Show $\sum_{n=0}^\infty\frac{1}{a^2+n^2}=\frac{1+a\pi\coth a\pi}{2a^2}$

Question

How to show the following equality? $$\sum_{n=0}^\infty\frac{1}{a^2+n^2}=\frac{1+a\pi\coth a\pi}{2a^2}$$

Answer 1

Note that

$$\sum_{n=-\infty}^\infty f(n)= -\sum_{j=1}^k \operatorname*{Res}_{z=j}\pi \cot (\pi z)f(z) $$

Assume $a \neq 0$.

To find the residues of $g(z) := \pi \cot (\pi z)\frac{1}{a^2+n^2}$, we see

$$\frac{1}{a^2+n^2} = \frac{1}{(n+ia)(n-ia)}$$

so $g$ has poles at $z_1 = ia$ and $z_2 = -ia$. Their respective residues, $b_1$ and $b_2$ can be found:

$$b_1 = \operatorname*{Res}_{z=ia}\,g(z) = \lim_{z \to ia} \pi \cot (\pi z)\frac{(z-ia)}{(z+ia)(z-ia)} = \pi \cot (\pi i a)\frac{1}{2ia} = -\frac{\pi \coth (\pi a)}{2a}$$

$$b_2 = \operatorname*{Res}_{z=-ia}\,g(z) = \lim_{z \to -ia} \pi \cot (\pi z)\frac{(z+ia)}{(z+ia)(z-ia)} = -\pi \cot (-\pi i a)\frac{1}{2ia} = -\frac{\pi \coth (\pi a)}{2a}$$

And finally:

$$\sum_{k=-\infty}^\infty \frac{1}{a^2+k^2} = -(b_1+b_2)=\frac{\pi \coth (\pi a)}{a}$$

To change the starting number from $-\infty$ to $0$, we divide the series, as it is symmetrical (i.e. $g(n)=g(-n)$):

$$ \sum_{k=-\infty}^\infty \frac{1}{a^2+k^2}= \frac{\pi \coth (\pi a)}{a}=\\ \sum_{k=-\infty}^{-1} \frac{1}{a^2+k^2}+\frac{1}{a^2}+\sum_{k=1}^\infty \frac{1}{a^2+k^2}=\\ \frac{1}{a^2}+2\sum_{k=1}^\infty \frac{1}{a^2+k^2}=\\ \frac{1}{a^2}+2\left(\sum_{k=0}^\infty \frac{1}{a^2+k^2}-\frac{1}{a^2}\right)=\\ 2\sum_{k=0}^\infty \frac{1}{a^2+k^2}-\frac{1}{a^2} $$

Thus

$$\sum_{k=0}^\infty \frac{1}{a^2+k^2} = \frac{\pi \coth (\pi a)}{2a}+\frac{1}{2a^2} = \frac{\pi a\coth (\pi a)+1}{2a^2}$$

Answer 2

Related problems: (I), (II). This problem is a direct application of Fourier transform and Poisson summation formula. Recalling the definition of Fourier transform and the Poisson summation formula respectively

$$ F(w) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} f(x) e^{-ixw} dx \,,$$

$$ \sum_{-\infty}^{\infty} f(n) = \sqrt{2\pi}\sum_{-\infty}^{\infty} F(2n\pi)\,, $$

where $F$ is the Fourier transform of $f$. Advancing with our problem, first, we compute the Fourier transform of $ f(x)=\frac{1}{x^2+a^2} $ which is equal to

$$ F(w) = \sqrt{\frac{\pi}{2}}\frac{1}{a}e^{-a|w|}\,.$$

Applying Poisson formula, we have

$$ \sum_{n=0}^{\infty}\frac{1}{n^2+a^2} = \frac{\pi}{a}\sum_{n=0}^{\infty}e^{-2an\pi} = \frac{\pi}{a} \sum_{n=0}^{\infty}r^{n}=\frac{\pi}{a}\frac{1}{1-r}\,,\quad r = e^{-2 \pi a} \,,$$

$$\Rightarrow \sum_{n=0}^{\infty}\frac{1}{n^2+a^2} = \frac{\pi}{a} \frac{1}{1-e^{-2a\pi}}=\frac{\pi}{a} \frac{e^{2a\pi}}{e^{2a\pi}-1} \,. $$

Now, I leave it to you to manipulate the above expression to reach the form

$$ \sum_{n=0}^\infty\frac{1}{a^2+n^2}=\frac{1+a\pi\coth a\pi}{2a^2} $$

You can use the identity

$$ \coth x = \frac{\cosh x}{\sinh x} = \frac {e^x + e^{-x}} {e^x - e^{-x}} = \frac{e^{2x} + 1} {e^{2x} - 1} \,. $$

Answer 3

This is what I have from an essay I wrote. I don't know if there's a more elementary way (or if it's completely correct).

Consider $f(z) = \dfrac{\cot{\pi z}}{z^2 + k}$. This will have residues at $z = \pm i \sqrt{k}$, and at $z = n$ for $n \in \mathbb{Z}$. At $z = n$, we can compute the residues as \begin{align*} \textrm{Res}_{z=n} f(z) & = \lim_{z \rightarrow n} \dfrac{(z-n) \cot{\pi z}}{z^2 + k} = \lim_{z \rightarrow n} \dfrac{(z-n)}{(z^2 + k) \tan{\pi z}} \\ & = \lim_{z \rightarrow n} \dfrac{1}{\pi (z^2 + k) \sec^2{\pi z} + 2z \tan{\pi z}} \\ & = \dfrac{1}{\pi (n^2 + k)}. \end{align*} We can calculate the residues at $z = \pm i \sqrt{k}$: $\displaystyle \textrm{Res}_{z=i\sqrt{k}} f(z) = \lim_{z\rightarrow i\sqrt{k}}\dfrac{(z-i\sqrt{k})\cot{\pi z}}{z^2 + k}$.

This equals:

$\lim_{z \rightarrow i\sqrt{k}} \dfrac{\cot{\pi z}}{z + i\sqrt{k}} = \dfrac{\cot{\pi i\sqrt{k}}}{2i\sqrt{k}}.$

It can be shown that the residue at $z = -i \sqrt{k}$ is the same, because $\cot{\pi z}$ is an odd function. And so the residue contribution from the two poles at $z = \pm i \sqrt{k}$ is

$-\dfrac{\cot{\pi i \sqrt{k}}}{i\sqrt{k}} = -\dfrac{1}{2\sqrt{k}} \dfrac{e^{2\pi \sqrt{k}} + 1}{e^{2\pi \sqrt{k}} - 1}$.

Hence, we have

$\displaystyle \int_\gamma f(z) dz = 2\pi i \left(\sum_{n \in \mathbb{Z}} \dfrac{1}{\pi(n^2 +k)} -\dfrac{1}{2\sqrt{k}} \dfrac{e^{2\pi \sqrt{k}} + 1}{e^{2\pi \sqrt{k}} - 1}\right)$.

It is tempting for the left-hand side to go to zero, which we can arrange. Take the large square contour centered at the origin with sidelength $2R$. Observe that since

$\cot{z} = i\dfrac{e^{2iz} + 1}{e^{2iz}-1}$,

in the limit as $|z| \geq R \rightarrow \infty$, we will have $|\cot{z}| \rightarrow 1$ since the numerator and denominator of $\cot{z}$ grow equally fast. Moreover, we have that:

$|z^2 + k| \geq |z^2| \geq R^2$,

and so the maximum modulus of $f(z)$ on $\gamma$ is $1/R^2$. By the ML-inequality, we have that

$\left|\displaystyle \int_\gamma f(z) dz\right| \leq 8R \cdot \dfrac{1}{R^2}$.

So as $R \rightarrow \infty$, the integral goes to zero. And thus, \begin{align*} \sum_{n \in \mathbb{Z}} \dfrac{1}{\pi(n^2 +k)} -\dfrac{1}{2\sqrt{k}} \dfrac{e^{2\pi \sqrt{k}} + 1}{e^{2\pi \sqrt{k}} - 1} & = 0\\ \sum_{n \in \mathbb{Z}} \dfrac{1}{\pi(n^2 +k)} & = \dfrac{1}{2\sqrt{k}} \dfrac{e^{2\pi \sqrt{k}} + 1}{e^{2\pi \sqrt{k}} - 1} \\ \sum_{n=1}^\infty \dfrac{1}{(n^2 +k)} & = \dfrac{\pi}{2\sqrt{k}} \dfrac{e^{2\pi \sqrt{k}} + 1}{e^{2\pi \sqrt{k}} - 1} - \dfrac{1}{2k}. \end{align*}

Taking $k = a^2$, this formula becomes

$\dfrac{a \pi \coth{\pi a} -1}{2a^2}$.

Hmm.. not sure about -1 or +1.