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I would like to estimate the main term of the integral $$\frac{1}{2\pi i} \int_{(c)} L(s) \frac{x^s}{s} ds$$ where $c > 0$, $\displaystyle L(s) = \prod_p \left(1 + \frac{2}{p(p^s-1)}\right)$.

Question: How to estimate the integral? In other words, is there any way to analytic continue this function?

The function as stated converges for $\Re s > 0$, but I'm not sure how to extend it past $y$-axis. Thanks!

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$$L(s)=\sum_{n=1}^\infty \frac{f(n)}{n^s}$$ $$f(n)=\frac{(-2)^{\omega(n)}\phi(n)}{n(\mu(n)n*1)}$$$$ f(n) \stackrel{}{=} \begin{cases} 1 & \text{if $n=1$} \\ \prod_{p\mid n}\frac{2}{p} & \text{if $n>1$} \end{cases} $$ –  Ethan Apr 30 '13 at 5:13

1 Answer 1

Let $\rho(d)$ count the number of solutions $x$ in $\frac{Z}{dZ}$, to $x^2\equiv \text{-1 mod d}$, then we have $$\sum_{n\leq x}d(n^2+1)=2x\sum_{n\leq x}\frac{\rho(n)}{n}+O(\sum_{n\leq x}\rho(n))$$ By multiplicative properties of $\rho(n)$ we have, $$\rho(n)=\chi(n)*|\mu(n)|$$

Where $\chi(n)$ is the non principal character modulo $4$

Which allows us to estimate, $$\sum_{n\leq x}\frac{\rho(n)}{n}=\sum_{n\leq x}\frac{\chi(n)*|\mu(n)|}{n}=\sum_{n\leq x}\frac{|\mu(n)|}{n}\sum_{k\leq \frac{x}{n}}\frac{\chi(k)}{k}=\sum_{n\leq x}\frac{|\mu(n)|}{n}(L(1,\chi)+O(\frac{n}{x}))=L(1,\chi)\sum_{n\leq x}\frac{|\mu(n)|}{n}+O(1)=\frac{\pi}{4}\sum_{n\leq x}\frac{|\mu(n)|}{n}+O(1)=\frac{\pi}{4}(\frac{6}{\pi^2}\ln(x)+O(1))$$ So that, $$\sum_{n\leq x}\frac{\rho(n)}{n}=\frac{3}{2\pi}\ln(x)+O(1)$$ Also note that, $$\sum_{n\leq x}\rho(n)=\sum_{n\leq x}{\chi(n)*|\mu(n)|}=\sum_{n\leq x}|\mu(n)|\sum_{k\leq \frac{x}{n}}\chi(k)\leq\sum_{n\leq x}|\mu(n)|\leq x$$ So we have, $$\sum_{n\leq x}\rho(n)=O(x)$$

Which gives, $$\sum_{n\leq x}d(n^2+1)=\frac{3}{\pi}x\ln(x)+O(x)$$

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Ah, good one. I still want to know what I should have done if I am to use Perron though. Thanks anyway! –  user27126 Apr 28 '13 at 6:15
    
@Sanchez Well, $$\sum_{n=1}^\infty \frac{\rho(n)}{n^s}=\frac{L(s,\chi)\zeta(s)}{\zeta(2s)}$$ –  Ethan Apr 28 '13 at 6:18
    
ah of course. Thanks! –  user27126 Apr 28 '13 at 6:29
    
sorry for unaccepting your answer, but I saw the gap of what I was thinking. The Dirichlet series I got above is actually wrong, which is why I couldn't solve the problem. I am now more interested in how the problem in my original question can be solved. –  user27126 Apr 28 '13 at 7:01
    
Ah nice. I just came back to this question to write up this answer (essentially), except based around $L(s, \chi_4) \zeta(s)/\zeta(2s)$ and Perron. Nice. +1 –  mixedmath Apr 29 '13 at 23:43

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