On the Mellin transform of logarithmic integral 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.
 A: $sx^{-s-1}\,dx=d(1-x^{-s})$ (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 $a\to 1^+$).
