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

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

• You might want to take a look at this link – EuYu Oct 6 '12 at 19:10

It is well known 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

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

• Another interesting thing to do with this sum is to start it from $k=0$ and then let $a \to 0$ to get $$\zeta(2) = \frac{\pi^2}{6}$$ – Argon Oct 7 '12 at 13:27
• +1, great answer, but why you didn't take the pole at z=0 when you calculate the residues ? also at wolfram they didnt take the pole at z=0 but why ? – mnsh Sep 14 '13 at 18:47
• @hmedan.mnsh The residue at zero is $\frac{1}{a^2+0^2}$, i.e. it is a term of the sum. – Argon Sep 15 '13 at 12:49
• A rigorous treatment should include the proof of the vanishing of the contour as it approaches infinity, which this proof lacks. – Hans Feb 7 '18 at 8:19
• @mugiwara Move the $k=0$ term to the RHS and let $a\to0$. The left side becomes the sum $1/1^2+1/2^2+\cdots$ and the right side becomes $\pi^2/6$. – Argon Dec 4 '20 at 13:07

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

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

• @Byron: Thanks for the edit. – Mhenni Benghorbal Feb 19 '13 at 17:24
• +1 Nice answer, by the way! – user940 Feb 19 '13 at 17:25
• @ByronSchmuland: Thanks for the comment. – Mhenni Benghorbal Feb 19 '13 at 17:41
• Excuse me, how could you compute that $F(w) = \sqrt{\frac{\pi}{2}}\frac{1}{a}e^{-a|w|}\,$ ? – user39843 Mar 12 '13 at 3:01
• @user39843: See here for a related problem. – Mhenni Benghorbal Mar 13 '13 at 22:29

Now, a real analytic proof. This one has no flaws (I hope).

Lemma 1. Integration by parts gives: $$\frac{1}{a}\int_{0}^{+\infty}\cos(n x)\,e^{-a x}\,dx = \frac{1}{a^2+n^2} = \int_{0}^{+\infty}\frac{\sin(n x)}{n}\,e^{-a x}\,dx.$$

Lemma 2. The series $$\sum_{n=1}^{+\infty}\frac{\sin(nx)}{n}$$ converges on $\mathbb{R}\setminus 2\pi\mathbb{Z}$ to the function: $$f(x) = \pi\left(\frac{1}{2}-\left\{\frac{x}{2\pi}\right\}\right).$$

Lemma 3. The dominated convergence theorem hence gives: $$\sum_{n=1}^{+\infty}\frac{1}{a^2+n^2}=\pi\int_{0}^{+\infty}\left(\frac{1}{2}-\left\{\frac{x}{2\pi}\right\}\right)e^{-ax}\,dx,$$ and by splitting $[0,+\infty)$ as $[0,2\pi)\cup[2\pi,4\pi)\cup\ldots$ we have:

$$\sum_{n=1}^{+\infty}\frac{1}{a^2+n^2}=\frac{e^{2a\pi}}{e^{2a\pi}-1}\int_{0}^{2\pi}\frac{\pi-x}{2}e^{-ax}dx=\frac{\pi a \coth(\pi a)-1}{2a^2}.$$

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.

• Thanks a lot. Its $-1$ if you start at $n=1$ and $+1$ if you start at $n=0$. – Spenser Oct 6 '12 at 19:13
• Ah, I didn't see that. Awesome then! – notes Oct 6 '12 at 21:11
• Your contour has some problems. For example "we will have $|\cot z|\to 1$" is not true when $R$ is a generic real number as $\cot(\pi z)$ blows up at $z = n$ where $n$ is an integer. If you restrict $R$ to be on the form $R = N + \frac{1}{2}$ with $N$ integer then $\cot(\pi z)$ is bounded (by $2$) on this contour and everything should be fine. – Winther Sep 12 '16 at 11:49

This one is a proof I gave when I was attending my high school, before studying complex analysis. It is a bit flawed, but just a little.

Step 1. If $p(x)$ is a real polynomial satisfying $p(0)=1$ and its roots are simple and real, $$\sum_{\xi:p(\xi)=0}\frac{1}{\xi}=-\frac{p'(0)}{p(0)}$$ follows from Vieta's theorem.

Step 2. All the roots of $\frac{\sin x}{x}$ are simple and real. Moreover, $$\frac{\sin x}{x}=\prod_{n=1}^{+\infty}\left(1-\frac{x^2}{\pi^2 n^2}\right)$$ holds. It is the Weierstrass product for the sine function.

Step 3. $\{a^2+1,a^2+2^2,a^2+3^2,\ldots\}$ is the zero set of the function: $$f(x)=\frac{\sinh\left(\pi\sqrt{a^2-x}\right)}{\pi\sqrt{a^2-x}}.$$

Step 4. Since $$f(0)=\frac{\sinh(\pi a)}{\pi a},\qquad f'(0)=-\frac{\cosh(\pi a)}{2a^2}+\frac{\sinh(\pi a)}{2\pi a^4},$$ Step 1 gives:

$$\sum_{n=1}^{+\infty}\frac{1}{n^2+a^2}=\frac{\pi a \coth(\pi a)-1}{2a^2}.$$

Known issues: the determination of the square root function and the fact that we can treat $\frac{\sin x}{x}$ like an "infinite degree" polynomial with known roots. Beyond the naif approach, this shows that the Vieta's theorem for polynomials and the residue theorem for meromorphic functions are very closely related.

• You used these tools while you were in High School? That is impressive. – Mark Viola Jun 17 '16 at 19:58
• @Dr.MV: at that time, this proof took me about a month, but I was overly proud of it :D – Jack D'Aurizio Jun 17 '16 at 20:14
• And rightfully so. – Mark Viola Jun 17 '16 at 20:47