# Contour integration problem, proving complex function is bounded

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 $N\in\Bbb{N}, \varepsilon \in (0,1)$

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.

• 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. Apr 3, 2018 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? Apr 3, 2018 at 14:55
• Not quite. See THIS ANSWER where I showed that the magnitude of the cotangent function, $|\cot(z)|$ is bounded on the circle $|z|=N+1/2$. Apr 3, 2018 at 15:18
• This helps, but just one clarification, how did you come to the fact, that... $$|\cot{\pi z}|=\sqrt{\frac{\cosh{2\pi y}+\cos{2\pi x}}{\cosh{2\pi y}-\cos{2\pi x}}}$$ Apr 3, 2018 at 15:53

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!

• 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. Apr 3, 2018 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. Apr 3, 2018 at 16:38
• Multiply numerator and denominator by $\frac12 e^{2y}$. Apr 3, 2018 at 17:14
• Yes! You have it now. Well done. Apr 3, 2018 at 19:28
• Michael, note that $\frac1{1-t}=1+t+t^2+\cdots=1+t+O(t^2)$. So, $\frac{1}{1-\frac13 x^2+O(x^4)}=1+\frac13x^2+O(x^4)$ Apr 3, 2018 at 20:19