方法能傳精神難，善用方法不簡單。

天生精神自己有，博學廣思勤鍛鍊。

Kline 教授說︰

$s_{2n} = \sum \limits_{\nu=1}^{\infty} \frac{1}{{\nu}^{2 n}} = {(-1)}^{n-1} \frac{{(2 \pi)}^{2 n}}{2 (2n)!} B_{2 n}$

此處 $B_{2n}$ 為白努利數。

是歐拉最好的勝利。

作者不知歐拉如何證明得到，不敢亂說。一位知道『留數定理』者

Residue theorem

In complex analysis, the residue theorem, sometimes called Cauchy’s residue theorem, is a powerful tool to evaluate line integrals of analytic functions over closed curves; it can often be used to compute real integrals as well. It generalizes the Cauchy integral theorem and Cauchy’s integral formula. From a geometrical perspective, it is a special case of the generalized Stokes’ theorem.

Illustration of the setting.

The statement is as follows:

Let $U$ be a simply connected open subset of the complex plane containing a finite list of points $a 1, \dots, a n$ , and $f$ a function defined and holomorphic on $U \ {a1,…,an}$ . Let $γ$ be a closed rectifiable curve in $U$ which does not meet any of the $a k$ , and denote the winding number of $γ$ around $a k$ by $I(γ, a k)$ . The line integral of $f$ around $γ$ is equal to $2π i$ times the sum of residues of $f$ at the points, each counted as many times as $γ$ winds around the point:

\oint _{\gamma }f(z)\,dz=2\pi i\sum _{k=1}^{n}\operatorname {I} (\gamma ,a_{k})\operatorname {Res} (f,a_{k}).

If $γ$ is a positively oriented simple closed curve, $I(γ, a k) = 1$ if $a k$ is in the interior of $γ$ , and 0 if not, so

\oint _{\gamma }f(z)\,dz=2\pi i\sum \operatorname {Res} (f,a_{k})

with the sum over those $a k$ inside $γ$ .

The relationship of the residue theorem to Stokes’ theorem is given by the Jordan curve theorem. The general plane curve $γ$ must first be reduced to a set of simple closed curves ${γ i}$ whose total is equivalent to $γ$ for integration purposes; this reduces the problem to finding the integral of $f dz$ along a Jordan curve $γ i$ with interior $V$ . The requirement that $f$ be holomorphic on $U0 = U \ {ak}$ is equivalent to the statement that the exterior derivative $d (f dz) = 0$ on $U 0$ . Thus if two planar regions $V$ and $W$ of $U$ enclose the same subset ${a j}$ of ${a k}$ , the regions $V \ W$ and $W \ V$ lie entirely in $U 0$ , and hence

\int _{V\backslash W}d(f\,dz)-\int _{W\backslash V}d(f\,dz)

is well-defined and equal to zero. Consequently, the contour integral of $f dz$ along $γ j = \partialV$ is equal to the sum of a set of integrals along paths $λ j$ , each enclosing an arbitrarily small region around a single $a j$ — the residues of $f$ (up to the conventional factor $2π i$ ) at ${a j}$ . Summing over ${γ j}$ , we recover the final expression of the contour integral in terms of the winding numbers ${I(γ, a k)}$ .

In order to evaluate real integrals, the residue theorem is used in the following manner: the integrand is extended to the complex plane and its residues are computed (which is usually easy), and a part of the real axis is extended to a closed curve by attaching a half-circle in the upper or lower half-plane, forming a semicircle. The integral over this curve can then be computed using the residue theorem. Often, the half-circle part of the integral will tend towards zero as the radius of the half-circle grows, leaving only the real-axis part of the integral, the one we were originally interested in.

倒是可以藉著

$\pi \cot(\pi z) = \lim \limits_{N \to \infty} \sum \limits_{n = -N}^{N} {(z-n)}^{-1}$

$= \frac{1}{z} + \sum \limits_{n = 1}^{\infty} \frac{2 z}{z^2 - n^2}$

Example 2

The fact that $π cot(π z)$ has simple poles with residue one at each integer can be used to compute the sum

\displaystyle \sum _{n=-\infty }^{\infty }f(n).

Consider, for example, $f (z) = z -2$ . Let $Γ N$ be the rectangle that is the boundary of $[- N - 1 / 2, N + 1 / 2] 2$ with positive orientation, with an integer $N$ . By the residue formula,

{\frac {1}{2\pi i}}\int _{\Gamma _{N}}f(z)\pi \cot(\pi z)\,dz=\operatorname {Res} \limits _{z=0}+\sum _{n=-N \atop n\neq 0}^{N}n^{-2}

The left-hand side goes to zero as $N \to \infty$ since the integrand has order $O (N -2)$ . On the other hand,^[1]

{\frac {z}{2}}\cot \left({\frac {z}{2}}\right)=1-B_{2}{\frac {z^{2}}{2!}}+\cdots ,\,B_{2}={\tfrac {1}{6}}.

(In fact, $z / 2 cot(z / 2) = iz / 1 - e - iz - iz / 2$ .) Thus, the residue $Res z = 0$ is $- π 2 / 3$ . We conclude:

\sum _{n=1}^{\infty }{\frac {1}{n^{2}}}={\frac {\pi ^{2}}{6}}

which is a proof of the Basel problem.

The same trick can be used to establish

\pi \cot(\pi z)=\lim _{N\to \infty }\sum _{n=-N}^{N}(z-n)^{-1}

that is, the Eisenstein series.

We take $f (z) = (w - z) -1$ with $w$ a non-integer and we shall show the above for $w$ . The difficulty in this case is to show the vanishing of the contour integral at infinity. We have:

\int _{\Gamma _{N}}{\frac {\pi \cot(\pi z)}{z}}\,dz=0

since the integrand is an even function and so the contributions from the contour in the left-half plane and the contour in the right cancel each other out. Thus,