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I lost in this proof of Riemann's paper: On the Number of Prime Numbers less than a Given Quantity.

If one now considers the integral $$ \int \frac{(-x)^{s-1}}{e^x-1} dx $$ from $\infty$ to $\infty$ taken in a positive sense around a domain which includes the value 0 but no other point of discontinuity of the integrand in its interior, then this is easily seen to be equal to

$$ (e^{-\pi si}-e^{\pi si}) \int_0^{\infty} \frac{x^{s-1}}{e^x-1}dx $$.

Jimmy Wang
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1 Answers1

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That contour of integration is equivalent to the contour consisting of a circle $C_0$ of radius $\epsilon$ around the origin and two half-lines $C_+=(\infty\rightarrow \epsilon)$ and $C_-=(\epsilon\rightarrow\infty)$.

  • The value of the integrand at the point $z=-\epsilon$ on the circle is real and negative.

  • Turning by $\pi$ counterclockwise around the origin produces a phase factor $e^{i\pi s}$, since on the corresponding demi-circle $x$ is parameterized as $x=\epsilon e^{i\varphi}$, $\varphi\in[\pi,2\pi]$. The integral over $C_-$ will therefore be given by $$-e^{i\pi s} \int_{\epsilon}^{\infty}\frac{x^{s-1}dx}{e^x-1}.\tag{1}$$

  • Similarly, turning by $\pi$ clockwise, the integral over $C_+$ gives $$e^{-i\pi s} \int_{\epsilon}^{\infty}\frac{x^{s-1}dx}{e^x-1}.\tag{2}$$

  • Now let $\epsilon\rightarrow 0$, then the integral over $C_0$ vanishes whereas (1) and (2) give the quoted result.

Start wearing purple
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  • Now here x is in a complex domain, I got your meaning. I did get this: Turning by π counterclockwise around the origin produces a phase factor $e^{i\pi s}$, – Jimmy Wang Jun 01 '13 at 14:24
  • @JimmyWang Sorry, your questions are not well-displayed (if they are indeed questions). Basically, what you need to do is to parameterize $C_0$, $C_+$ and $C_-$ and remember that the integrated function is continuous on the union of these three contours. This function is multivalued and we fix its branch by requiring for it to be real negative on the negative part of the real axis. – Start wearing purple Jun 01 '13 at 14:30
  • Yes, there is a error in latex command. Integral over $C_{_}$ will be equation (1). I did not realize how the factor $-e^{i\pi s}$ comes from. – Jimmy Wang Jun 01 '13 at 14:34
  • @JimmyWang I suggest you to start from a simpler example. Consider what happens to the function $z^s$ when you analytically continue it from the positive part of the real axis to the negative part by turning around zero counterclockwise. If we write $z=re^{i\phi}$, then $\phi=0$ and $\phi=\pi$ correspond to the real positive and negative part. Now since $z^s=r^s e^{is(\phi-\pi)+i\pi s}$, the complex argument of $z^s$ on the negative part of the real axis will be $\pi s$. But if we turned clockwise, this argument would be $-\pi s$. – Start wearing purple Jun 01 '13 at 14:50
  • Thanks for your answer so details. I got that you advance $(-1)^{s-1}$ ahead in the form of $-e^{i\pi s}$. – Jimmy Wang Jun 01 '13 at 15:13
  • @JimmyWang Yes, very roughly, all we have to do is to write $(-1)^s$ correctly. But some care is needed because $-1$ can be $e^{i\pi}$ but also $e^{-i\pi}$. By the way I have just noticed that you have not accepted any answer to your previous questions (and probably didn't upvote them). I think this is unfair to people who tried to help you (I'm not talking about myself). – Start wearing purple Jun 01 '13 at 15:20
  • Sorry,I didn't realize that that to upvote is to accept. I thought that is for other people to evaluate or tell the answer correctly or important. – Jimmy Wang Jun 01 '13 at 15:26