# Half-values at discontinuities in Riemann's prime counting function

Wikipedia (here) and MathWorld (here) disagree about the behaviour of Riemann's prime counting function $$\Pi(x)$$ at points of discontinuity. The function is defined by

$$\Pi(x)=\sum _{p \text{ prime} \\ p^{\alpha }

In Riemann's original paper, he stipulates that at points of discontinuity the value increases by half the value of the jump; i.e. with $$0<\delta<1$$,

$$\Pi(x)=\frac{1}{2} \left(\sum _{p^{\alpha }

Wikipedia follows this convention, MathWorld does not (as evidenced by its list of the functions first few values).

What I want to know is, does this matter? Is it simply that the function is otherwise undefined at jump-points and therefore Riemann is following the standard convention, or do the half-values have real significance to the rest of Riemann's paper?

The key finding I am thinking of is

$$\frac{\log (\zeta (s))}{s}=\int_0^{\infty } \Pi (x) x^{-s-1} \, dx$$

(which I found here, eqn. 42). Perhaps, since It's an integral, the value at the discontinuity is irrelevant?

From $$\frac{\log (\zeta (s))}{s}=\int_0^{\infty } \Pi (x) x^{-s-1} \, dx$$ the Fourier-Laplace-Mellin inversion theorem gives that $$\Pi(x)= \frac1{2i\pi} \int_{2-i\infty}^{2+i\infty} \frac{\log \zeta(s)}{s} x^sds$$ At the points of discontinuity the RHS is the half value. Then (for $$x > 1$$) some kind of residue theorem expresses the contour integral as a series over the singularities of $$\log \zeta(s)$$ (at the pole and zeros of $$\zeta(s)$$, this is called the explicit formula)
Thus you define $$\Pi(x)$$ as you want, but when needing Mellin inversion and the explicit formula you'll choose the half-value definition.
When dealing with Dirichlet convolutions it is more convenient to choose the $$\Pi(x)=\sum_{p^k\le x} 1/k$$ definition. Of course this doesn't affect $$\frac{\log (\zeta (s))}{s}=\int_0^{\infty } \Pi (x) x^{-s-1} \, dx$$.
• Hi @reuns. One further question if I may: is the integral in $\Pi(x)= \frac1{2i\pi} \int_{2-i\infty}^{2+i\infty} \frac{\log \zeta(s)}{s} x^sds$ to be considered as a contour integral or an antiderivative? Commented Oct 19, 2020 at 9:50
• (for $f$ analytic) $\int_C f(s)ds$, the contour integral of $f(s)$ along a curve $C=\{\gamma(t),t\in [0,1]\}$ means by definition $\int_0^1 f(\gamma(t))\gamma'(t)dt$. The point is that we can change the parametrization $\gamma$ and even the curve $C$ (keeping its endpoints) without changing the value of the integral. To compute those integrals we refer to an antiderivative of $f$. Here $C$ has an infinite length and $f$ has many singularities, so it's a bit more involved. Commented Oct 20, 2020 at 6:58