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

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.

Thanks,

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\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 $0$.
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