Abel–Plana formula

Remarks

Proof idea: the kernels $1/(e^{\mp 2\pi i z} - 1)$ have simple poles at the integers. Apply the residue theorem to $f$ times such a kernel on large rectangles in the upper and the lower quarter planes, with one side on the imaginary axis. The residues at the integers produce the sum, the decay of $f$ kills the far sides, and the sides on the imaginary axis give the last integral.

Example: $f(z) = 1/(z + 1)^2$ satisfies the hypotheses with $\varepsilon = 1$, since $\abs{z + 1} \ge \abs{z}$ when $\operatorname{Re} z \ge 0$. Here $f(it) - f(-it) = -4it/(1 + t^2)^2$, so the formula reads $$\frac{\pi^2}{6} = 1 + \frac12 + 4 \int_0^\infty \frac{t}{(1 + t^2)^2 (e^{2\pi t} - 1)}\,dt.$$ Applied to $f(z) = (z + a)^{-s}$ it gives an integral formula for the Hurwitz zeta function.

Some decay or growth condition is needed for both sides to make sense: for $f = 1$ the series and the integral $\int_0^\infty f$ both diverge.

Going deeper

For a student

The Abel–Plana formula is an exact alternative to the Euler–Maclaurin formula for functions that extend holomorphically to the right half-plane. Euler–Maclaurin gives an asymptotic series with Bernoulli numbers, while Abel–Plana gives one convergent integral. Expanding $f(it) - f(-it)$ in powers of $t$ and integrating term by term recovers the Euler–Maclaurin terms, since $\int_0^\infty t^{2k-1} / (e^{2\pi t} - 1)\,dt$ is a multiple of a Bernoulli number. It is a standard tool for the zeta functions and for the analytic continuation of series in physics, such as the computation of Casimir energies.