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In this post, @reuns worked this out from the left hand side:

$$ \log{1\over s-1}=s\int_1^\infty\operatorname{li}(x)x^{-s-1}\mathrm dx $$

To reinforce my understanding, I tried attacking it from the right (where it is assumed that $\Re(s)>1$):

$$ \begin{aligned} s\int_1^\infty\operatorname{li}(x)x^{-s-1}\mathrm dx &=\int_1^\infty\operatorname{li}(x)\mathrm d(x^{-s}-1) \\ &=\operatorname{li}(x)(x^{-s}-1)|_1^\infty-\int_1^\infty{x^{-s}-1\over\log x}\mathrm dx \\ \end{aligned} $$

I currently have no knowledge about the growth of $\operatorname{li}(x)$ near $x=1$, so I jump to work on the second integral:

$$ \begin{aligned} -\int_1^\infty{x^{-s}-1\over\log x}\mathrm dx &=-\int_1^\infty\int_0^{-s}x^r\mathrm dr\mathrm dx \\ &=\int_{-s}^0\int_1^\infty x^r\mathrm dr\mathrm dx \\ &=\int_{-s}^0\left[x^{r+1}\over r+1\right]_0^\infty\mathrm dr \end{aligned} $$

I do not think I can continue from here because of convergence problem. I wonder whether it is possible to proceed here, or if it is possible with modifications on previous steps.

TravorLZH
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1 Answers1

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$sx^{-s-1}\,dx=d(1-x^{-s})$ (note the sign) doesn't help because $(1-x^{-s})\operatorname{li}(x)\underset{x\to\infty}{\longrightarrow}\infty$ as well.

Rather, if we define $f(s)=\int_1^\infty sx^{-s-1}\operatorname{li}(x)\,dx$ for $\Re s>1$, then $$f(s)-f(t)\underset{\big[\text{IBP}\big]}{=}\int_1^\infty\frac{x^{-s}-x^{-t}}{\log x}\,dx\underset{\big[x=e^y\big]}{=}\log\frac{t-1}{s-1}$$ using a Frullani integral, thus (as noted by @reuns) it suffices to show that $f(2)=0$.

Here is a possible way to do this. For $a>1$ we have \begin{align*} \operatorname{li}(a)&=\lim_{\epsilon\to 0^+}\left(\int_{1+\epsilon}^a\frac{dt}{\log t}+\int_0^{1-\epsilon}\frac{dt}{\log t}\right) \\&=\lim_{\epsilon\to 0^+}\left(\int_{1+\epsilon}^a\frac{dx}{\log x}-\int_{1/(1-\epsilon)}^\infty\frac{x^{-2}\,dx}{\log x}\right) \\&=\lim_{\epsilon\to 0^+}\left(\int_{1+\epsilon}^a\frac{1-x^{-2}}{\log x}\,dx+\int_{1+\epsilon}^{1/(1-\epsilon)}\frac{x^{-2}\,dx}{\log x}-\int_a^\infty\frac{x^{-2}\,dx}{\log x}\right) \\&=\int_1^a\frac{1-x^{-2}}{\log x}\,dx-\int_a^\infty\frac{x^{-2}\,dx}{\log x}, \end{align*} so that, integrating by parts (again), \begin{align*} f(2)&=\lim_{a\to 1^+}\int_a^\infty 2x^{-3}\operatorname{li}(x)\,dx \\&=\lim_{a\to 1^+}\left(a^{-2}\operatorname{li}(a)+\int_a^\infty\frac{x^{-2}\,dx}{\log x}\right) \\&=\lim_{a\to 1^+}\left((a^{-2}-1)\operatorname{li}(a)+\int_1^a\frac{1-x^{-2}}{\log x}\,dx\right). \end{align*}

And both terms tend to zero (use $\operatorname{li}(a)=\log(a-1)+O(1)$ as $a\to 1^+$).

metamorphy
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