# On a alternate series representation of Riemann xi function

In a recent paper Dan Romik proved the following alternating infinite series representation for Riemann xi function:

I may be wondering if we can transform this infinite alternating sum into Abel Plana alternate Summation Formula as :

$$\sum_{k=0}^∞ (-1)^nf(k) = (1/2)f(0) + i\int_0^∞\frac{f(iy)−f(−iy)}{2\sinh(πy))} dy$$ ?

(Coefficients seems to follow the growth condition for Abel Plana)

Your formula has nothing special. The Hermite functions $$h_n(t)=e^{-x^2/2} H_n(t),H_n(t)=e^{t^2}\frac{d^n}{dt^n} e^{-t^2}$$ are an orthogonal basis of $$L^2$$ thus $$e^{-x^2/2}\Xi(t)=\sum_n \frac{c_n}{\|h_n\|_2^2} h_n(t), \qquad c_n= \int_{-\infty}^\infty e^{-t^2/2}\Xi(t) h_n(t)dt$$ Since $$e^{-t^2/2} \Xi(t)$$ is even and $$H_{2n+1}$$ is odd then $$c_{2n+1}=0$$.
$$\Xi(t)$$ is the Fourier transform of $$\Phi$$ and the Fourier transform of $$\frac{d^{2n}}{dt^{2n}} e^{-t^2}$$ is $$\sqrt{\pi}(ix)^{2n}e^{-x^2/4}$$ thus $$c_{2n}=\int_{-\infty}^\infty \Xi(t)\frac{d^{2n}}{dt^{2n}} e^{-t^2}dt=\sqrt{\pi}\int_{-\infty}^\infty \Phi(x) \sqrt{\pi}(ix)^{2n}e^{-x^2/4}dx$$ No it can't be turned into an Abel-Plana sum.