Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm trying to give at least some partial answers for one of my own questions (this one). There the following arose:

$\hskip1.7in$ Why is $\lim_{x \to 0} {\rm li}(n^x)-{\rm li}(2^x)=\log\left(\frac{\log(n)}{\log(2)}\right)$?

Expanding at $x=0$ doesn't look reasonable to me since ${\rm li}(1)=-\infty$ and Wolfram only helps for concrete numbers, see here for example. Would a "$\infty-\infty$" version of L'Hospital work? Any help appreciated.


share|cite|improve this question
up vote 6 down vote accepted

$$ \begin{align} \lim_{x\to0}\int_{2^x}^{n^x}\frac{\mathrm{d}t}{\log(t)} &=\int_{x\log(2)}^{x\log(n)}\frac{e^u}{u}\mathrm{d}u\\ &=\lim_{x\to0}\left(\color{#C00000}{\int_{x\log(2)}^{x\log(n)}\frac{e^u-1}{u}\mathrm{d}u} +\color{#00A000}{\int_{x\log(2)}^{x\log(n)}\frac{1}{u}\mathrm{d}u}\right)\\ &=\color{#C00000}{0}+\lim_{x\to0}\big(\color{#00A000}{\log(x\log(n))-\log(x\log(2))}\big)\\ &=\log\left(\frac{\log(n)}{\log(2)}\right) \end{align} $$ Note Added: since $\lim\limits_{u\to0}\dfrac{e^u-1}{u}=1$, $\dfrac{e^u-1}{u}$ is bounded near $0$, therefore, its integral over an interval whose length tends to $0$, tends to $0$.

share|cite|improve this answer
thx again, would it also work for $x\in \Bbb C$? – draks ... Jan 23 '13 at 23:17
@draks...: I see no problem allowing $x\in\mathbb{C}$. – robjohn Jan 23 '13 at 23:20

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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