= Solution
Use the <Fourier transform> convention $\widehat F(k)=\int_{\mathbb R}F(x)e^{-ikx}\,dx$. First compute the <Fourier coefficients> of the periodized sum. Absolute integrability gives
$$
\sum_{j\in\mathbb Z}\int_0^{2\pi}|F(2\pi j+\tau)|\,d\tau=\int_{\mathbb R}|F(x)|\,dx<\infty.
$$
Thus <Fubini's theorem> justifies termwise integration. For integer $m$, the change of variable $x=2\pi j+\tau$ and $e^{2\pi imj}=1$ give
$$
c_m=\frac1{2\pi}\int_0^{2\pi}f(\tau)e^{-im\tau}\,d\tau
=\frac1{2\pi}\sum_j\int_{2\pi j}^{2\pi(j+1)}F(x)e^{-imx}\,dx
=\boxed{\frac{\widehat F(m)}{2\pi}}.
$$
Reindexing a convergent periodization gives periodicity; the <integral> argument always gives its periodic representative almost everywhere. This proves the Fourier-coefficient content of the <Poisson summation formula>.
There is a genuine qualification to the printed pointwise claim: the <periodization of an integrable function> need not agree pointwise with its <Fourier series> under the stated hypotheses alone. Set $F(0)=1$ and $F(x)=0$ for $x\ne0$. This is absolutely integrable, and on one closed period the periodization has only two possibly nonzero summands, so the series is uniformly convergent. Yet $f(0)=1$, while $\widehat F(m)=0$ for every $m$. The claimed pointwise equality at zero is therefore false. This counterexample also works for Riemann integrability, not only for an almost-everywhere convention.
The coefficient identity gives a rigorous general replacement: the weighted <Fejér sums>
$$
\sigma_Nf(\tau)=\frac1{2\pi}\sum_{|m|\leq N}\left(1-\frac{|m|}{N+1}\right)\widehat F(m)e^{im\tau}
$$
converge to the periodization in the <integral> norm, and at every <continuity> point. Indeed they are its convolution with the normalized <Fejér kernel>; that kernel is nonnegative, has <integral> one, and has <integral> tending to zero outside every fixed neighbourhood of zero. For convergence in the <L1 norm>, split the convolution error into small translations, controlled by <continuity> of translation in the <integral> norm, and the remaining tail. The same split gives <pointwise convergence> at a <continuity> point. If the periodization is piecewise continuously differentiable, the ordinary <Fourier series> has the usual midpoint limit; if it is continuous and its coefficients are absolutely summable, the displayed unweighted formula holds pointwise. These are sufficient interpretations of the intended identity.
For the specified exponential example all the needed <Fourier series> convergence properties do hold. Split its <Fourier transform> at zero to obtain
$$
\widehat F(k)=\int_0^\infty e^{-(1+ik)x}\,dx+\int_0^\infty e^{-(1-ik)x}\,dx
=\frac{2}{1+k^2}.
$$
Its periodization is continuous, with uniformly geometrically bounded tails, and the sampled coefficients are absolutely summable. The absolutely summable coefficients define a uniformly convergent <Fourier series> with the computed coefficients. Its continuous sum equals the continuous periodization: their difference has every <Fourier coefficient> zero, hence is zero in the <L2 norm> by Fourier completeness, and continuity upgrades that to equality everywhere. Therefore its ordinary <Poisson summation formula> is valid. Evaluate it at zero:
$$
\sum_{j\in\mathbb Z}e^{-2\pi|j|}=1+\frac{2e^{-2\pi}}{1-e^{-2\pi}}=\coth\pi
=\frac1\pi\sum_{n\in\mathbb Z}\frac1{1+n^2}.
$$
Thus the fully justified requested sum is
$$
\boxed{\sum_{n=-\infty}^{\infty}\frac1{1+n^2}=\pi\coth\pi}.
$$
Back to article page
