Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I want to understand why assuming that $\sum_{n \ge 1} \frac{a_n}{n^s}$ converges uniformly for $\mathrm{Re}(s) > \sigma > 0$ with $c > \sigma$ implies that $$ \sum_{n \le x} \, \!\!^* a_n = \frac 1{2\pi i}\int_{c-i\infty}^{c+i\infty} \sum_{n \ge 1} \frac{a_n}{n^s} \frac{x^s}{s} \, ds. $$ I've managed to show that for $c > 0$, $$ \frac 1{2\pi i}\int_{c-i \infty}^{c + i \infty} \frac{y^s}s \, ds = \begin{cases} 0 & \text{ if } 0 < y < 1 \\ 1/2 & \text{ if } y = 1 \\ 1 & \text{ if } y > 1 \\ \end{cases} $$ so we can write $$ \sum_{n \le x} \, \!\!^* a_n = \frac 1{2\pi i} \sum_{n \ge 1} \int_{c-i\infty}^{c+i\infty} a_n \left( \frac xn \right)^s \frac{ds}s \overset{!}{=} \frac 1{2\pi i}\int_{c-i\infty}^{c+i\infty} \sum_{n\ge 1} \frac{a_n}{n^s} \frac{x^s}s \, ds. $$ But that $!$ that I put there means I don't understand why the sum can go under the integral sign. Any ideas about that part? Thanks.

share|improve this question
    
I must say that $$\frac 1{2\pi i}\int_{c-i \infty}^{c + i \infty} \frac{y^s}s \, ds$$ doesn't not converge for $y=1$, but his Cauchy principal value $$\lim_{T \to \infty} \frac 1{2\pi i}\int_{c-i T}^{c + i T} \frac{y^s}s \, ds$$ exist and it's equal to $1/2$. You can find here detail version of proof, but there is small problem: it is on Serbian, but there is a lot formula so you should made it :-) –  Cortizol Aug 30 '13 at 19:27
add comment

1 Answer

up vote 1 down vote accepted

Even the basic identity, about integrating $y^s/s$ on a vertical line, requires qualification to be truly sensible, since the integral is certainly not absolutely convergent. One way to be completely up-front about it is to compute $\int_{c-iT}^{c+iT} {y^s\over s}\,ds$ and keep track of the error (from the ideal answers you give) in terms of $y$ and $T$. A finite-extent integral can certainly be interchanged with the sum over $n$. Then summing the Dirichlet series gives an estimable error, which goes to $0$ as $T$ goes to $+\infty$.

share|improve this answer
    
So you're saying I should just re-work out the proof I did for the integral $\int_{c-i\infty}^{c+i \infty}$ y^s/s \, ds$ and keep track of the error when I'm summing? I thought it would be more easy than that, I was too lazy to try it that way. This approach does seem hard though. –  Patrick Da Silva May 8 '13 at 2:29
    
Yes, probably you should just redo the intuitive version more scrupulously. Yes, it requires some labor, but it's not profoundly difficult. That it requires something is not unreasonable, given the not-absolute-convergence. In some scenarios, integrating against $y^s/s(s-1)(s-2)...(s-\ell)$ with $\ell\ge1$ is sufficient, and does give better convergence. But, in fact, often this more-smoothed "sample" gives substantially-less-interesting results, unfortunately. Thus, all the more with hindsight, the trouble it takes to be sure of the simpler-but-not-absolutely-convergent case is warranted. –  paul garrett May 8 '13 at 2:41
    
Um, you seem to put $y^2$ all the time, I meant $y^s$, you know that right? Or is it just a frequent typo of yours :P –  Patrick Da Silva May 8 '13 at 2:42
    
Heh... "2" or "s"... kinda similar? :) –  paul garrett May 8 '13 at 2:43
    
So should I try to bound $$ \left| \int_{c-iT}^{c+iT} \sum_{ n \ge 1} \frac{a_n}{n^s} \frac{x^s}s \, ds - \sum_{n \le x} \,\!\!^* a_n \right| $$ by some function of $T$ that goes to zero as $T$ goes to infinity? –  Patrick Da Silva May 8 '13 at 2:50
show 1 more comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.