# Asymptotic expansion of $\int_0^x \exp\big( \frac{1}{\log(t)} \big)~dt$

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?

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

where

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

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*}

where

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

• 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 ? Dec 14 '20 at 5:57
• @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 $x \to 1^-$. Dec 14 '20 at 6:22
• Thank you for your answer. Cheers :-) Dec 14 '20 at 6:23