Could the circle method be used in the average order problem of the sum-of-divisors function?

Question

The problem of the average order of the sum-of-divisors function $\sigma(n)=\sum_{d|n} d$ is that of estimating the element $R(x)$ in the formula $S(x)=\sum_{n\leq x} \sigma(n)= \zeta(2)x^2/2+xR(x)$ ........(1).

There has been no improvement for some sixty years on Walfisz' result $R(x) =O(\log^{2/3}x)$. I have often wondered whether an approach via Hardy's circle method might be effective and if any great expert could advise on whether this is at all possible.

The initial moves at least, are straightforward. Thus if $F(z)= \sum_{n=1}^{\infty}\sigma(n)z^n$ with |z|<1 then $S(n) =\frac{1}{2\pi i}\int_{\gamma}\frac{F(z)dz}{(1-z)z^{n+1}}$ where $\gamma$ is a circle |z|=R<1. A Farey dissection of the contour $\gamma$ of order N=N(n) yields $S(n)=\frac{1}{2\pi i}\sum_{p,q}\int_{\xi_{p,q}}\frac{F(z)dz}{(1-z)z^{n+1}}$ where 1<=q<=N, (p,q)=1. If q=1, p=0 while if q>1 then 1<=p<q. If on $\xi_{p,q}$ we have $z=e_{q}(p)e^{-y}$ where $y=\frac{1}{n}+i\theta$ and $e_{q}(p)=e^{2\pi i p/q}$ then the previous integral is $I=I_{p,q}= \frac{1}{2\pi i}e_{q}(-np)\int_{\frac{1}{n}-i\theta_{p,q}}^{\frac{1}{n}+i\theta'_{p,q}}\frac{e^{ny}F_{p,q}(y)dy}{(1-e_{q}(p)e^{-y})}$ where $F_{p,q}(y)=F(e_{q}(p)e^{-y})$ and$\frac{\pi}{qN}\le \theta_{p,q},\theta'_{p,q}\le \frac{2\pi}{qN}$.

By the Mellin transform, $F_{p,q}(y)= \frac{1}{2\pi i}\int_{\alpha-i\infty}^{\alpha+i\infty}y^{-s}\Gamma(s)Z_{p,q}(s)ds$ where $ Z_{p,q}(s)=\sum_{n=1}^{\infty}\frac{e_{q}(np)\sigma(n)}{n^s}$ and initially $\alpha>2$. To evaluate $Z_{p,q}(s)$ we note that $\sigma(\omega^{n+1})=\sigma(\omega^{n})\sigma(\omega)-\omega\sigma(\omega^{n-1})$ for prime $\omega$. It can be shown that $\sum_{n=1}^{\infty} \frac{\chi_{i}(n)\sigma(n)}{n^{s}}=L(s,\chi_{i})L(s-1,\chi_{i})$ where the L-fns and characters-$\chi$ are to modulus q and so $\sum_{l=0}^{\infty}\frac{\sigma(lq+m)}{(lq+m)^{s}}=\frac{1}{\phi(q)}\sum_{i=1}^{\phi(q)}\overline{\chi_{i}}(m)\sum_{n=1}^{\infty}\frac{\chi_{i}(n)\sigma(n)}{n^{s}}$ as $\sum_{i=1}^{\phi(q)}\overline{\chi_{i}}(m)\chi_{i}(n) =\phi(q)$ if $m\equiv n$ mod q and 0 otherwise. Finally, as $\sum_{n=1}^{\infty}\frac{e_{q}(np)\sigma(n)}{n^s}=\sum_{m=1}^{q}e_{q}(mp) \sum_{l=0}^{\infty}\frac{\sigma(lq+m))}{(lq+m)^s}$ it follows that $Z_{p,q}(s)=\sum_{i=1}^{\phi(q)}C_{i}L(s,\chi_i)L(s-1,\chi_i)$ where $C_{i}=C_{i}(p,q)=\frac{1}{\phi(q)}\sum_{m=1}^{q} e_{q}(mp)\overline\chi_i(m)$, $C_{1}=\frac{\mu(q)}{\phi(q)},|C_{i}|\le \frac{q^{\frac{1}{2}}}{\phi(q)}$ for all p with (p,q)=1.

Moving the line in the Mellin transform left to $\alpha=\delta<1$ there are simple poles at s=2 and s=1 from the product of the two principal L-fns $L(s,\chi_{1})L(s-1,\chi_{1})=\Pi_{\omega|q}(1-\frac{1}{\omega^{s}})(1-\frac{1}{\omega^{s-1}})\zeta(s)\zeta(s-1)$ The apparent pole at s=1 only gives a non-zero residue when q=1. The residue at s=1 when q>1 vanishes as then $L(0,\chi_{1})=0$. (If we were to shift the line of integration as far left as $\alpha=\delta<0$ there would be a pole at s=0 from the gamma-function with residue $Z_{0,1}(0)$)

When q=1,$F_{0,1}(y)=\frac{\zeta(2)}{2y^{2}}-\frac{1}{2y}+\frac{1}{2\pi i}\int_{\delta-i\infty}^{\delta+i\infty}y^{-s}\Gamma(s)Z_{0,1}(s)ds$ as $\zeta(0)=-\frac{1}{2}$. Call the last integral $R_{0,1}(y)$ Thus the integral $I=I_{1}+I_{2}$ where $I_{1}= \frac{1}{2\pi i}\int_{\frac{1}{n}-i\theta_{0,1}}^{\frac{1}{n}+i\theta'_{0,1}}\frac{e^{ny}}{(1-e^{-y})}(\frac{\zeta(2)}{2y^2}-\frac{1}{2y})dy$ and $I_{2}= \frac{1}{2\pi i}\int_{\frac{1}{n}-i\theta_{0,1}}^{\frac{1}{n}+i\theta'_{0,1}}\frac{e^{ny}}{(1-e^{-y})}R_{0,1}(y)dy$ which should yield $\frac{1}{2}\zeta(2)n^{2} +O(n)+$ error of lower n-order from $I_{1}$+an error of unknown n-order from $I_{2}$

For general q>1 $L(0,\chi_{1})=0$ so $F_{p,q}(y)=\frac{\mu(q)}{\phi(q)}\Pi_{\omega|q}(1-\frac{1}{\omega})L(2,\chi_{1})y^{-2}+\frac{1}{2\pi i}\int_{\delta-i\infty}^{\delta+i\infty}y^{-s}\Gamma(s)Z_{p,q}(s)ds$. Call the last integral $R_{p,q}(y)$. (If we were to shift the line of integration as far left as $\alpha=\delta<0$ there would be a pole at s=0 from the gamma-function with residue $Z_{p,q}(0)$) As $ \Pi_{\omega|q}(1-\frac{1}{\omega}) = \frac{\phi(q)}{q}$ and $L(2,\chi_{1}) = \Pi_{\omega|q}(1-\frac{1}{\omega^{2}})\zeta(2)$ this reduces to $F_{p,q}(y)=\frac{\mu(q)}{q}\zeta(2)\Pi_{\omega|q}(1-\frac{1}{\omega^{2}})y^{-2}+R_{p,q}(y)$

The integral $I_{p,q}$ becomes $ \zeta(2)\frac{\mu(q)}{q}\Pi_{\omega|q}(1-\frac{1}{\omega^{2}})\frac{e_q(-np)}{2\pi i}\int_{\frac{1}{n}-i\theta_{p,q}}^{\frac{1}{n}+i\theta'_{p,q}}\frac{e^{ny}dy}{y^{2}(1-e_{q}(p)e^{-y})}+\frac{e_{q}(-np)}{2\pi i}\int_{\frac{1}{n}-i\theta_{p,q}}^{\frac{1}{n}+i\theta'_{p,q}}\frac{e^{ny}R_{p,q}(y)dy}{(1-e_{q}(p)e^{-y})}$ which should yield $\zeta(2)n\frac{\mu(q)}{q}e_{q}(-np)\Pi_{\omega|q}(1-\frac{1}{\omega^{2}})\frac{1}{(1-e_{q}(p))} +$ error of lower n-order+an error of unknown n-order from the second integral. The hope would be that summing these terms for $0\leq p\lt q $ with$(p,q)=1$ and $1\leq q \leq N$ would give an estimate of the original $R(n)$ in equation (1).

Gathering all the terms of order n, there is a total multiplier equal to $\zeta(2)\sum_{2\leq q \leq N}\sum_{(p,q)=1}\frac{e_{q}(-np)}{1-e_{q}(p)}\frac{\mu(q)}{q}\Pi_{\omega|q}(1-\frac{1}{\omega^{2}}) -\frac{1}{2}$ This is as far as I think I can get. The following facts might be of relevance in any attempt to estimate this sum. The inner p-sum $\sum_{(p,q)=1}\frac{e_{q}(-np)}{1-e_{q}(p)}$ can be shown to be $\sum_{r=0}^{n}c_{q}(r)-\frac{\phi(q)}{2}$ ($c_{q}(r)$ are Ramanujan's sums) or $\sum_{r=1}^{n}c_{q}(r)+\frac{\phi(q)}{2}$ as $c_{q}(0)=\phi(q)$. If M is an integer, any sum of the form $\sum_{Mq}^{(M+1)q-1}c_{q}(r)$ vanishes. Also $c_{q}(Mq+r) =c_{q}(r)$. Thus $\sum_{r=0}^{n}c_{q}(r)=\sum_{r=0}^{n-q[\frac{n}{q}]}c_{q}(r)$ When q is quadratfrei the Dirichlet series $\sum_{r=1}^{\infty}\frac{c_{q}(r)}{r^{s}}$ has sum $\mu(q)\zeta(s)\Pi_{\omega|q}(1-\frac{1}{\omega^{s-1}})$ for $Re (s)>1$, which tends to $\frac{\phi(q)}{2}$ as s tends to 1.

Answer 1

Prof. Tao's suggestion prompts the following. Notification of any fallacious step welcomed ! The 'singular series' in his case will, if the previous machinery is not in error, be $n\sum_{q=1}^{N}\frac{\mu(q)}{q}\sum_{(p,q)=1}e_{q}(-np)\Pi_{\omega|q}(1-\frac{1}{\omega^{2}})=n\sum_{q=1}^{N}\frac{\mu(q)}{q}c_{q}(n)\Pi_{\omega|q}(1-\frac{1}{\omega^{2}})$

$c_{\omega}(n) = \omega-1$ if $\omega|n$ and $c_{\omega}(n) = -1$ otherwise. $\sum_{q=1}^{\infty}\frac{\mu(q)}{q^{s}}c_{q}(n)\Pi_{\omega|q}(1-\frac{1}{\omega^{2}})=\Pi_{\omega|n}(1-\frac{(\omega-1)}{\omega^{s}}(1-\frac{1}{\omega^2}))\Pi_{\omega \dagger n}(1+\frac{1}{\omega^{s}}(1-\frac{1}{\omega^2}))$

This is $\Pi_{\omega|n}(1-\frac{(\omega-1)}{\omega^{s}}(1-\frac{1}{\omega^2}))/(1+\frac{1}{\omega^{s}}(1-\frac{1}{\omega^2}))\Pi_{\omega}(1+\frac{1}{\omega^{s}}(1-\frac{1}{\omega^2}))$ =R(s,n)F(s) where F(s) converges if Re(s)>1 and R(s,n) is a finite rational product. $F(s)= \Pi_{\omega}(1+\frac{1}{\omega^{s}}(1-\frac{1}{\omega^2}))/(1+\frac{1}{\omega^{s}})\Pi_{\omega}(1+\frac{1}{\omega^{s}})= \Pi_{\omega}(1+\frac{1}{\omega^{s}}(1-\frac{1}{\omega^2}))/(1+\frac{1}{\omega^{s}})\frac{\zeta(s)}{\zeta(2s)}=G(s)\frac{\zeta(s)}{\zeta(2s)}$

where G(s) is regular in Re(s)>0. A Perron-type argument will give $\sum_{q=1}^{N}\frac{\mu(q)}{q}c_{q}(n)\Pi_{\omega|q}(1-\frac{1}{\omega^{2}})= \frac{1}{2\pi i}\int_{c-i\infty}^{c +i\infty}\frac{N^{w}}{w}\frac{\zeta(w+1)}{\zeta(2w+2)}R(w+1,n)G(w+1)dw$

The residue at w=0 is $ R(1,n)G(1)\log N/\zeta(2)$

$R(1,n)= \Pi_{\omega|n}(1-\frac{(\omega-1)}{\omega}(1-\frac{1}{\omega^2}))/(1+\frac{1}{\omega}(1-\frac{1}{\omega^2}))$ is clearly <=1 in absolute value.

If not in error, this would seem to show that the circle method gives $\sigma(n)=O(n\log N)$. If N could be chosen to be as small as a power of $\log n$ the optimal result would follow. Of course I have no idea how large N must be to ensure the singular series term is dominant.