What is the asymptotic expansion of $\int_0^x \exp\big( \frac{1}{\log(t)} \big)~dt$? As $x\to 1$.

I took the series expansion of the integrand and got $e^{\frac{1}{\log(x)}}\approx 1+\frac{1}{\log(x)}+\frac{1}{2\log(x)}+ \cdot\cdot\cdot$

and then I integrated term by term...

I got...

$\int_0^x \exp\big( \frac{1}{\log(t)} \big)~dt \approx x+li(x)+\frac{1}{2}\big(li(x)-\frac{x}{\log(x)}\big)+ \cdot\cdot\cdot$

where $li(x)$ is the logarithmic integral.

How right is this?

  • $\begingroup$ @SangchulLee looking at the asymptotic expansion as $x \to 1$ $\endgroup$
    – geocalc33
    Dec 13 '20 at 23:21
  • $\begingroup$ why you do not add this fundamental clarification in your post !? $\endgroup$
    – G Cab
    Dec 13 '20 at 23:24
  • $\begingroup$ @GCab I've updated the post accordingly. Thanks for your support! $\endgroup$
    – geocalc33
    Dec 13 '20 at 23:43

Write $\varphi(x) = \exp(1/\log x)$ and note that $\varphi^{-1} = \varphi$. So by substituting $y = \varphi(t)$, or equivalently $t = \varphi(y)$,

\begin{align*} I(x) := \int_{0}^{x} \exp\left(\frac{1}{\log t}\right) \, \mathrm{d}y &= \int_{1}^{\varphi(x)} y \varphi'(y) \, \mathrm{d}y \\ &= \left[ y \varphi(y) \right]_{1}^{\varphi(x)} + \int_{\varphi(x)}^{1} \varphi(y) \, \mathrm{d}y \\ &= c + x \varphi(x) - \int_{0}^{\varphi(x)} \varphi(y) \, \mathrm{d}y, \end{align*}


$$c := \int_{0}^{1} \varphi(y) \, \mathrm{d}y \approx 0.27973176.$$

Using the expansion $ \varphi(y) = \sum_{n=0}^{\infty} \frac{1}{n!(\log y)^n} $ which converges uniformly for $0 < y \leq \varphi(x)$, we can perform term-wise integration, obtaining

\begin{align*} I(x) &= c + x \varphi(x) - \sum_{n=0}^{\infty} \int_{0}^{\varphi(x)} \frac{1}{n!(\log y)^n} \, \mathrm{d}y \\ &= c - (1-x) \varphi(x) - \sum_{n=0}^{\infty} \frac{1}{(n+1)!} \int_{0}^{\varphi(x)} \frac{\mathrm{d}y}{(\log y)^{n+1}} \end{align*}

Now by using the integral formula

$$ \int \frac{\mathrm{d}y}{(\log y)^{n+1}} = \frac{1}{n!} \left( \operatorname{li}(y) - \sum_{k=1}^{n} \frac{(k-1)! y}{(\log y)^k} \right), $$

where $\operatorname{li}(x) = \int_{0}^{x} \frac{\mathrm{d}t}{\log t}$ is the logarithmic integral function, we get

\begin{align*} I(x) &= c - (1-x) \varphi(x) - \sum_{n=0}^{\infty} \frac{1}{n!(n+1)!} \left( \operatorname{li}(\varphi(x)) - \varphi(x) \sum_{k=1}^{n} (k-1)! (\log x)^k \right) \\ &= \boxed{ c - a_0 \operatorname{li}(\varphi(x)) + \varphi(x) \sum_{k=1}^{\infty} (k-1)! a_k (\log x)^k }, \tag{*} \end{align*}


$$ a_k := \sum_{n=0}^{\infty} \frac{1}{(n+k)!(n+k+1)!}. $$

  • $\begingroup$ Why not to use $\int \frac{\mathrm{d}y}{(\log y)^{n+1}}=(-1)^{n+1} \Gamma (-n,-\log (y))$ ? Would this make a problem ? $\endgroup$ Dec 14 '20 at 5:57
  • $\begingroup$ @ClaudeLeibovici, I have no doubt that using the incomplete gamma function gives an identity, although it is not quite suited for studying the asymptotic behavior. $\text{(*)}$ is more suited for this purpose, since truncating the sum at $k=N$ gives $$I(x)=c - a_0 \operatorname{li}(\varphi(x)) + \varphi(x) \sum_{k=1}^{N} (k-1)! a_k (\log x)^k+\mathcal{O}\bigl(\varphi(x)(\log x)^{N+1}\bigr)$$ as $x \to 1^-$. $\endgroup$ Dec 14 '20 at 6:22
  • 1
    $\begingroup$ Thank you for your answer. Cheers :-) $\endgroup$ Dec 14 '20 at 6:23

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