Question: Taking the differences between terms of the sequence generated with the expression below (I'm taking the floor of each integral, and then finding the differences between the integers that I get) $$\left \lfloor \int_e^{ne} \phi(x)~dx \right \rfloor$$
where $\phi(x)=\exp(\frac{1}{\ln(x)}),$ for $n=2,3,4,...$
seems to only take on values $5,4,3.$ Is this actually true? Why does this happen? I've calculated up to $n=47.$ The first term is $5$ and that occurs once. Then the oscillations between $3$ and $4$ start. Maybe after calculating more terms there will be a limit.
Let $\phi(x)=\exp(1/\ln x)$, $F(n)= \int_e^{ne} \phi(x)~dx$, and $G(n)=\left \lfloor F(n) \right \rfloor$. As $\exp$ is increasing and $1/\ln x$ is decreasing for $x>1$, its composition $\phi=\exp\circ(1/\ln x)$ is also decreasing for $x>1$.
If $0 < x < 1$, Taylor series gives the lower bound $e^x> 1 + x$ and the upper bound \begin{align}e^x&=\sum_{n=0}^{\infty} \dfrac{x^n}{n!}=1+x+\sum_{n=2}^{\infty} \dfrac{x^n}{n!}=1+x+x^2\sum_{n=2}^{\infty} \dfrac{x^{n-2}}{n!}\\ &=1+x+x^2\sum_{n=0}^{\infty} \dfrac{x^{n}}{(n+2)!}<1+x+x^2(e-2).\\ \end{align}
Therefore $1+\dfrac1{\ln(x)} \lt \phi(x) \lt 1+ \dfrac1{\ln(x)}+\dfrac{e-2}{\ln^2(x)} $.
Moreover, since $0<1/\ln x\le1$ and Taylor series gives $e^{-t}>1-t$ for $t\in(0,1)$, we have $e^t>1/(1-t)$ so $$\phi(x)=\exp\left(\frac1{\ln x}\right)<\frac1{1-\frac1{\ln x}}=\frac1{\ln x-1}.$$ Thus
\begin{align}F(n)&= \int_e^{ne} \phi(x)~dx< \int_e^{ne}\left(1+ \dfrac1{\ln(x)-1}\right)dx\\ &= \int_e^{ne} dx+\int_e^{ne} \dfrac1{\ln(x)-1}~dx\\ \end{align} and invoking the upper bound of $\phi$ derived earlier, the difference is no more than \begin{align}F(n+1)-F(n)&= \int_{ne}^{(n+1)e} \phi(x)~dx\\ &\lt \int_{ne}^{(n+1)e} \left( 1+ \dfrac1{\ln(x)}+\dfrac{e-2}{\ln^2(x)}\right)~dx\\ &= \int_{ne}^{(n+1)e}~dx+ \int_{ne}^{(n+1)e}\dfrac1{\ln(x)}\,dx+\int_{ne}^{(n+1)e}\dfrac{e-2}{\ln^2(x)}\,dx\\ &= e+ \int_{ne}^{(n+1)e}\dfrac1{\ln(x)}\,dx+\int_{ne}^{(n+1)e}\dfrac{e-2}{\ln^2(x)}\,dx\\ &< e+ \dfrac{e}{\ln(n)}+\dfrac{e-2}{\ln^2(n)}=b(n).\end{align} Similarly, invoking the lower bound of $\phi$, the difference is no less than \begin{align}F(n+1)-F(n) &= \int_{ne}^{(n+1)e} \phi(x)~dx\\ &\gt \int_{ne}^{(n+1)e} \left( 1+ \dfrac1{\ln(x)}\right)~dx\\ &= \int_{ne}^{(n+1)e}~dx+ \int_{ne}^{(n+1)e}\dfrac1{\ln(x)}~dx\\ &= e+ \int_{ne}^{(n+1)e}\dfrac1{\ln(x)}~dx\\ &> e+ \dfrac{e}{\ln(n)+1}=a(n).\end{align} Hence the difference of the differences is no more than \begin{align}b(n)-a(n) &=\left( e+ \dfrac{e}{\ln(n)}+\dfrac{e-2}{\ln^2(n)}\right)-\left(e+ \dfrac{e}{\ln(n)+1}\right)\\ &=e\left(\dfrac1{\ln(n)}-\dfrac1{\ln(n)+1}\right)+\dfrac{e-2}{\ln^2(n)}\\ &=\dfrac{e}{\ln(n)(\ln(n)+1)}+\dfrac{e-2}{\ln^2(n)}.\\ \end{align} The problem will be solved if $b(n)-a(n)<1$ for $n>N$ for some $N>1$, and if $G(n+1)-G(n)\ge3$ for $n>N$, as the cases $n\le N$ will be finite. The first condition holds for $n>4$, since $b(n)-a(n)$ is evidently a decreasing function of $n$ (as all the terms are positive). Furthermore, since $a(n)$ is also decreasing and $3<a(n)<4$ for $n>4$, it follows that $G(n+1)-G(n)\ge3$ as $a(n)$ is a lower bound of $F(n+1)-F(n)$.
Checking the cases $n=2,3,4$ yields $G(n+1)-G(n)=5,4,4$ respectively so the result follows.