3

I've got a complex function: $$f(z)=\frac{\pi\cot{\pi z}}{z^2}$$ I want to integrate it around a contour $\Gamma_N$ such that the poles at $$-N,-N+1,-N+2\cdots-1,0,1,\cdots N-2, N-1, N$$ are contained in the contour, so i choose a circle with radius $N+\varepsilon$ where $N\in\Bbb{N}, \varepsilon \in (0,1)$ The contour

My goal is to show that the integral for $N\to\infty$ $$\oint_{\Gamma_N}f(z)dz=0$$ In fact, it must equal to zero because by residue theorem $$\oint_{\Gamma_N}f(z)dz=\operatorname{Res}(f,0)+\sum_{n=-N,n\neq0}^Nn^{-2}$$ the residuum at $0$ is equal to $-\pi^2/3$ and for $N\to\infty$ this solves the basel problem. Returning to the contour integral: $$\oint_{\Gamma_N}f(z)dz=\int_0^{2\pi}f((N+\varepsilon)e^{i\varphi})d(N+\varepsilon)e^{i\varphi}=$$ $$=\int_0^{2\pi}\frac{\pi\cot({\pi(N+\varepsilon)e^{i\varphi})}}{(N+\varepsilon)e^{i\varphi}}id\varphi$$ Now i would prove that the limit as $N\to\infty$ of the integrand goes to zero, namely: $$\mathcal L= \lim_{N\to\infty}\frac{\cot{(\pi(N+\varepsilon))}}{(N+\varepsilon)}=\lim_{N\to\infty}\frac{\cos{(\pi(N+\varepsilon))}}{(N+\varepsilon)\sin{(\pi(N+\varepsilon)})}$$ and the argument is: $$\lim_{N\to\infty}\frac{\cos{(N+\varepsilon)}} {N+\varepsilon}=0$$ by squeeze theorem and $$\sin(\pi(N+\varepsilon))$$ is never equal to zero, because $$(N+\varepsilon)\notin\Bbb{N}$$ This would prove that integrand is zero therefore the contour integral is equal to zero. My question is, is my argument correct? Thanks for any advice.

Mark Viola
  • 179,405
Michal Dvořák
  • 2,885
  • The argument is not valid. You need to show that $\frac{\cot((N+1/2)e^{i\phi})}{(N+1/2)}\to 0$. So, you need to analyze more carefully the behavior of the complex cotangent on the circle. – Mark Viola Apr 03 '18 at 14:45
  • So, i used the definition of cotangent: $$\cot{z}=i\frac{e^{2iz}+1}{e^{2iz}-1}$$ and obtained an integral where the numerator is a fraction of the form $$\frac{e^{2i\pi(N+\varepsilon)\cos{\varphi}}e^{-2\pi(N+\varepsilon)\sin{\varphi}}+1}{e^{2i\pi(N+\varepsilon)\cos{\varphi}}e^{-2\pi(N+\varepsilon)\sin{\varphi}}-1}$$ this entire thing in absolute value, can i now say that the part with $e^{i...}$ doesn't change the value and $$e^{-2\pi...}$$ is bounded and for $N\to\infty$ goes to zero? – Michal Dvořák Apr 03 '18 at 14:55
  • 1
    Not quite. See THIS ANSWER where I showed that the magnitude of the cotangent function, $|\cot(z)|$ is bounded on the circle $|z|=N+1/2$. – Mark Viola Apr 03 '18 at 15:18
  • This helps, but just one clarification, how did you come to the fact, that... $$|\cot{\pi z}|=\sqrt{\frac{\cosh{2\pi y}+\cos{2\pi x}}{\cosh{2\pi y}-\cos{2\pi x}}}$$ – Michal Dvořák Apr 03 '18 at 15:53
  • Please see the edited answer. – Mark Viola Apr 03 '18 at 16:42

1 Answers1

2

In THIS ANSWER, I showed that for $|z|+N+1/2$, the magnitude of the complex cotangent function, $|\cot(z)|$, satisfies the bound

$$|\cot(z)|=\sqrt{\frac{\cosh(2y)+\cos(2x)}{\cosh(2y)-\cos(2x)}}\le \sqrt{1+\frac{16}{3\pi^2}}$$

Hence, we have

$$\begin{align} \left|\,\oint_{|z|=N+1/2} \frac{\pi\cot(\pi z)}{z^2}\,dz\,\right|&\le \oint_{|z|=N+1/2}\frac{\pi |\cot(z)|}{|z|^2}\,|dz|\\\\ &\le \,2\pi (N+1/2)\left(\frac{\pi\sqrt{1+\frac{16}{3\pi^2}}}{(N+1/2)^2}\right)\\\\ &=\frac{2\pi^2\sqrt{1+\frac{16}{3\pi^2}}}{N+1/2} \end{align}$$

which approaches $0$ as $N\to \infty$.

Finally, using the residue theorem we find that

$$\begin{align} \sum_{n\ne 0}\frac1{n^2}=-\text{Res}\left(\frac{\pi \cot(\pi z)}{z^2}, z=0\right)&=\lim_{z\to 0}\frac{d^2}{dz^2}\left(\pi z\cot(\pi z)\right)\\\\ &=-\frac12\lim_{z\to 0}\frac{d^2}{dz^2}\left(\frac{1-\frac12(\pi z)^2+O(\pi z)^4}{1-\frac16(\pi z)^2+O(\pi z)^4}\right)\\\\ &=-\frac12\lim_{z\to 0}\frac{d^2}{dz^2}\left(1-\frac13 (\pi z)^2+O(\pi z)^4\right)\\\\ &=\frac{\pi^2}{3} \end{align}$$

Appendix:

In this appendix we address the question that the OP asked in a comment regarding the equality $|\cot(z)|=\sqrt{\frac{\cosh(2y)+\cos(2x)}{\cosh(2y)-\cos(2x)}}$. Proceeding we have

$$\begin{align} |\cot(z)|&=\left|\frac{e^{i2z}+1}{e^{i2z}-1}\right|\\\\ &=\left|\frac{(1+\cos(2x)e^{-2y})+i\sin(2x)e^{-2y}}{(1+\cos(2x)e^{-2y})-i\sin(2x)e^{-2y}}\right|\\\\ &=\sqrt{\frac{1+2\cos(2x)e^{-2y}+\cos^2(2x)e^{-4y}+\sin^2(2x)e^{-4y}}{1-2\cos(2x)e^{-2y}+\cos^2(2x)e^{-4y}+\sin^2(2x)e^{-4y}}}\\\\ &=\sqrt{\frac{1+e^{-4y}+2\cos(2x)e^{-2y}}{1+e^{-4y}-2\cos(2x)e^{-2y}}}\\\\ &=\sqrt{\frac{\cosh(2y)+\cos(2x)}{\cosh(2y)-\cos(2x)}} \end{align}$$

as was to be shown!

Mark Viola
  • 179,405
  • I got the part with the residue myself already, was just looking to prove the boundeness of the cotangent function, which is practically in the answer you posted a link to, anyways, asked for a brief explanation here. – Michal Dvořák Apr 03 '18 at 16:21
  • @MichalDvořák I edited to address the question you had asked in a comment under your posted question. Let me know if you have any further questions. Really hope all of this helped. – Mark Viola Apr 03 '18 at 16:38
  • I see it now, deleted it, because i realised it fast enough, i was blind. I guess im done with the part for $|\cot{\pi z}|$, gonna now focus on the proof of boundeness part. – Michal Dvořák Apr 03 '18 at 16:56
  • Sounds as if you're making good progress. Pleased to hear. – Mark Viola Apr 03 '18 at 16:58
  • And feel free to up vote both this and the referenced question and accept this answer, all as you see fit. – Mark Viola Apr 03 '18 at 16:59
  • Got stuck on the third (fourth) line again :/ what do i factor out to get the $\cosh{2y}$ in there? namely expression: $$\sqrt{\frac{1+e^{-4y}+2\cos{(2x)}e^{-2y}}{...}}$$ – Michal Dvořák Apr 03 '18 at 17:11
  • 1
    Multiply numerator and denominator by $\frac12 e^{2y}$. – Mark Viola Apr 03 '18 at 17:14
  • Great! Ok, now just to be sure, i don't have to explore the case 2, where you prove that $|\cot{\pi z}|$ is bounded by $1,24$. It should be enough when i show that $|\cot{\pi z}|\le 1$ and therefore say that my integrand goes to zero, is that correct? – Michal Dvořák Apr 03 '18 at 17:18
  • No, you need to show that it is bounded on the circle. – Mark Viola Apr 03 '18 at 17:23
  • I see, well, this is going to take me some more time and will leave it for another day i guess. Anyways, i might reconsider choosing the path of integration, i think, if i choose the contour to be rectangle with bounds $N+\varepsilon,-N-\varepsilon$ it might be easier to prove that it goes to zero with $N\to\infty$... anyways, thanks for clarifications and help, i apreciate it, i still have lot to learn about complex analysis. – Michal Dvořák Apr 03 '18 at 17:48
  • @MichalDvořák Don't give up. You're almost there. In my referenced answer, I showed that $|\cot(z)|$ is bounded on the circle $|z|=N+1/2$. – Mark Viola Apr 03 '18 at 18:29
  • I'm struggling with the part of the proof that $|\cot{\pi z}|$ is bounded. I got the first part (the Case 1). We show that $|\cot{\pi z}|$ is bounded above by $1$ in the strip $N+1/4\le|\operatorname{Re}(z)|\le N+1/2$. That is 100% clear to me – Michal Dvořák Apr 03 '18 at 18:35
  • Okay, to the line where you say: $$\cosh(2\pi y) \ge 1+\frac{1}{2}(2\pi y^2)$$ that's just fine,...and then say that's equal to $$1+2\pi ^2((N+1/2)^2-x^2)$$ Isn't there a typo or am i missing something? By the way, where did this $x^2$ come from :S? – Michal Dvořák Apr 03 '18 at 19:01
  • For the other part, we are on the circle $|z|+N+1/2$ for $|x|\le N+1/4$. We can find a lower bound for $\cosh(2y)$ on that part of the circle. We can choose to use the crude estimate $\cosh(2\pi y)\ge 1+\frac12 (2\pi y)^2$. But recall $x^2+y^2=(N+1/2)^2$ and so $y^2=(N+1/2)^2-x^2\ge (N+1/2)^2-(N+1/4)^2= \frac12N+\frac3{16}\ge \frac3{16}$. – Mark Viola Apr 03 '18 at 19:02
  • Okay, got that part, now how do i exactly return back to the (1) equation (as you reference it) because for $|x| \le N+1/4$ we do not have any bounds set for $\cos{2\pi x}$ – Michal Dvořák Apr 03 '18 at 19:16
  • Michael, we always have $|\cos(2\pi x)|\le 1$. – Mark Viola Apr 03 '18 at 19:19
  • Ah, yes, forgot about that, true. Seems like i have it. Just for clarification, because this way of proof seems really weird to me. First, we proved that in the vertical strip where $N+1/4 \le \operatorname{Re}(z) \le N+1/2$ the cotangent function is bounded. In the second step, by setting some other bound for cosh, we found, that in the rest of the plane, that means for $ N+1/4\ge \operatorname{Re}(z)$ we found that the cotangent is bounded by this weird transcendental number, which is approximately 1,24. Therefore we proved that cotangent is bounded on the entire circle. Am i right? – Michal Dvořák Apr 03 '18 at 19:27
  • 1
    Yes! You have it now. Well done. – Mark Viola Apr 03 '18 at 19:28
  • Thanks, may i also ask about the limit, as you solved it. I personally went through the second derivative and got to same result after some trigonometric limit. I haven't really come across this big O notation. I just know that it means something like "we don't care about those terms". But how do you get from $$\lim\bigg(\frac{1-\frac{1}{2}x^2+...}{1-\frac{1}{6}x^2+...}\bigg)$$ to $$\lim \bigg(1-\frac{1}{3}x^2\bigg)$$? – Michal Dvořák Apr 03 '18 at 19:42
  • 1
    Michael, note that $\frac1{1-t}=1+t+t^2+\cdots=1+t+O(t^2)$. So, $\frac{1}{1-\frac13 x^2+O(x^4)}=1+\frac13x^2+O(x^4)$ – Mark Viola Apr 03 '18 at 20:19
  • $\displaystyle\lim_{\Im\left(z\right) \to \pm\infty}\left\vert\cot\left(z\right)\right\vert = 1$ and $\displaystyle\left\vert\oint_{\left\vert z\right\vert = R}{\mathrm{d}z \over z^{2}}\right\vert \sim {2\pi \over R}$ as $\displaystyle R \to \infty$. – Felix Marin Apr 18 '18 at 04:13