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Let $$\psi(x;q,a)=\sum_{n\leq x\atop n\equiv a\pmod q}\Lambda(n)$$

where $\Lambda$ denotes the von Mangoldt function and $\phi$ to be Euler's totient function. Then the Siegel-Wafisz theorem states that given any real number $N$ there exists a positive constant $C_N$ depending only on $N$ such that

$$\psi(x;q,a)=\frac{x}{\varphi(q)}+O\left(x\exp\left(-C_N(\log x)^\frac{1}{2}\right)\right),$$

whenever $(a, q) = 1$ and $q\le(\log x)^N$.

Let $\omega$ be some smooth function, and consider $$ \sum_{n\leq x\atop n\equiv a\pmod q}\omega(n) \Lambda(n) $$ instead. Are there any results out there where we can obtain the result with a better range for $q$ by adding a smooth weigh to $\psi$? Thank you very much!

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  • $\begingroup$ This is not a situation where smoothing is of use; so, no. $\endgroup$ – Lucia Jun 8 '17 at 0:01
  • $\begingroup$ @Lucia Would it be possible to add some explanation on why this is a situation where smoothing is not helpful in the answer? $\endgroup$ – Johnny T. Jun 8 '17 at 16:13
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Smoothing often removes technical issues in a problem, but fundamental obstructions cannot be removed merely by smoothing. It may be helpful to consider some examples.

  1. The divisor problem. Here one asks for the error term with a sharp cutoff $\sum_{n\le x} d(n)$, and this error is supposed to be $O(x^{\frac 14+\epsilon})$. While we don't know how to prove that, the error term can be given in terms of an oscillating sum, and smoothing will get rid of that. For example, consider the smoothed sum $$ \sum_{n} d(n) e^{-n/x} = \frac{1}{2\pi i} \int_{(c)} \zeta(s)^2 x^s \Gamma(s) ds, $$ and here you can move the line of integration to say the line Re$(s)=-1.5$ and obtain an asymptotic with an error term of $O(x^{-1.5})$. Clearly smoothing makes a big difference in this problem.

  2. The prime number theorem. With a classical zero free region $\beta < 1- c/\log \gamma$, one obtains the prime number theorem in the form $\psi(x)=x+O(x\exp(-c\sqrt{\log x}))$. Consider instead the smoothed sum $$ \sum_n \Lambda(n) e^{-n/x} = x - \sum_{\rho} x^{\rho} \Gamma(\rho) + O(1). $$ Since the $\Gamma$-function decays exponentially on vertical lines, the classical zero-free region now gives an error term here of $$ O\Big(x \exp\Big(-c \frac{\log x}{\log\log x}\Big)\Big). $$ This is much better than the unsmoothed case, but note that the fundamental obstruction from zeros of zeta persists here. We cannot obtain a power saving in the error term without proving a quasi Riemann hypothesis. One recognizes that the error term in the prime number theorem has a more fundamental number theoretic obstruction than the error term in the divisor problem, which I would view more as a problem of analysis.

  3. Your question on primes in progressions. Again let us consider the exponential smoothing as above. Then, with negligible error,
    $$ \sum_{\substack{n\equiv a\pmod q}} \Lambda(n) e^{-n/x} = \frac{x}{\phi(q)} - \frac{1}{\phi(q)} \sum_{\chi \pmod q} \overline{\chi(a)} \sum_{\rho_{\chi}} x^{\rho_{\chi}} \Gamma(\rho_{\chi}). $$ The $\Gamma$-function helps a lot with zeros having large imaginary part, as in example 2, but it doesn't help with zeros near the real axis. For example, if there were a Siegel zero at $\beta$, it would still make a contribution of size $O(x^{\beta }/\phi(q))$, exactly as in the unsmoothed case. Thus the range of $q$ cannot be improved. If you assume that there is no Siegel zero, then as in Example 2, in the smoothed case you would obtain an improvement in the error term, but the fundamental obstruction of a Siegel zero cannot simply be ``smoothed away."

Put differently, a Siegel zero corresponds to a distribution of primes in progressions where half the residue classes get almost twice as many primes as expected, and the other half get hardly any primes. Smoothing makes no difference to this statement.

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  • $\begingroup$ Very interesting and helpful! Thank you very much! $\endgroup$ – Johnny T. Jun 9 '17 at 14:35
  • $\begingroup$ Do you happen to know a reference for the explicit formulas with smooth weights? I am currently interested in the explicit formula for $\sum \Lambda(n) \chi(n)$... (I asked here mathoverflow.net/questions/282913/…) $\endgroup$ – Johnny T. Oct 7 '17 at 9:44

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