Limit involving Chebyshev function

Question

Let $\psi(x)=\sum_{n\leq x} \Lambda(n)$ be the weighted prime counting function/Chebyshev prime counting function. I am trying to evaluate the limit $$\lim_{n\to\infty}\frac{1}{n}\int_{1}^{\infty}\frac{\psi(x)-x}{x^{2}} L_{n-1}^{(2)}(\log x) \, \mathrm{d}x$$ where $L_{n}^{(\alpha)}(x)=\sum_{k=0}^n (-1)^k \binom{n+\alpha}{k+\alpha}\frac{x^k}{k!}$ is the Generalized Laguerre polynomial.

I tried the following code in Mathematica:

p[x_] := Sum[MangoldtLambda[n], {n, 2, x}]
f[x_] := 1/n*(p[x] - x)/x^2 * LaguerreL[n - 1, 2, Log [x]]
AsymptoticIntegrate[f[x], {x, 1, Infinity}, n -> Infinity] 

I have also asked the question on Mathematica.SE and the limit is numerically close to $-1$.

Any help would be highly appreciated.

Edit: Denote $$I_{n+1}=\int_{1}^{\infty}\frac{\psi(x)-x}{x^{2}} L_{n}^{(2)}(\log x) \, \mathrm{d}x$$ So we need to prove, $$\lim_{n\to\infty} \frac{I_{n+1}}{n+1}=-1$$ Using the generating function of Generalized Laguerre polynomial we have $$\sum_{n=0}^\infty L_{n}^{(2)}(t) z^{n}=\frac{e^{-\frac{z}{1-z}x}}{(1-z)^{3}},\ \ \text{where}\ \ |z|<1$$ We define $$S(z)=\sum_{n=1}^\infty I_{n+1} z^{n} \ \ \text {where}\ \ \left|z-\frac{1}{2}\right|<\frac{1}{2}$$ Using Dominated convergence theorem for $\{z\in\mathbb{C}:\left|z-\frac{1}{2}\right|<\frac{1}{2}\}$, $$S(z)=\frac{1}{(1-z)^3}\int_{1}^{\infty}\frac{\psi(x)-x}{x^{2+\frac{z}{1-z}}} \, \mathrm{d}x$$ This can be written as $$S(z)=\frac{1}{(1-z)^3}\int_{1}^{\infty}\frac{\psi(x)-x}{x^{1+\frac{1}{1-z}}} \, \mathrm{d}x$$ By Mellin transform of Chebyshev function $$\int_1^\infty \frac{\psi(x)}{x^{s+1}}\ dx=\frac{-1}{s}\frac{\zeta'(s)}{\zeta(s)}\ \ \text{where}\ \ \Re(s)>1$$ So that for $s=\frac{1}{1-z}$, $\Re (s)>1$ gives $|z-\frac{1}{2}|<\frac{1}{2}$ and we obtain $$S(z)=-\frac{1}{z(1-z)^2}-\frac{1}{(1-z)^2}\frac{\zeta'\left(\frac{1}{1-z}\right)}{\zeta\left(\frac{1}{1-z}\right)}$$ Now let $$S(z)=A(z)+B(z),$$ $$A(z)= -\frac{1}{z(1-z)^2}$$ $$B(z)=-\frac{1}{(1-z)^2}\frac{\zeta'\left(\frac{1}{1-z}\right)}{\zeta\left(\frac{1}{1-z}\right)}$$ Let, [$z^n$] $f(z)$ be the coefficient of $z^n$ in $f(z)$. Then $$I_{n+1}=[z^n] S(z)$$ We have $$[z^n]A(z)=-(n+2)$$ Hence we get $$I_{n+1}=-(n+2)+[z^n]B(z)$$ So we obtain $$\lim_{n\to\infty} \frac{I_{n+1}}{n+1}=-1+\lim_{n\to\infty}\frac{[z^n]B(z)}{n+1}$$ How do we find the remaining limit?

Answer 1

It seems that analytically, some terms are controlled by the same singularities that encode the nontrivial zeros of $\zeta(s)$, making this a Riemann-Hypothesis dependant statement. A solution to this limit would conversely imply positively, or negatively, the RH. To show this let us start with the fact that we have $$ I_{n+1}=-(n+2)+b_n $$ with $b_n=[z^n]B(z)$. Divide by $n+1$ to get $$ \frac{I_{n+1}}{n+1}=-\frac{n+2}{n+1}+\frac{b_n}{n+1} =-1-\frac{1}{n+1}+\frac{b_n}{n+1} $$ Assume $$ \lim_{n\to\infty}\frac{I_{n+1}}{n+1}=-1 $$ Taking limits in the identity above and using $\lim_{n\to\infty}\frac{1}{n+1}=0$ gives $$ -1=\lim_{n\to\infty}\frac{I_{n+1}}{n+1} =-1+ \lim_{n\to\infty}\frac{b_n}{n+1} $$ hence $$ \lim_{n\to\infty}\frac{b_n}{n+1}=0. $$

It is obvious that $\zeta'(s)/\zeta(s)$ has poles precisely at zeros $\rho$ of $\zeta(s)$. Under the change of variables $$ s=\frac1{1-z} $$ a zero $\rho$ corresponds to $$ z_\rho=1-\frac1\rho=\frac{\rho-1}{\rho} $$ and $B(z)$ has a pole at $z=z_\rho$. Writing $\rho=\beta+i\gamma$, we have $$ |z_\rho|^2=\left|\frac{\rho-1}{\rho}\right|^2=\frac{(\beta-1)^2+\gamma^2}{\beta^2+\gamma^2} $$ so $|z_\rho|<1$ iff $\beta>\tfrac12$. Hence, if there exists a zero $\rho$ with $\Re(\rho)>\tfrac12$, then $B(z)$ has a singularity in $|z|<1$, and therefore the Taylor series $B(z)=\sum_{n\ge0}b_n z^n$ has radius of convergence $R<1$. By Cauchy-Hadamard one has, $$ \limsup_{n\to\infty}|b_n|^{1/n}=\frac1R>1 $$ so $b_n$ cannot satisfy $b_n=o(n)$, therefore a contradiction. Therefore there are no zeros with $\Re(\rho)>\tfrac12$. By the functional equation symmetry $\rho\mapsto 1-\overline{\rho}$, there are also no zeros with $\Re(\rho)<\tfrac12$. Hence every nontrivial zero satisfies $\Re(\rho)=\tfrac12$, implying the RH holds if the limit is $-1$

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For interest:

$$ I_{n} = -\sum_{\rho}\frac{1}{\rho}\left(n+\rho-\frac{\rho^{n+1}}{(\rho-1)^n}\right)-n\log(2\pi) +\frac12\sum_{k\ge1}\frac{1}{k}\left(n-2k-\frac{(-2k)^{n+1}}{(-2k-1)^n}\right) $$