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Use the Fourier transform convention $\hat{F}(k)={\int }_{\mathbb{R}}F(x)e^{-ikx}dx$. First compute the Fourier coefficients of the periodized sum. Absolute integrability givesThus Fubini's theorem justifies termwise integration. For integer $m$, the change of variable $x=2\pi j+\tau$ and $e^{2\pi imj}=1$ giveReindexing 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 $x\ne 0$. 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 $\hat{F}(m)=0$ 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 sumsconverge 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 obtainIts 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:Thus the fully justified requested sum is

Use the Fourier transform convention $\tilde{F}(\omega )={\int }_{\mathbb{R}}F(x)e^{-i\omega x}dx$. Reindexing the convergent periodization of an integrable function gives $g(\tau +2\pi )=g(\tau )$; absolute integrability also ensures that the defining sum is absolutely convergent almost everywhere in a period. To calculate its Fourier series coefficients, uniform convergence permits passing the limit of the partial sums through the finite-interval integral. Changing variables $x=\tau +2\pi k$ and using $e^{2\pi ink}=1$ givesThe sum of these integrals is absolutely convergent, bounded by ${\int }_{\mathbb{R}}|F|$, so the last rearrangement is justified. This proves the proposed Fourier expansion at the level of coefficients.

There is a genuine missing hypothesis in the pointwise assertion as printed. Take $F(x)=1_{\{0\}}(x)$. It is absolutely integrable with integral zero, and its periodization has only finitely many nonzero terms on $[0,2\pi ]$, so the series converges uniformly there. Yet $g(0)=1$ while $\tilde{F}(n)=0$ for every $n$. Thus the displayed pointwise identity does not follow from the stated assumptions. Additional regularity ensuring Fourier convergence to the prescribed values of $g$, for example a continuous piecewise continuously differentiable periodization, makes the coefficient identity into the intended Poisson summation formula.

The particular function $F(x)=e^{-|x|}$ satisfies that stronger condition. Direct integration yieldsOn $0\le \tau \le 2\pi$, summing the two geometric tails gives $g(\tau )=\cosh (\tau -\pi )/\sinh \pi$, a continuous periodic function which is piecewise smooth. Its Fourier series has absolutely summable coefficients and converges to $g$. At $\tau =0$,Consequently

The Poisson summation formula equates sample sums of a function and its Fourier transform. Exchanging broad and narrow sums can save work. Use the conventionFor every Schwartz function $f\in \mathcal{S}(\mathbb{R})$,Both series have absolute convergence. A sufficient hypothesis is that $f$ is smooth and ${\sup }_x(1+|x|{)}^M|f^{(r)}(x)|<\infty$ for all nonnegative integers $M,r$; its Fourier transform is then also a Schwartz function.^[ref] This assumption controls both decay and regularity; mere membership of an L1 space does not control values at individual sampling points, as periodization of an integrable function illustrates.

Here $\xi$ counts cycles per unit length. In the angular convention used elsewhere in Fourier analysis, $\tilde{f}(\omega )=\int f(x)e^{-i\omega x}dx$, so $\hat{f}(\xi )=\tilde{f}(2\pi \xi )$ and the right-hand side becomes ${\sum }_k\tilde{f}(2\pi k)$. Keeping the Fourier transform convention explicit prevents stray $2\pi$ factors. Changing the sample spacing leads to the Poisson summation formula for a Euclidean lattice.