Ytukyg

I've already see a proof by Marko Riedel which I list it follows:

The standard way to treat these sums is to integrate
$$ f(z) = frac{1}{(z+alpha)^2} pi cot(pi z)$$
along a contour consisting of a circle of radius $R$ and with $R$ going to infinity and hence being larger than $alpha$, where the circle does not pass through the poles on the real axis.
Now along the semicircle in the upper half plane we have
$$|f(z)| le frac{1}{(R-|alpha|)^2}pi
left|frac{e^{ipi Rexp(itheta)} + e^{-ipi Rexp(itheta)}}
{e^{ipi Rexp(itheta)} - e^{-ipi Rexp(itheta)}}right|=
frac{1}{(R-|alpha|)^2} pi
left|frac{e^{2ipi Rexp(itheta)}+1}{e^{2ipi Rexp(itheta)}-1}right| <
frac{1}{(R-|alpha|)^2} pi
frac{1+e^{-2pi Rsin(theta)}e^{2ipi Rcos(theta)}}
{1-e^{-2pi Rsin(theta)}e^{2ipi Rcos(theta)}}$$
This last term is clearly $O(1/R^2)$ as $R$ goes to infinity as the quotient of the two exponentials goes to one since $exp(-R)$ vanishes and there is no singularity when $theta = 0$ or $theta = pi$ as $Rcostheta = pm R$, which is not an integer by the assumption that the circle avoids the poles and hence cannot be one.

My question:

It is clear that for each $z=Re^{i theta}$ , $f(z)=O(1/R^2)$ ，but can this guarantee that the last term on RHS is uniform bounded ? Since $theta$ varies on a compact interval $[0, pi]$ , I want to show that for each $ theta in [0, pi]$ , and a fixed positive number $M gt 1 $ there exist an open ball contains $theta$ and a fixed $R_{theta}gt 0$ , for every $R ge R_{theta}$ and $x in $ the open ball $$frac{1+e^{-2pi Rsin(x)}e^{2ipi Rcos(x)}}
{1-e^{-2pi Rsin(x)}e^{2ipi Rcos(x)}} le M$$ For $theta neq 0,pi$ , it is easy to find the desired $R_{theta}$ and the open ball , but if $theta = 0$ or $ theta = pi$ , for every open ball containing $theta$ , it behave erratically near $theta$ , I have no idea how to deal with this .

With the quoted proof being unsatisfactory we try again. With the
goal of evaluating

$$sum_{n=-infty}^infty frac{1}{(u+n)^2}$$

where $u$ is not an integer we study the function

$$f(z) = frac{1}{(u+z)^2} picot(pi z).$$

which has the property that with $S$ being our sum,

$$S = sum_n mathrm{Res}_{z=n} f(z) = sum_n frac{1}{(u+n)^2}.$$

We examine what happens when we integrate $f(z)$ along the rectangle
$$Gamma =pm (N+1/2) pm i N$$ with $N$ a large positive
integer.

There are no poles on this contour and seeing that
$frac{1}{(u+z)^2}inTheta(1/N^2)$ on the contour, the integral
$$int_Gamma f(z) dz$$ goes to zero as $N$ goes to infinity. This
will be shown below.

This implies that the sum of the residues at the poles of $f(z)$ is
zero, giving

$$S + mathrm{Res}_{z=-u} frac{1}{(u+z)^2} picot(pi z) = 0.$$

The residue at the double pole at $z=u$ is given by

$$left.(pi cot(pi z))'right|_{z=-u} =
left. - frac{pi^2}{sin(pi z)^2} right|_{z=-u}$$

so that we have

$$bbox[5px,border:2px solid #00A000]{
sum_n frac{1}{(u+n)^2} = frac{pi^2}{sin(pi u)^2}.}$$

To convince yourself that the integral really does vanish consider the
two lines $Gamma_1$ which is $N+1/2pm iN$ (right vertical) and
$Gamma_2$ which is $pm N+1/2+iN$ (top horizontal). With $N$ going to
infinity we may suppose that $Ngt 1$ and also $Ngt max(|Re(u)|,
|Im(u)|).$

We parameterize $Gamma_1$ with $z=N+1/2+it$ so that
$$left|int_{Gamma_1} f(z) dz right|
= left|int_{-N}^N f(N+1/2+it) i dtright|.$$

The norm of the fractional term attains its maximum at $t=0$ when we
cross the real axis where $|u+z|$ is minimized, giving an upper bound
on the norm which is

$$frac{1}{(Re(u)+N+1/2)^2}
= frac{1}{N^2} frac{1}{(1+(Re(u)+1/2)/N)^2}.$$

For the norm of the trigonometric term we get
$$|picot(pi (N+1/2) + pi it)|
=pileft|frac{e^{ipi (N+1/2) - pi t}+e^{-ipi (N+1/2) + pi t}}
{e^{ipi (N+1/2) - pi t}-e^{-ipi (N+1/2) + pi t}}right|
\ = pileft|frac{i(-1)^N e^{- pi t} - i(-1)^N e^{pi t}}
{i(-1)^N e^{- pi t} + i(-1)^N e^{pi t}}right|
= pi|tanh(pi t)|.$$

Observe that with $t$ real $pi tanh(pi t)$ has no poles and its
norm is bounded by $pi$. Therefore the norm of the integral along
$Gamma_1$ is bounded by

$$2N times frac{pi}{(Re(u)+N+1/2)^2}
in 2N times Theta(1/N^2) = Theta(1/N)$$

and the integral vanishes as $N$ goes to infinity as claimed.

For $Gamma_2$ we parameterize with $z = t + i N$ so that
$$left|int_{Gamma_2} f(z) dz right|
= left|int_{-(N+1/2)}^{N+1/2} f(t+iN) dtright|.$$

The norm of the fractional term is minimized when we cross the
imaginary axis, giving an upper bound on the norm which is

$$frac{1}{(Im(u)+N)^2}
= frac{1}{N^2} frac{1}{(1+Im(u)/N)^2}.$$

For the norm of the trigonometric term we get
$$|picot(pi t + pi i N)|
= pileft|frac{e^{ipi t - pi N} + e^{-ipi t + pi N}}
{e^{ipi t - pi N} - e^{-ipi t + pi N}}right|
le pileft|frac{e^{pi N}+e^{-pi N}}{e^{pi N}-e^{-pi N}}right|
=pi|coth(pi N)|.$$

There aren't any poles here either and this term is bounded above by
$picoth(pi)$ because $N>1.$ This gives the following bound on the
norm of the integral along $Gamma_2:$

$$(2N+1) times frac{picoth(pi)}{(Im(u)+N)^2}
in (2N+1) times Theta(1/N^2) = Theta(1/N)$$

and this integral also vanishes as $N$ goes to infinity as claimed.

The other two line segments can be bounded by the same technique.