For positive integer \(n\), find all poles of \[ \frac{\pi}{z^n\tan(\pi z)} \] and the residues of all simple poles.
Prove that for positive even \(n\), \[ \frac{\sum_{m=1}^{\infty}m^{-n}}{\pi^n} \] is rational.
Solution
For a positive integer \(n\), set \[ F_n(z) \coloneqq \frac{\pi}{z^n\tan(\pi z)} = \frac{\pi\cot(\pi z)}{z^n}. \]
1. The poles of \(F_n\) are exactly the integers. Every nonzerointeger is a simple pole, while \(0\) is a pole of order \(n+1\).
The function \(\cot(\pi z)\) has simple poles exactly at the integers and zeros at the half-integers. Near an integer \(m\), \[ \sin(\pi z) = (-1)^m\pi(z-m)+O((z-m)^3), \] while \[ \cos(\pi z) = (-1)^m+O((z-m)^2). \] Hence \[ \pi\cot(\pi z) = \frac1{z-m}+O(z-m). \]
If \(m\ne0\), the factor \(z^{-n}\) is holomorphic and nonzero at \(m\), so the pole remains simple. At \(m=0\), \[ \pi\cot(\pi z) = \frac1z+O(z), \] and division by \(z^n\) gives a pole of order \(n+1\). There are no other poles.
2. For every nonzero integer \(m\),\[ \boxed{ \operatorname{Res}_{z=m}F_n(z)=\frac1{m^n}. } \]
By the expansion in step 1, \[ F_n(z) = \frac1{z^n} \left( \frac1{z-m}+O(z-m) \right). \] Since \(z^{-n}\) is holomorphic at \(m\), the coefficient of \((z-m)^{-1}\) is its value at \(m\), namely \(m^{-n}\). This completes part (a).
3. Assume from now on that \(n\) is positive and even. For\(N\ge1\), put\[ R_N\coloneqq N+\frac12 \]and let \(\Gamma_N\) be the positively oriented boundary of the square\[ \{z:|\operatorname{Re}z|\le R_N,\ |\operatorname{Im}z|\le R_N\}. \]There is a constant \(C>0\), independent of \(N\), such that\[ |\cot(\pi z)|\le C \]for every \(z\in\Gamma_N\).
On a vertical side, \[ \operatorname{Re}z=\pm\left(N+\frac12\right). \] Writing \(z=x+iy\) and using the elementary formulas for sine and cosine of a complex argument gives \[ |\cot(\pi z)| = |\tanh(\pi y)| \le1. \]
On a horizontal side, \[ |\operatorname{Im}z|=R_N\ge\frac12. \] For \(z=x+iy\), \[ |\cot(\pi z)|^2 = \frac{\cos^2(\pi x)+\sinh^2(\pi y)} {\sin^2(\pi x)+\sinh^2(\pi y)} \le \coth^2(\pi|y|) \le \coth^2\left(\frac\pi2\right). \] Thus one may take \[ C=\max\left\{1,\coth\left(\frac\pi2\right)\right\}. \]
4. One has\[ \int_{\Gamma_N}F_n(z)\,dz \longrightarrow 0. \]
Every point of \(\Gamma_N\) satisfies \[ |z|\ge R_N. \] By step 3, \[ |F_n(z)| \le \frac{\pi C}{R_N^n} \] on \(\Gamma_N\). The perimeter of the square is \(8R_N\), so the \(ML\)-estimate gives \[ \left| \int_{\Gamma_N}F_n(z)\,dz \right| \le 8\pi C R_N^{1-n}. \] Because \(n\ge2\), the right-hand side tends to \(0\).
5. If\[ r_n\coloneqq\operatorname{Res}_{z=0}F_n(z), \]then\[ r_n+2\sum_{m=1}^{\infty}\frac1{m^n}=0. \]
The poles inside \(\Gamma_N\) are precisely \[ -N,-N+1,\ldots,-1,0,1,\ldots,N. \] The residue theorem and step 2 give \[ \frac1{2\pi i} \int_{\Gamma_N}F_n(z)\,dz = r_n + \sum_{\substack{-N\le m\le N\\m\ne0}} \frac1{m^n}. \] Since \(n\) is even, \[ \frac1{(-m)^n}=\frac1{m^n}, \] so \[ \frac1{2\pi i} \int_{\Gamma_N}F_n(z)\,dz = r_n + 2\sum_{m=1}^{N}\frac1{m^n}. \] Now let \(N\to\infty\). Step 4 sends the left-hand side to \(0\), and the series converges because \(n\ge2\). This proves the identity.
6. The residue \(r_n\) is a rational multiple of \(\pi^n\).
Define \[ h(w)\coloneqq w\cot w. \] The functions \[ \cos w \qquad\text{and}\qquad \frac{\sin w}{w} \] have power series at \(0\) with rational coefficients, and the second has constant term \(1\). Therefore its reciprocal also has a power series with rational coefficients. Hence \[ h(w) = \frac{\cos w}{\sin w/w} \] has a power series with rational coefficients.
Moreover, \(h\) is even, so there are rational numbers \(c_j\) such that \[ h(w) = \sum_{j=0}^{\infty}c_jw^{2j} \] near \(0\). Since \[ \pi\cot(\pi z) = \frac{h(\pi z)}z, \] we obtain \[ F_n(z) = \frac{h(\pi z)}{z^{n+1}} = \sum_{j=0}^{\infty} c_j\pi^{2j}z^{2j-n-1}. \] The residue is the coefficient of \(z^{-1}\). Because \(n\) is even, this occurs exactly when \[ 2j=n. \] Thus \[ r_n = c_{n/2}\pi^n \] with \(c_{n/2}\in\QQ\).
7. Therefore\[ \boxed{ \frac{\sum_{m=1}^{\infty}m^{-n}}{\pi^n}\in\QQ. } \]
By step 5, \[ \sum_{m=1}^{\infty}\frac1{m^n} = -\frac{r_n}{2}. \] By step 6, \[ r_n=c_{n/2}\pi^n \] with \(c_{n/2}\in\QQ\). Hence \[ \frac{\sum_{m=1}^{\infty}m^{-n}}{\pi^n} = -\frac12c_{n/2} \in \QQ. \] This proves part (b).
8. Q.E.D.
Steps <1>1–<1>2 prove part (a), and steps <1>3–<1>7 prove part (b).