= Solution
Put $h(x)=(\alpha^2+x^2)^{-1}$ and form its period-one <periodization of an integrable function>
$$
H(x)=\sum_{k\in\mathbb Z}h(x+k).
$$
The <function> $h$ is not a <Schwartz function>, so justify the needed <Poisson summation formula> directly. For every $j\geq0$, its <derivative> satisfies $h^{(j)}(x)=O(|x|^{-2-j})$ as $|x|\to\infty$. Thus the <Weierstrass M-test> gives <uniform convergence> on $[0,1]$ of this <series> and each differentiated <series>. It follows that $H$ is a <smooth> period-one <periodic function>.
Absolute integrability and <Fubini's theorem> allow its <Fourier coefficients> to be computed by splitting the real line into unit intervals:
$$
\begin{aligned}
b_n&=\int_0^1H(x)e^{-2\pi inx}\,dx
=\sum_k\int_k^{k+1}h(y)e^{-2\pi iny}\,dy\\
&=\widehat h(2\pi n)=\frac\pi\alpha e^{-2\pi\alpha|n|}.
\end{aligned}
$$
These <Fourier coefficients> are absolutely summable. Their <Fourier series> defines a continuous <periodic function> with the same <Fourier coefficients> as $H$; <uniqueness of Fourier coefficients in L1> identifies the two <almost everywhere>, and continuity then identifies them everywhere. Evaluating at zero gives the rigorously justified <Poisson summation formula>
$$
\frac1{\alpha^2}+2\sum_{n=1}^\infty\frac1{\alpha^2+n^2}
=H(0)=\frac\pi\alpha\left(1+2\sum_{n=1}^\infty e^{-2\pi\alpha n}\right).
$$
Rearranging proves
$$
\boxed{\sum_{n=1}^\infty\frac1{\alpha^2+n^2}
=\frac\pi{2\alpha}-\frac1{2\alpha^2}+\frac\pi\alpha\sum_{n=1}^\infty e^{-2\pi\alpha n}\qquad(\alpha>0).}
$$
The last <series> is a <geometric series> with sum $(e^{2\pi\alpha}-1)^{-1}$. Equivalently, the left side is $\pi\coth(\pi\alpha)/(2\alpha)-1/(2\alpha^2)$, where $\coth$ is the <hyperbolic cotangent>.
Back to article page
