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What is the asymptotic growth rate of $$f(x) = \int_e^\infty e^{ - x t / \log t} dt$$ as $x \to 0$?

As an example of what is meant by "growth rate" consider $$g(x) = \int_e^\infty e^{-x t} dt = \frac {e^{-ex}} x \approx x^{-1}.$$

By comparing to $g$, it is easy to see that $f(x) \approx x^{-1 - o(1) }$, but we would like to know what the lower order correction is.

We are guessing that $f(x) \approx \log^p(1/x)/x$ for some $p >0$, but unsure how to show this or find $p$.

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  • $\begingroup$ numerically $f(x)\simeq x^{-1}\log(1/x)$ for comes close for $x\rightarrow 0$ $\endgroup$ Jun 3, 2022 at 17:28
  • $\begingroup$ Thanks. That does seem like a highly plausible answer. There ought to be an analytic way to see it. $\endgroup$ Jun 3, 2022 at 17:33

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Denote $t/\log t=y$. Then $y$ increases from $e$ to $\infty$ when $x$ goes from $e$ to $\infty$, and $dt(1/\log t-1/\log^2 t)=dy$, thus $dt\sim \log t\cdot dy\sim \log y\cdot dy$ for large $t$ (since $\log y=\log t-\log\log t\sim \log t$). Over any finite interval, the integral is uniformly bounded, thus we get $$f(x)\sim \int_e^\infty e^{-xy}\log y dy.$$ Here denote $xy=\tau$ to get $$ \int_e^\infty e^{-xy}\log y dy=x^{-1}\int_{ex}^\infty e^{-\tau}\log (\tau/x) d\tau=x^{-1}\left(\int_{ex}^\infty e^{-\tau}\log \tau d\tau-\log x \int_{ex}^\infty e^{-\tau}d\tau\right)\\ =-\frac{\log x+O(1)}x. $$

Therefore, $f(x)\sim -\frac{\log x}x$.

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Take any real $a>e$. Then \begin{equation*} f(x)=f_{1,a}(x)+f_{2,a}(x), \tag{1}\label{1} \end{equation*} where \begin{equation*} f_{1,a}(x):=\int_e^a dt\,e^{-xg(t)},\quad f_{2,a}(x):=\int_a^\infty dt\,e^{-xg(t)}, \tag{2}\label{2} \end{equation*} \begin{equation*} g(t):=\frac t{\ln t}. \tag{3}\label{3} \end{equation*} Note that \begin{equation*} 0\le f_{1,a}(x)\le a. \tag{4}\label{4} \end{equation*} The function $g\colon[e,\infty)\to[e,\infty)$ is continuous and increasing from $g(e)=e$ to $g(\infty-)=\infty$, with \begin{equation*} g'(t)=\frac{\ln t-1}{\ln^2 t}>0 \end{equation*} for $t\in(e,\infty)$. So, $g$ has the inverse $h:=g^{-1}$, which is continuous and increasing from $h(e)=e$ to $h(\infty-)=\infty$, and also $h$ is differentiable on $(e,\infty)$. So, making the substitution $s=g(t)\iff t=h(s)$, we have \begin{equation*} f_{2,a}(x)=\int_{g(a)}^\infty ds\,h'(s)e^{-xs}=\int_{g(a)}^\infty ds\,\frac{\ln^2 h(s)}{\ln h(s)-1}e^{-xs}. \tag{5}\label{5} \end{equation*} Note that $h(s)\sim s\ln s$ as $s\to\infty$. So, there exists a function $A$ from a right-neighborhood of $0$ to $(e,\infty)$ such that \begin{equation*} A(x)\to\infty,\quad A(x)=o\Big(\frac1x\ln\frac1x\Big),\quad xg(A(x))\to0, \tag{6}\label{6} \end{equation*} and \begin{equation*} \text{$h(s)\sim s\ln s$, and hence $\ln h(s)\sim\ln s$, } \\ \text{uniformly in $s\in[g(A(x)),\infty)$;}\tag{7}\label{7} \end{equation*} all limits here and in what follows are taken for $x\downarrow0$.

It follows by \eqref{5}, \eqref{6}, \eqref{7} that \begin{equation*} \begin{aligned} f_{2,A(x)}(x)&\sim\int_{g(A(x))}^\infty ds\,\ln s\,e^{-xs} \\ &=\int_{xg(A(x))}^\infty \frac{du}x\,\Big(\ln u+\ln\frac1x\Big)\,e^{-u} \\ &=\frac1x\,\int_{xg(A(x))}^\infty du\,\ln u\,e^{-u} +\frac1x\,\ln\frac1x\,\int_{xg(A(x))}^\infty du\,e^{-u} \\ &=\frac{1+o(1)}x\,\int_0^\infty du\,\ln u\,e^{-u} +\frac{1+o(1)}x\,\ln\frac1x\,\int_0^\infty du\,e^{-u} \\ &\sim\frac1x\,\ln\frac1x. \end{aligned} \tag{8}\label{8} \end{equation*} It also follows from \eqref{4} and \eqref{6} that \begin{equation} f_{1,A(x)}(x)=o\Big(\frac1x\,\ln\frac1x\Big). \tag{9}\label{9} \end{equation}

Finally, by \eqref{1}, \eqref{8}, \eqref{9}, \begin{equation} f(x)\sim\frac1x\,\ln\frac1x. \end{equation}

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    $\begingroup$ I have only noticed Fedor Petrov's answer after posting mine. The two answers are essentially the same, perhaps with more details in mine. So, I'll let it be. $\endgroup$ Jun 3, 2022 at 18:30
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THIS ANSWER IS INCORRECT, SEE THE COMMENTS.

According to Maple, $$ f(x) = -\frac{\gamma+\log x}{x} +\tfrac14 e^2 x - \tfrac19 e^3x^2 + \tfrac{1}{32}e^4x^3 - \tfrac{1}{150}e^5x^4 + O(x^5).$$

A convergent series is available. The exact value of the integral is $\frac{1}{x}(e^{-ex} + \mathrm{Ei}(1,ex))$ where Ei is the exponential integral. Using the known Taylor series of the exponential integral, we get $$f(x) = -\frac{\gamma+\log x}{x} + \sum_{j=2}^\infty \frac{(-1)^j(j-1)}{j\,j!} e^j x^{j-1}.$$

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  • $\begingroup$ neat result; is this the output of the asympt routine in Maple? $\endgroup$ Jun 4, 2022 at 15:42
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    $\begingroup$ @CarloBeenakker Maple evaluates the exact integral as $\frac{1}{x}(e^{-ex} + \mathrm{Ei}(1,ex))$ where Ei is the exponential integral. This can also be written in terms of the incomplete gamma function $\frac{1}{x}(e^{-ex} + \Gamma(0,ex))$. The "series" function knows how to expand both of those special functions. $\endgroup$ Jun 4, 2022 at 23:56
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    $\begingroup$ @CarloBeenakker Your question prompted me to look up the taylor expansion of the exponential integral, which is quite simple. $\endgroup$ Jun 5, 2022 at 0:24
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    $\begingroup$ Mathematica cannot evaluate $f(x)$ in closed form. However, the numerical values of $f(x)$ that it produces are different from the corresponding values of $F(x):=\frac1x\,(e^{-e x}+\Gamma(0,e x))$. For instance, Mathematica says that $f(1)=0.31\ldots$ and $F(1)=0.08\ldots$. In fact, since $g(t):=t/\ln t$ is increasing in $t>e$, we have $f(1)>\int_0^5 dt\,e^{-g(t)}>5e^{-g(5)}=0.22\ldots>0.08\ldots$. So, the expression $F(x)$ for $f(x)$ cannot be true. $\endgroup$ Jun 5, 2022 at 13:47
  • $\begingroup$ @IosifPinelis YIKES. You are correct. I used Fedor's first displayed equation without noticing that it is just an approximation as $x\to 0$. Back to the drawing board. Sorry for the confusion. $\endgroup$ Jun 5, 2022 at 14:00

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