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On page 188.

Lemma 10.2.3 is $\sum_{\substack{\delta \leq x \\ 2 \nmid \delta}}\frac{\mu^{2}(\delta)}{\phi_{1}(\delta)} = A_{1}log(x) +A_{2} + O(\frac{1}{x^{1/4}})$ for positive $A_{1}$, $A_{2}$.

And $I = \int_{3}^{x} (\sum_{\substack{\delta \leq t \\ 2 \nmid \delta}}\frac{\mu^{2}(\delta)}{\phi_{1}(\delta)}) \frac{1}{t (1+log(\frac{x}{t}))} dt$

Now my question is how did they get the following equation by lemma 10.2.3?

$I = -A_{1} \int_{3}^{x} \frac{dt}{t} + (A_{1}log(x)+A_{1}+A_{2})\int_{3}^{x} \frac{dt}{t(1+log(x)-log(t)} + O(\frac{1}{x^{1/4}})$

I have been stuck on this for a long while. Any help is appreciated. :)

Edited

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    $\begingroup$ You should definitely replace the '$x$' appearing as an upper bound for the variable $\delta$ in the expression of the integral $I$ with '$t$'. $\endgroup$
    – Salvo Tringali
    Commented Apr 30, 2014 at 14:03
  • $\begingroup$ Plugging in the lemma, you'd get an $A_1 \log t$ term. This is the same as $-A_1(1+\log(x/t))+A_1\log x + A_1$. Is that your question? $\endgroup$
    – rlo
    Commented May 1, 2014 at 17:32
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    $\begingroup$ Actually, wait. Unless I'm missing something, their claimed error term is wrong, and it should be $O(1/\log x)$ rather than $O(1/x^{1/4})$ (this is fine for their purposes, as the error on the next line is $O(1/\log x)$ anyway). To see it has to be at least this big, just look at the integral of the error from 3 to 4. To see that it's no bigger, split the integral into two parts, integrating from 3 to $x^{1/2}$, say, and from $x^{1/2}$ to $x$. The latter decays like $x^{-1/8}$, while the former can be bounded by $O(1/\log x)$. $\endgroup$
    – rlo
    Commented May 1, 2014 at 18:22

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I am not really understanding the question, probably, but why can't you just substitute the estimate into the integral? if you omit the error term, you get:

$$ \log \left(\log \left(\frac{x}{3}\right)+1\right) (A_1 \log (x)+A_2). $$ Doing the error term separately, you get:

$$ \log \left(\log \left(\frac{x}{3}\right)+1\right) O(1/x^{1/4}). $$

Are you saying this is different from what the authors claim?

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  • $\begingroup$ Sorry, I had a typo. The sum inside the integral should be a function with respects to t. Thanks. $\endgroup$
    – Doorbell
    Commented May 1, 2014 at 8:30

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