Veit Elser: h(z)=e^(-2^(2z))-e^(-2^(2z+1)) Make a plot of h. Convince yourself it vanishes exponentially for z -> -\infty and super-exponentially for z -> +\infty. Now look back at the formula for f(y) as a semi-infinite sum of h(m-y) and convince yourself that f(y) approaches a periodic function of y for large y. In fact, the periodic limit is exactly the periodic function you get by extending the sum to be doubly-infinite: p(y)=\sum_{m=-\infty}^\infty h(m-y) This is where you use the Poisson summation formula, to express p(y) in terms of its Fourier amplitudes: p(y)=\sum_{k=-\infty}^\infty e^(i 2pi k y) a_k a_k=\int_{-\infty}^{+\infty} e^(i 2pi k y) h(y) So the Fourier amplitudes of the periodic limit (in the logarithmically stretched variable y) is just the Fourier transform of h(y) evaluated at integers. Apart from a_0=1/2, these turn out to be very small: a_1=0.0000789616 + 0.00137234 i --WDS: brilliant. Now, change variables u=4^z and z=log_4(u) and du/u=(ln4)dz then h(z) =e^(-u)-e^(-2u) and then a_k=\int_{-\infty}^{+\infty} e^(i 2pi k y) h(y) dy = (ln4) \int_{-\infty}^{+\infty} u^(i 2pi k/ln4) [exp(-u) - exp(-2u)] h(y) du/u which now can be expressed in terms of gamma functions, this is just Euler's integral definition of the gamma function with a few minor tweaks. So Elkies was right that all the a_k can be expressed in closed form using gamma function of imaginary arguments.