AlexKontorovich/PrimeNumberTheoremAnd
Source indexedtheorem · leanprover/lean4:v4.32.0
Kadiri.kadiri_thm_3_1_q1_gamma_symmetrization
PrimeNumberTheoremAnd.IEANTN.Kadiri · PrimeNumberTheoremAnd/IEANTN/Kadiri.lean:1292 to 1323
Mathematical statement
Exact Lean statement
@[blueprint
"kadiri-thm-3-1-q1-gamma-symmetrization"
(title := "$\\Gamma'/\\Gamma$ symmetrization on the critical line")
(statement := /-- For every $s \in \mathbb{C}$ with $\Re s = 1/2$,
$$ \frac{1}{2}\!\left\{
\frac{\Gamma'}{\Gamma}\!\Big(\frac{s}{2}\Big)
+ \frac{\Gamma'}{\Gamma}\!\Big(\frac{1-s}{2}\Big)
\right\}
\;=\; \Re\!\left[\frac{\Gamma'}{\Gamma}\!\Big(\frac{s}{2}\Big)\right]. $$
Used to identify the integrand of $I_3$ after shifting to the critical line
(\cite[p.~13]{Kadiri2005}, displayed equation between (14) and (15)). -/)
(proof := /-- On $\Re s = 1/2$, $1 - s = \bar s$, hence $(1 - s)/2 = \overline{s/2}$.
Since $\Gamma'/\Gamma$ has real Taylor coefficients away from its poles, it commutes
with complex conjugation: $\Gamma'/\Gamma((1-s)/2) = \overline{\Gamma'/\Gamma(s/2)}$.
Then $\tfrac{1}{2}(z + \bar z) = \Re z$ with $z = \Gamma'/\Gamma(s/2)$. To be
formalised. -/)
(latexEnv := "sublemma")
(discussion := 1544)]
theorem kadiri_thm_3_1_q1_gamma_symmetrization {s : ℂ} (_hs : s.re = 1 / 2) :
(1 / 2 : ℂ) * (digamma (s / 2) + digamma ((1 - s) / 2)) =
((digamma (s / 2)).re : ℂ)Complete declaration
Lean source
Full Lean sourceLean 4
@[blueprint "kadiri-thm-3-1-q1-gamma-symmetrization" (title := "$\\Gamma'/\\Gamma$ symmetrization on the critical line") (statement := /-- For every $s \in \mathbb{C}$ with $\Re s = 1/2$, $$ \frac{1}{2}\!\left\{ \frac{\Gamma'}{\Gamma}\!\Big(\frac{s}{2}\Big) + \frac{\Gamma'}{\Gamma}\!\Big(\frac{1-s}{2}\Big) \right\} \;=\; \Re\!\left[\frac{\Gamma'}{\Gamma}\!\Big(\frac{s}{2}\Big)\right]. $$ Used to identify the integrand of $I_3$ after shifting to the critical line (\cite[p.~13]{Kadiri2005}, displayed equation between (14) and (15)). -/) (proof := /-- On $\Re s = 1/2$, $1 - s = \bar s$, hence $(1 - s)/2 = \overline{s/2}$. Since $\Gamma'/\Gamma$ has real Taylor coefficients away from its poles, it commutes with complex conjugation: $\Gamma'/\Gamma((1-s)/2) = \overline{\Gamma'/\Gamma(s/2)}$. Then $\tfrac{1}{2}(z + \bar z) = \Re z$ with $z = \Gamma'/\Gamma(s/2)$. To be formalised. -/) (latexEnv := "sublemma") (discussion := 1544)]theorem kadiri_thm_3_1_q1_gamma_symmetrization {s : ℂ} (_hs : s.re = 1 / 2) : (1 / 2 : ℂ) * (digamma (s / 2) + digamma ((1 - s) / 2)) = ((digamma (s / 2)).re : ℂ) := by have h1s : 1 - s = (starRingEnd ℂ) s := by apply Complex.ext · rw [Complex.sub_re, Complex.one_re, Complex.conj_re, _hs] norm_num · rw [Complex.sub_im, Complex.one_im, Complex.conj_im] ring have hconj : (1 - s) / 2 = (starRingEnd ℂ) (s / 2) := by rw [map_div₀, map_ofNat, h1s] rw [hconj, digamma_conj, Complex.add_conj] push_cast ring