 Solution

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