zeros of the asymptotic function for \( \exp(x)\ln(x+1) \) at MSE, along with an approximation.

It would be interesting if this equation is related to the integral solution posted on Mathoverflow, but I don't understand that solution (yet).