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Old 2020-02-23, 17:31   #1
wildrabbitt
 
Jul 2014

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Default something I don't understand to do with Dirichlets theorem

Hi,

the following is something I've been reading.





He started from the power series



\(\sum_{n=1}^\infty \big(\frac{n}{q}\big)x^n=\frac{1}{1-x^q}\sum_{m=1}^{q-1}\big(\frac{m}{q}\big)x^m=\frac{xf(x)}{1-x^q}\)


say, and by putting this in the formula


\(\Gamma(s)n^{-1}=\int_0^1 x^{n-1}(\log x^{-1})^{s-1} \mathrm{d}x\)


he obtained


\(\Gamma(s)L(s)=-\int_0^1\frac{f(x)}{x^q-1}(\log x^{-1})^{s-1}\mathrm{d}x\)


I'm stuck because I can't see how he put what he put in the formula.


Can anyone explain it step by step?

Last fiddled with by wildrabbitt on 2020-02-23 at 17:32
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Old 2020-02-23, 20:17   #2
R.D. Silverman
 
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Nov 2003

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Default

Quote:
Originally Posted by wildrabbitt View Post
Hi,

the following is something I've been reading.





He started from the power series



\(\sum_{n=1}^\infty \big(\frac{n}{q}\big)x^n=\frac{1}{1-x^q}\sum_{m=1}^{q-1}\big(\frac{m}{q}\big)x^m=\frac{xf(x)}{1-x^q}\)


say, and by putting this in the formula


\(\Gamma(s)n^{-1}=\int_0^1 x^{n-1}(\log x^{-1})^{s-1} \mathrm{d}x\)


he obtained


\(\Gamma(s)L(s)=-\int_0^1\frac{f(x)}{x^q-1}(\log x^{-1})^{s-1}\mathrm{d}x\)


I'm stuck because I can't see how he put what he put in the formula.


Can anyone explain it step by step?
I am getting "math processing error". Your post is not being parsed correctly.
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Old 2020-02-23, 23:58   #3
Dr Sardonicus
 
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Default

Quote:
Originally Posted by wildrabbitt View Post
Hi,

the following is something I've been reading.





He started from the power series



\(\sum_{n=1}^\infty \big(\frac{n}{q}\big)x^n=\frac{1}{1-x^q}\sum_{m=1}^{q-1}\big(\frac{m}{q}\big)x^m=\frac{xf(x)}{1-x^q}\)


say, and by putting this in the formula


\(\Gamma(s)n^{-1}=\int_0^1 x^{n-1}(\log x^{-1})^{s-1} \mathrm{d}x\)


he obtained


\(\Gamma(s)L(s)=-\int_0^1\frac{f(x)}{x^q-1}(\log x^{-1})^{s-1}\mathrm{d}x\)


I'm stuck because I can't see how he put what he put in the formula.


Can anyone explain it step by step?
I took the liberty of TEXing your equations. That makes them a whole lot easier to read, for me at least. You can probably drop the big parens.

Is that supposed to be (log(x)-1^(s-1)? That would be log(x)1-s, yes?
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Old 2020-02-24, 07:52   #4
wildrabbitt
 
Jul 2014

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I just wrote what was written in the book.


I'm pretty sure it's supposed to be


\(\log \frac{1}{x} \)


otherwise it would have been written the way you wrote.
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Old 2020-02-24, 09:51   #5
wildrabbitt
 
Jul 2014

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Default clarification

I aught to mention that


\( L(s)=\sum_{n=1}^{\infty}\big(\frac{n}{q}\big)n^{-s}\)


for which


\(\big(\frac{n}{q}\big)\) is the Legendre symbol as is it is the first post.



Don't know what to do if my latex is not parsing correctly. It comes out right when I preview the post.


I can give more info if necessary.

Last fiddled with by wildrabbitt on 2020-02-24 at 09:52
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Old 2020-02-24, 12:49   #6
Dr Sardonicus
 
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Quote:
Originally Posted by wildrabbitt View Post
I just wrote what was written in the book.


I'm pretty sure it's supposed to be


\(\log \frac{1}{x} \)


otherwise it would have been written the way you wrote.
Ah, log(1/x), not 1/log(x). OK, now I see a Gamma-function integral in prospect in your second equation. (x = exp-u, scribble scribble, u = t/n, scribble scribble...)

Your second equation is wrong. It should be

\Gamma(s)n^{-s}=\int_0^1 x^{n-1}(\log x^{-1})^{s-1} \mathrm{d}x

(It is n-s on the left side, not n-1)
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Old 2020-02-24, 20:40   #7
wildrabbitt
 
Jul 2014

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Thanks.


Are you substituting \(e^{-u}\) for \(x\) ? What are you doin with \(u\)?




Also, in case there's an obvious remedy to my latex not parsing properly, here's an example of what I wrote in the first post without the \\( and \\).


\sum_{n=1}^\infty \big(\frac{n}{q}\big)x^n=\frac{1}{1-x^q}\sum_{m=1}^{q-1}\big(\frac{m}{q}\big)x^m=\frac{xf(x)}{1-x^q}

Last fiddled with by wildrabbitt on 2020-02-24 at 20:44 Reason: needed to access page source
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Old 2020-02-25, 08:21   #8
Nick
 
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Quote:
Originally Posted by wildrabbitt View Post
Also, in case there's an obvious remedy to my latex not parsing properly, here's an example of what I wrote in the first post without the \\( and \\).

\sum_{n=1}^\infty \big(\frac{n}{q}\big)x^n=\frac{1}{1-x^q}\sum_{m=1}^{q-1}\big(\frac{m}{q}\big)x^m=\frac{xf(x)}{1-x^q}
LaTeX: just put backslash then open-square-bracket before it, and backslash then close-square-bracket afterwards:

\[\sum_{n=1}^\infty \big(\frac{n}{q}\big)x^n=\frac{1}{1-x^q}\sum_{m=1}^{q-1}\big(\frac{m}{q}\big)x^m=\frac{xf(x)}{1-x^q}\]
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Old 2020-02-25, 13:32   #9
Dr Sardonicus
 
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Quote:
Originally Posted by wildrabbitt View Post
Thanks.


Are you substituting \(e^{-u}\) for \(x\) ? What are you doin with \(u\)?




Also, in case there's an obvious remedy to my latex not parsing properly, here's an example of what I wrote in the first post without the \\( and \\).
<snip>
My problem with LaTex in posts is mine alone. Viewing on other computers, everything looks OK. The fact that everything looked OK on my system when I simply slapped tex tags around the equations says the expressions parse OK.

If you don't know how to make substitutions in integrals, it's high time you learn. Unfortunately, I'm probably not the one to teach you.
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Old 2020-02-25, 17:01   #10
wildrabbitt
 
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I didn't really get the answer I was hoping for so I thought I'd just ask the first things that it occurred to me to ask.


I do know how to make substitutions in integrals I just wasn't sure if that's what you were doing and it looked like you were making two substitutions.


Thanks though. So you're substiting for x. I don't know what you're doing with u.
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