I was studying some zero-density results for $\zeta(s)$, mostly from Titchmarsh's book "The Theory of the Riemann zeta function", Chapter 9. In one place, as per the literature, a mollifier of $\zeta(s)$ is defined as $M_{X}(s) = \sum_{n\leq X}\frac{\mu(n)}{n^s}$, and subsequently a function $f_{X}(s)$ defined as $f_{X}(s) = \zeta(s)M_{X}(s) -1 = \sum_{m\geq 1}\frac{1}{m^s}\sum_{n\leq X}\frac{\mu(n)}{n^s} -1 = \sum_{n > X} \frac{\sum_{d\vert n, d\leq X}\mu(d)}{n^s} =\sum_{n\geq 1}\frac{a_{X}(n)}{n^s}$, being the series representation of $f_{X}(s)$, where $a_{X}(n) = 0$ if $n\leq X$ and $a_{X}(n) = \sum_{d\vert n, d \leq X} \mu(d)$ if $n>X$. My question is if I define a mollifier in similar lines for the Dirichlet $L$-function as $M_{X}(s, \chi) = \sum_{n\leq X} \frac{\mu(n)\chi(n)}{n^s}$ and subsequently define $f_{X}(s, \chi) = M_{X}(s,\chi)L(s, \chi) - 1$, what can be the series representation for $f_{X}(s, \chi)$? What can be the expression of the general term?
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1$\begingroup$ What did you try? Please see Quick beginner guide for asking a well-received question $\endgroup$– Lucky ChouhanCommented Nov 22, 2023 at 8:10
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1$\begingroup$ My apologies @Lucky Chouhan. Intuitively, it seems to me that the general term should be $a_{X}(n) = 0$ if $n \leq X$ and $= \sum_{d\vert n, d \leq X} \mu(d) \chi(d)$ for $n > X$, but I am not able to explain it. Should the character term play some other role in it? That is where I am getting stuck actually. $\endgroup$– djangounchained0716Commented Nov 22, 2023 at 8:29
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$\begingroup$ Dirichlet characters are completely multiplicative, so they should appear outside the divisor sum. $\endgroup$– TravorLZHCommented Nov 23, 2023 at 10:09
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$\begingroup$ @TravorLZH, yes, and in that case, I believe what you meant is that generally, $a_{X}(n) = \sum_{a\leq X, ab = n} \mu(a) \chi(a)\chi(b) - 1$? In that case, for $n \leq X$, we should have $a_{X}(n) = \chi(n)\sum_{ab = n} \mu(a) -1 = 0$. Does this seem correct? $\endgroup$– djangounchained0716Commented Nov 25, 2023 at 20:26
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