1
$\begingroup$

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?

$\endgroup$
3
  • $\begingroup$ @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
2
$\begingroup$

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

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

$\endgroup$
3
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.