# Why is Lower Integration Bound 00 Versus 2−ϵ2-\epsilon in ∫∞0x−sψ′[x]dx\int_{0}^{\infty}x^{-s}\psi'[x]dx

Assume the following definitions:
ζ[s]\zeta[s] – Riemann’s zeta function.
ψ′[x]\psi'[x] – First-order derivative of second Chebyshev function.

The following integral converges to −ζ′[s]/ζ[s]-\zeta'[s]/\zeta[s] for Re[s]>1Re[s]>1.
∫∞0x−sψ′[x]dx=−ζ′[s]/ζ[s]\int_{0}^{\infty}x^{-s}\psi'[x]dx=-\zeta'[s]/\zeta[s]

The Fourier series representation for ψ′[x]\psi'[x], which consist of cosine terms, can be used to derive an infinite series representation for -ζ′[s]/ζ[s]\zeta'[s]/\zeta[s] where each term in the latter series is of the following form:
−ix−s(Γ[1−s,−ixan,k](−ixan,k)s−Γ[1−s,ixan,k](ixan,k)s)2an,k-\frac{ix^{-s}(\Gamma[1-s,-ixa_{n,k}](-ixa_{n,k})^s-\Gamma[1-s,ixa_{n,k}](ixa_{n,k})^s)}{2a_{n,k}}
Note this term cannot be evaluated at x=0x=0, and also note ψ′[x]\psi'[x] evaluates to 00 for 0\leq x<20\leq x<2, so my question is: Why is the lower integration bound 00 versus 2-\epsilon2-\epsilon in \int_{0}^{\infty}x^{-s}\psi'[x]dx\int_{0}^{\infty}x^{-s}\psi'[x]dx? My reason for asking this question was I wanted to confirm there are no hidden assumptions with respect to the lower evaluation bound of x=0x=0, since I can't evaluate the term defined above at x=0x=0. =================      ok.. \psi'(x)\psi'(x) is the distribution \sum_{p^k} \delta(x-p^k) \ln p\sum_{p^k} \delta(x-p^k) \ln p . If f \in C^\inftyf \in C^\infty then \int_0^\infty f(x) \psi'(x)dx= \sum_{p^k} f(p^k) \ln p\int_0^\infty f(x) \psi'(x)dx= \sum_{p^k} f(p^k) \ln p – user1952009 2 days ago      Is \int_{2-\epsilon}^{\infty}f(x)\psi'(x)dx=\sum_{p^k}f(p^k)Log(p)\int_{2-\epsilon}^{\infty}f(x)\psi'(x)dx=\sum_{p^k}f(p^k)Log(p) not also valid? – Steve C. 2 days ago      Of course \int_a^b f(x) \psi'(x)dx = 0\int_a^b f(x) \psi'(x)dx = 0 whenever [a-2,b-2] \cap \mathbb{N}=\emptyset[a-2,b-2] \cap \mathbb{N}=\emptyset. See the definition of the Dirac delta distribution \delta(x)\delta(x) – user1952009 2 days ago      That was my point. I take it the lower limit of 00 is perhaps used for convenience, but the lower limit 2-\epsilon2-\epsilon is mathematically more precise. – Steve C. 2 days ago      No. The point is that you have to look at the Dirac delta and the distributional derivative, also at the Fourier transform (and all the convergence problems), and stop with the "Fourier series of \psi'(x)\psi'(x)" – user1952009 2 days ago ================= 1 Answer 1 ================= (this is an answer to all your previous questions) There is what you have to know before playing with formal Fourier series "converging" to a non-periodic function.. (eluding the convergence problems) Let \chi(s) = s \int_0^\infty \sin(2 \pi x)x^{-s-1} dx= 2^s \pi ^{s-1} \sin(\pi s / 2) \Gamma(1-s)\chi(s) = s \int_0^\infty \sin(2 \pi x)x^{-s-1} dx= 2^s \pi ^{s-1} \sin(\pi s / 2) \Gamma(1-s). \zeta(s)\zeta(s) is special since \zeta(s) =\chi(s) \zeta(1-s)\zeta(s) =\chi(s) \zeta(1-s). For an arbitrary Dirichlet series F(s)F(s), all you'll get is F(s) = \frac{s}{s-b}\chi(s-b) G(s-b)F(s) = \frac{s}{s-b}\chi(s-b) G(s-b) where G(s)G(s) is a Mellin/Laplace transform : Let \displaystyle F(s) = \sum_{n=1}^\infty a_n n^{-s}\displaystyle F(s) = \sum_{n=1}^\infty a_n n^{-s}. By the Abel summation formula, with A(x) = \sum_{n < x} a_nA(x) = \sum_{n < x} a_n : F(s) = s \int_0^\infty A(x) x^{-s-1}dxF(s) = s \int_0^\infty A(x) x^{-s-1}dx Choose bb such that |A(x)x^{-b}|x^{1+\epsilon} \to 0|A(x)x^{-b}|x^{1+\epsilon} \to 0, and let h(x) = A(x)|x|^{-b}-A(-x)|x|^{-b}h(x) = A(x)|x|^{-b}-A(-x)|x|^{-b}. Note that h(x) = -h(-x)h(x) = -h(-x), and consider its Fourier transform \hat{h}(\xi) = \int_{-\infty}^\infty h(x) e^{-2 i \pi \xi x}dx = 2i\int_0^\infty A(x) x^{-b}\sin(2\pi \xi x)dx\hat{h}(\xi) = \int_{-\infty}^\infty h(x) e^{-2 i \pi \xi x}dx = 2i\int_0^\infty A(x) x^{-b}\sin(2\pi \xi x)dx By inverse Fourier transform h(x) = \int_{-\infty}^\infty \hat{h}(\xi) e^{2i \pi \xi x}d\xi h(x) = \int_{-\infty}^\infty \hat{h}(\xi) e^{2i \pi \xi x}d\xi and for x > 0x > 0 :

A(x) x^{-b} = -2i\int_0^\infty \hat{h}(\xi)\sin(2 \pi \xi x)d\xiA(x) x^{-b} = -2i\int_0^\infty \hat{h}(\xi)\sin(2 \pi \xi x)d\xi

So that
F(s) = s \int_0^\infty (-2i\int_0^\infty \hat{h}(\xi)\sin(2 \pi \xi x)d\xi)x^{b-s-1}dxF(s) = s \int_0^\infty (-2i\int_0^\infty \hat{h}(\xi)\sin(2 \pi \xi x)d\xi)x^{b-s-1}dx =-2i s \int_0^\infty \hat{h}(\xi)\int_0^\infty \sin(2 \pi \xi x) x^{b-s-1}dxd\xi
=-2i s \int_0^\infty \hat{h}(\xi)\int_0^\infty \sin(2 \pi \xi x) x^{b-s-1}dxd\xi
= -2i s \int_0^\infty \hat{h}(\xi)\xi^{1-s+b}\int_0^\infty \sin(2 \pi x) x^{b-s-1}dxd\xi = -2i s \int_0^\infty \hat{h}(\xi)\xi^{1-s+b}\int_0^\infty \sin(2 \pi x) x^{b-s-1}dxd\xi = \frac{-2i s}{s-b}\chi(s-b)\int_0^\infty \hat{h}(\xi)\xi^{1-s+b}d\xi
= \frac{-2i s}{s-b}\chi(s-b)\int_0^\infty \hat{h}(\xi)\xi^{1-s+b}d\xi
= \frac{s}{s-b}\chi(s-b) G(b-s) = \frac{s}{s-b}\chi(s-b) G(b-s)

Where G(s) = -2i\int_0^\infty x^{s-1}\hat{h}(x)dxG(s) = -2i\int_0^\infty x^{s-1}\hat{h}(x)dx