The main term when counting $d(n)$ in arithmetic progressions

Question

For $q,a\in \mathbb N$ write $d=(q,a)$. Why might be $$\sum _{h|q}\frac {c_h(a)\log (d/h)}{h}=-\frac {q'}{\phi (q')}\sum _{h|q}\frac {c_h(a)}{h}\sum _{h'|q'}\frac {\mu (h')\log h'}{h'}+\sum _{h|d}\frac {\phi (q/h)}{q/h}\log h?$$ I can see that this is true for $d=q$ or $d=1$ but in general, don't know.

My motivation for the question is quite long - if you really want it see below. But my question as it stands is just that above.

Here $c_h(a)$ is Ramanujan's sum \[ c_h(a)=\sum _{n=1\atop {(n,h)=1}}^he^{2\pi ina/h}=\frac {\mu (h/(h,a))\phi (h)}{\phi (h/(h,a))}.\]

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By https://www.impan.pl/en/publishing-house/journals-and-series/acta-arithmetica/all/168/4/82214/the-divisor-function-on-residue-classes-i (stated in the abstract, no access necessary) we have $$\sum _{n\leq X\atop {n\equiv a(q)}}d(n)\approx \frac {X}{q}\sum _{h|q}\frac {c_h(a)}{h}\Big (\log (X/h^2)+2\gamma -1\Big ).$$ On the other hand the main term is provided by a residue which I think should be, writing $d=(q,a)$ and $q'=q/d$, $$\sum _{n\leq X\atop {n\equiv a(q)}}d(n)\approx \frac {1}{\phi (q')}Res\left \{ \frac {(X/d)^{s}}{s}\sum _{n=1\atop {(n,q')=1}}^\infty \frac {d(nd)}{n^s}\right \} =\frac {1}{\phi (q')}Res\left \{ \frac {X^{s}}{s}\sum _{n=1\atop {(n,q)=d}}^\infty \frac {d(n)}{n^s}\right \} =:\frac {\mathcal R}{\phi (q')}$$ and as a plausability check it's possible to compare with Theorem 1 of http://matwbn.icm.edu.pl/ksiazki/aa/aa47/aa4713.pdf (there it's stated for $d_3(n)$ but I'm hoping it should also be the case for $d(n)$). So these two main terms should be equal, i.e. we should have $$\frac {\mathcal R}{\phi (q')}=\frac {X}{q}\sum _{h|q}\frac {c_h(a)}{h}\Big (\log (X/h^2)+2\gamma -1\Big )\hspace {10mm}(1)$$ and my issue is I can't see this. Here's my reasoning: As $$\sum _{n=1\atop {(n,q)=1}}^\infty \frac {1}{n^s}=\prod _{p\not |q}\left (1-p^{-s}\right )^{-1}=\zeta (s)\prod _{p|q}(1-p^{-s})=\zeta (s)\sum _{h|q}\frac {\mu (h)}{h^s}$$ the series in $\mathcal R$ is $$\sum _{n,m=1\atop {(nm,q)=d}}^\infty \frac {1}{(nm)^s}=\sum _{\delta |d}\sum _{n,m=1\atop {d|nm\atop {(nm,q)=d\atop {(n,d)=\delta }}}}^\infty \frac {1}{(nm)^s}=\sum _{\delta |d}\frac {1}{\delta ^s}\sum _{n,m=1\atop {d/\delta |m\atop {(n\delta m,q)=d\atop {(n,d/\delta )=1}}}}^\infty \frac {1}{(nm)^s}=\frac {1}{d^s}\sum _{\delta |d}\sum _{n,m=1\atop {(ndm,q)=d\atop {(n,d/\delta )=1}}}^\infty \frac {1}{(nm)^s}=\frac {1}{d^s}\left (\sum _{m=1\atop {(m,q')=1}}^\infty \frac {1}{m^s}\right )\left (\sum _{\delta |d}\sum _{n=1\atop {(n,q/\delta )=1}}^\infty \frac {1}{n^s}\right )=\frac {\zeta (s)^2}{d^s}\left (\sum _{h|q'}\frac {\mu (h)}{h^s}\right )\left (\sum _{\delta |d\atop {h|q/\delta }}\frac {\mu (h)}{h^s}\right )=:\frac {\zeta (s)^2}{d^s}\Delta (s)$$ so \[ \mathcal R=Res_{s=1}\Bigg \{ \frac {(X/d)^{s}}{s}\zeta (s)^2\Delta (s)\Bigg \} =:Res\Bigg \{ \lambda (s)\zeta (s)^2\Delta (s)\Bigg \} .\] And here I get stuck - I don't see how the $\log $'s in the residue can reduce to something so simple as in (1). Below is calculating the residue explicitly.

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Here's my work in detail which gets it to the point above I'm stuck at. We have \begin{align} \mathcal R&=\lim _{s\rightarrow 1}\frac {d}{ds}\lambda (s)(s-1)^2\zeta (s)^2\Delta (s) \\ &=\lim _{s\rightarrow 1}\left (\Delta (s)\frac {d}{ds}\lambda (s)(s-1)^2\zeta (s)^2+\lambda (s)(s-1)^2\zeta (s)^2\frac {d}{ds}\Delta (s)\right ) \\ &=\Delta (1)\lim _{s\rightarrow 1}\left (\lambda (s)\frac {d}{ds}(s-1)^2\zeta (s)^2+(s-1)^2\zeta (s)^2\frac {d}{ds}\lambda (s)\right )+\lambda (1)\Delta '(1) \\ &=\Delta (1)\Big (2\gamma \lambda (1)+\lambda '(1)\Big )+\lambda (1)\Delta '(1) \\ &=\frac {X}{d}\Bigg (\left (\log (X/d)+2\gamma -1\right )\Delta (1)+\Delta '(1)\Bigg ). \end{align} Then \[ \sum _{\delta h|q\atop {\delta |d}}\frac {\mu (h)}{h}=\sum _{\delta |h|q\atop {\delta |d}}\frac {\mu (h/\delta )}{h/\delta }=\sum _{h|q}\frac {1}{h}\sum _{\delta |h,d}\mu (h/\delta )\delta =\sum _{h|q}\frac {c_h(a)}{h}\] and similarly \[ \sum _{\delta h|q\atop {\delta |d}}\frac {\mu (h)\log h}{h}=\sum _{h|q}\frac {1}{h}\sum _{\delta |h,d}\mu (h/\delta )\delta \log (h/\delta )=\sum _{h|q}\frac {c_h(a)\log h}{h}-\sum _{\delta |d\atop {\delta h|q}}\frac {\mu (h)\log \delta }{h}=\sum _{h|q}\frac {c_h(a)\log h}{h}-\sum _{\delta |d}\frac {\phi (q/\delta )}{q/\delta }\log \delta \] so \[ \Delta (1)=\frac {\phi (q')}{q'}\sum _{\delta h|q\atop {\delta |d}}\frac {\mu (h)}{h}=\frac {\phi (q')}{q'}\sum _{h|q}\frac {c_h(a)}{h}\] and \begin{align} \Delta '(1)&=\frac {d}{ds}\vline _{s=1}\left (\sum _{\delta |d\atop {h|q/\delta \atop {h'|q'}}}\frac {\mu (h)\mu (h')}{(hh')^s}\right ) \\ &=-\left (\sum _{\delta |d\atop {h|q/\delta }}\frac {\mu (h)\log h}{h}\right )\left (\sum _{h'|q'}\frac {\mu (h')}{h'}\right )-\left (\sum _{\delta |d\atop {h|q/\delta }}\frac {\mu (h)}{h}\right )\left (\sum _{h'|q'}\frac {\mu (h')\log h'}{h'}\right ) \\ &=-\frac {\phi (q')}{q'}\left (\sum _{h|q}\frac {c_h(a)\log h}{h}-\sum _{\delta |d}\frac {\phi (q/\delta )}{q/\delta }\log \delta \right )-\left (\sum _{h|q}\frac {c_h(a)}{h}\right )\left (\sum _{h'|q'}\frac {\mu (h')\log h'}{h'}\right ) \end{align} so \begin{align} \mathcal R=\frac {X}{d}\sum _{h|q}\frac {c_h(a)}{h}\left (\left (\log (X/d)+2\gamma -1\right )\frac {\phi (q')}{q'}-\frac {\phi (q')\log h}{q'}-\sum _{h'|q'}\frac {\mu (h')\log h'}{h'}\right ) \\ +\frac {X\phi (q')}{dq'}\sum _{\delta |d}\frac {\phi (q/\delta )}{q/\delta }\log \delta \end{align} so, comparing with (1) we should have \begin{align} \sum _{h|q}\frac {c_h(a)}{h}\Big (\log (X/h^2)+2\gamma -1\Big )&=\sum _{h|q}\frac {c_h(a)}{h} \\ &\left (\left (\log (X/d)+2\gamma -1\right )-\log h-\frac {q'}{\phi (q')}\left (\sum _{h'|q'}\frac {\mu (h')\log h'}{h'}\right )\right ) \\ &+\sum _{\delta |d}\frac {\phi (q/\delta )}{q/\delta }\log \delta \end{align} and therefore $$\sum _{h|q}\frac {c_h(a)\log (d/h)}{h}=-\frac {q'}{\phi (q')}\sum _{h|q}\frac {c_h(a)}{h}\sum _{h'|q'}\frac {\mu (h')\log h'}{h'}+\sum _{\delta |d}\frac {\phi (q/\delta )}{q/\delta }\log \delta $$ and that's the identity I want to establish at the start of this post.

Answer 1

I feel I'm answering my own questions too much here but I guess I should post the answer if I posted the question and found the answer in the meantime.

Write $\Delta (n)=\phi (n)/n$ and write $\Delta =1-1/p$. Let $$R(q,d)=\sum _{h|q}\frac {c_h(d)}{h}\hspace {6mm}\text { and }\hspace {6mm}S(q,d)=\sum _{h|d}\Delta (q'h).$$ Then $$\sum _{h|q}\frac {c_h(d)\log h}{h}=R(q,d)\left (-\sum _{p|q'}\frac {\log p}{p\Delta }+\sum _{p^\alpha |d\atop {p|q'}}\frac {\log p^\alpha }{1+D}+\sum _{p^\alpha |d\atop {p\nmid q'}}\frac {\log p^\alpha \Delta }{1+\Delta D}\right )$$ and $$\sum _{h|d}\Delta (q'h)\log h=R(q,d)\left (\sum _{p^\alpha |d,q'}\frac {\log p^\alpha }{1+D}+\sum _{p^\alpha |d\atop {p\nmid q'}}\frac {\log p^\alpha \Delta }{1+\Delta D}\right ).$$ The claim in my question then follows from these claims and this question How to prove $\sum_{d\mid q}\frac{\mu(d)\log d}{d}=-\frac{\phi(q)}{q}\sum_{p\mid q}\frac{\log p}{p-1}$? Proof The first sum in question is \begin{align} \sum _{p^\alpha |q}\frac {\log p^\alpha c_{p^\alpha }(d)}{p^\alpha }\sum _{h|q\atop {p\nmid h}}\frac {c_h(d)}{h}=:\sum _{p^\alpha |q}\frac {\log p^\alpha c_{p^\alpha }(d)}{p^\alpha }R_p(q,d). \end{align} Then \begin{align} R_p(q,d)&=&\prod _{P|q\atop {P\nmid d\atop {P\not =p}}}\sum _{h|q}\frac {c_h(d)}{h}\prod _{P|d,q'\atop {P\not =p}}\sum _{h|q}\frac {c_h(d)}{h}\prod _{P|d\atop {P\nmid q'\atop {P\not =p}}}\sum _{h|q}\frac {c_h(d)}{h} \\ &=&\prod _{P|q\atop {P\nmid d\atop {P\not =p}}}\Delta \prod _{P|d,q'\atop {P\not =p}}\left (\Delta (1+D)\right )\prod _{P|d\atop {P\nmid q'\atop {P\not =p}}}\left (1+\Delta D\right ) \\ &=&R(q,d)\underbrace {\Delta ^{-1}}_{p|q'}\underbrace {(1+D)^{-1}}_{p|d,q'}\underbrace {(1+\Delta D)^{-1}}_{p|d\atop {p\not |q'}} \end{align} so our sum has the correct factor and the rest becomes \begin{align} &&\sum _{p^\alpha |q}\frac {\log p^\alpha c_{p^\alpha }(d)}{p^\alpha }\underbrace {\Delta ^{-1}}_{p|q'}\underbrace {(1+D)^{-1}}_{p|d,q'}\underbrace {(1+\Delta D)^{-1}}_{p|d\atop {p\not |q'}} \\ &&\hspace {10mm}=\hspace {4mm}\sum _{p^\alpha |q\atop {p\not |d}}\frac {\log p^\alpha \mu (p^\alpha )}{p^\alpha \Delta }+\sum _{p^\alpha |q\atop {p|d,q'}}\frac {\log p^\alpha c_{p^\alpha }(d)}{p^\alpha \Delta (1+D)}+\sum _{p^\alpha |q\atop {p|d\atop {p\not |q'}}}\frac {\log p^\alpha c_{p^\alpha }(d)}{p^\alpha (1+\Delta D)} \\ &&\hspace {10mm}=\hspace {4mm}-\sum _{p|q\atop {p\not |d}}\frac {\log p}{p\Delta }+\sum _{p^\alpha |d\atop {p|d,q'}}\frac {\log p^\alpha }{1+D}-\sum _{p|d,q'}\frac {\log p^{D+1}}{p\Delta (1+D)}+\sum _{p^\alpha |d\atop {p\not |q'}}\frac {\log p^\alpha \Delta }{1+\Delta D} \\ &&\hspace {10mm}=\hspace {4mm}-\sum _{p|q'}\frac {\log p}{p\Delta }+\sum _{p^\alpha |d\atop {p|q'}}\frac {\log p^\alpha }{1+D}+\sum _{p^\alpha |d\atop {p\nmid q'}}\frac {\log p^\alpha \Delta }{1+\Delta D} \end{align} The second sum in question is \begin{align} \sum _{p^\alpha |d}\log p^\alpha \sum _{h|d\atop {p\nmid h}}\Delta (q'hp^\alpha )=:\sum _{p^\alpha |d}\log p^\alpha S_p(q,d). \end{align} Then \begin{align} S_p(q,d)&=&\prod _{P|q\atop {P\nmid d\atop {P\not =p}}}\Delta (q')\prod _{P|d,q'\atop {P\not =p}}\sum _{h|d}\Delta (q'h)\prod _{P|d\atop {P\nmid q'\atop {P\not =p}}}\sum _{h|d}\Delta (h)\cdot \Delta (q'P) \\ &=&R(q,d)\underbrace {\Delta ^{-1}}_{p|q'}\underbrace {(1+D)^{-1}}_{p|d,q'}\underbrace {(1+\Delta D)^{-1}}_{p|d\atop {p\nmid q'}}\cdot \Delta \end{align} so our sum has the right factor and the rest is \begin{align} &&\sum _{p^\alpha |d}\log p^\alpha \underbrace {\Delta ^{-1}}_{p|q'}\underbrace {(1+D)^{-1}}_{p|d,q'}\underbrace {(1+\Delta D)^{-1}}_{p|d\atop {p\nmid q'}}\cdot \Delta \\ &&\hspace {10mm}=\hspace {4mm}\sum _{p^\alpha |d,q'}\frac {\log p^\alpha }{1+D}+\sum _{p^\alpha |d\atop {p\nmid q'}}\frac {\log p^\alpha \Delta }{1+\Delta D} \end{align}