Question: What is the definition of $Li(x)$ in the MathOverflow questions linked below and their related answers? $Li(x)=\int\limits_0^x\frac{1}{\log(t)}\,dt$, $Li(x)=\int\limits_1^x\frac{1}{\log(t)}\,dt$, $Li(x)=\int\limits_2^x\frac{1}{\log(t)}\,dt$, or $Li(x)=\int\limits_0^{1-\epsilon}\frac{1}{\log(t)}\,dt+\int\limits_{1+\epsilon}^x\frac{1}{\log(t)}\,dt$?

(a) On the integral $I_s =\int_{1}^{\infty} (\pi(x)-Li(x))x^{-s-1} dx$

(b) On the integral $I_s = \int_{1}^{\infty} (\pi(x)-Li(x))x^{-s-1} \mathrm{d}x$-follow up question

(c) Error bounds for $\pi(x)-Li(x)$ and convergence of the associated Dirichlet integral

The related questions, answers, and comments to the following Math StackExhange questions seem to use various definitions. My motivation for asking this question is an attempt to understand the exact formula associated with the evaluation of $I_s$ in order to gain insight into the questions, answers, and comments associated with these questions.

(d) A probably wrong proof of the Riemann Hypothesis, but where is the mistake?

(e) Possible progress on the Riemann hypothesis?

The remainder of this question is an attempt to further clarify my motivation for asking this question and is based on the definitions in formulas (1) to (3) below where $P(s)$ is the prime zeta function.

(1) $\quad \pi(x)=\sum\limits_{p\le x}1$

(2) $\quad P(s)=s\int\limits_1^\infty \pi(x)\,x^{-s-1}\,ds=\sum\limits_p\frac{1}{p^s}\,,\quad\Re(s)>1$

(3) $\quad P(s)=\sum\limits_{m=1}^\infty\frac{\mu(m)}{m}\log\zeta(m\,s)\,,\qquad\qquad\Re(s)>0$

The questions at the links (d) and (e) above use the definition in (4) below, the claimed identity in (5) below, and the equivalence in (6) below to derive the claimed equality in (7) below.

(4) $\quad Li(x)=\lim_{\epsilon\rightarrow 0^{+}}\Big(\int_{0}^{1-\epsilon}+\int_{1+\epsilon}^{x}\Big)\frac{dt}{\log t}$

(5) $\quad s\int_{1}^\infty Li(x)\,x^{-s-1}\,dx=-\log(s-1)\,,\quad\Re(s)>1$

(6) $\quad\sum\limits_p\frac{1}{p^s}=\sum\limits_{m=1}^\infty\frac{\mu(m)}{m}\log\zeta(m\,s)\,,\quad\Re(s)>1$

(7) $\quad s\int_{1}^\infty (\pi(x)-Li(x))\,x^{-s-1}\,dx-\log((s-1)\,\zeta(s))=\sum\limits_{m=2}^{\infty}\frac{\mu(m)}{m}\log\zeta(m\,s)\,,\quad Re(s)>1$

The questions at links (d) and (e) above initially state the claimed equality illustrated in (7) above is valid for $Re(s)>1$, but then claim the equality can be extended via analytic continuation to $Re(s)>\Theta$ where $\Theta$ is the supremum of the real parts of the zeros of $\zeta(s)$.

Assuming correctness of the claimed identity in (5) above, evaluation of the integral in (7) above leads to (8) below which is valid for $\Re(s)>1$, but I don't understand the claim with respect to analytic continuation to $Re(s)>\Theta$. The equivalence illustrated in (6) above can be used to rewrite (8) below as (9) below which leads to (10) below which is valid for $\Re(s)>0$, but this isn't useful.

(8) $\quad \sum\limits_p\frac{1}{p^s}+\log(s-1)-\log((s-1)\,\zeta(s))=\sum\limits_{m=2}^{\infty}\frac{\mu(m)}{m}\log\zeta(m\,s)\,,\quad Re(s)>1$

(9) $\quad \sum\limits_{m=1}^{\infty}\frac{\mu(m)}{m}\log\zeta(m\,s)+\log(s-1)-\log((s-1)\,\zeta(s))=\sum\limits_{m=2}^{\infty}\frac{\mu(m)}{m}\log\zeta(m\,s)\,,\\$ $\quad Re(s)>1$

(10) $\quad \sum\limits_{m=2}^{\infty}\frac{\mu(m)}{m}\log\zeta(m\,s)=\sum\limits_{m=2}^{\infty}\frac{\mu(m)}{m}\log\zeta(m\,s)\,,\\$ $\quad Re(s)>0$

The following figure illustrates the right-hand side of (8) above (which is strictly real in the interval $\frac{1}{2}<s<1$) in blue and the real and imaginary parts of the left-hand side of (8) above in orange and green respectively where the left-hand side sum is taken over the first $10,000$ primes. The figure below clearly illustrates the fallacy of the equivalence of the left and right hand sides of (8) above for $\Theta<Re(s)<1$.

Illustration of right and left hand sides of formula (8)

Figure (1): Illustration of RHS of (8) (blue) and LHS of (8) ($\Re$ orange and $\Im$ green)

  • $\begingroup$ en.wikipedia.org/wiki/Logarithmic_integral_function $\endgroup$ – Wojowu Sep 14 '19 at 15:51
  • 1
    $\begingroup$ All your proposed definitions are wrong... $\endgroup$ – Unit Sep 14 '19 at 15:56
  • $\begingroup$ @Unit Thanks for point out the error. I corrected the upper limits from $\infty$ to $x$ in the integrals related to the definition of $Li(x)$ above. $\endgroup$ – Steven Clark Sep 14 '19 at 16:14
  • 3
    $\begingroup$ there is a standard definition (see Wikipedia) and then changing lower bounds makes it differ by a constant which is utterly irrelevant when one does Mellin transforms as in the post since differences are entire functions, while convergence issues are always at infinity; it is more interesting when one does the complex Li - usually the integral definition in Edwards book on RZ is best to work as it excludes the negative reals and in particular it doesn't affect non-trivial RZ zeroes $\endgroup$ – Conrad Sep 14 '19 at 16:18
  • 2
    $\begingroup$ The difference between definitions is the mellin transform of a constant so kind of trivial; trying to actually estimate the integral, so involving $Li(x^\rho)$ for non trivial zeta roots is what you want $\endgroup$ – Conrad Sep 14 '19 at 16:47

$$li(x)=pv(\int_0^x \frac1{\log(t)}dt), \qquad x > 0$$

The problem is much less the definition of $Li$ than the possible Mellin transforms. There are mostly 3 things to try :

  • $\int_0^\infty li(x)x^{-s-1}dx$ never converges.

  • $\int_0^\infty 1_{x>2}(li(x)-li(2))x^{-s-1}dx$ approximates $\frac{-\log(s-1)}{s}$ very well around $s=1$ and since $1_{x>2}(li(x)-li(2))$ is continuous and piecewise $C^1$ its Mellin transform is $L^1$ on vertical lines.

  • Let $f(x)= li(x)1_{x > 1}\in L^1_{loc}$, due to the huge discontinuity at $x=1$ its distributional derivative is complicated, but if $\phi \in C^\infty_c(0,\infty),\phi(1)=0$ then $$<f',\phi>\ =\ \int_1^\infty \frac{\phi(x)}{\log x}dx \quad \implies \quad f' \log x = 1_{x > 1}$$ Which means $$F(s)=\int_0^\infty f(x)x^{-s-1}dx,\qquad sF(s)=\int_0^\infty f'(x)x^{-s}dx$$ $$-(sF(s))' =\int_0^\infty f'(x)\log(x)x^{-s}dx=\int_0^\infty 1_{x > 1}x^{-s}dx= \frac1{s-1}$$ $$sF(s)=-\log (s-1)+A$$

  • Finding $A$ is a pain : around $x=1$ we have $li(x) = B+log |x-1|+O(x-1)$ where the $O(x-1)$ term is $C^1$ thus as $s \to +\infty$

$$F(s) = \int_1^\infty log |x-1|x^{-s-1}dx+\frac{B}s+o(\frac{1}{s})$$

$$\int_1^\infty log |x-1|x^{-s-1}dx=\frac1s\int_1^\infty \frac{x^{-s}-1}{x-1}dx=\frac1s\int_1^\infty \frac{x^{-s-1}-x^{-1}}{1-x^{-1}}dx$$ $$= \frac1s \sum_{m=0}^\infty \int_1^\infty (x^{-s-1-m}-x^{-1-m})dx= \frac1s \sum_{m=0}^\infty(\frac{1}{s+m}-\frac{1}{1+m})= \frac{-\psi(s)-\gamma}{s}$$ And hence from the asymptotic expansion of $\psi$ and $B=\int_0^1 (\frac1{\log (t)}-\frac1{t-1})dt=\gamma $ we obtain $ A=0$ and $$\int_1^\infty li(x)x^{-s-1}dx=F(s) = \frac{-\log(s-1)}{s}$$

  • $\begingroup$ You seem to have confirmed the claimed identity in formula (5) of my question above which is valid for $\Re(s)>1$. The question at link (e) above originally used the definition $Li(x)=\int\limits_1^x\frac{1}{\log(t)}\,dt$, but the definition was later revised to $Li(x)=\lim_{\epsilon\rightarrow 0^{+}}\Big(\int_{0}^{1-\epsilon}+\int_{1+\epsilon}^{x}\Big)\frac{dt}{\log t}$ which is consistent with your derivation. $\endgroup$ – Steven Clark Sep 15 '19 at 16:30
  • $\begingroup$ In the comments associated with the question at link (e) above (which have since been moved to chat), there was an argument made based on the answers to the question at link (a) above. I wasn't sure if the answers to the question at link (a) assumed the same definition of $Li(x)$, but the comments on my question above seem to indicate it really doesn't matter. $\endgroup$ – Steven Clark Sep 15 '19 at 16:30

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.