k=0.76422365358922066299069873125009232811679054139340951472168667374;

P=Select[Table[Prime[i],{i,1,30000}],Mod[#,4]==3&]; (*List of primes = 3 mod 4*)
PT=Table[Prime[i],{i,1,3000000}]; (*List of regular primes*)
PB[s_]:=Times@@((1-P^(-2(s+1)))) (*Euler product in B(s)*)
B[s_]:=(DirichletL[4,2,s+1]*(1-2^(-s-1))*PB[s])^(-1/2)
A[s_]:=1-2^(-s)+2^(-2s)
M[s_]:=A[s]*B[s]
PBLD[s_]:=2Total[P^(-2(s+1))Log[P]/(1-P^(-2(s+1)))] (*Logarithmic Derivative of the PB function*)
LDS[s_]:=Log[2]/(2^(s+1)-1)
BLD[s_]:=(-1/2)(LLD[s+1]+PBLD[s]+LDS[s]) (*Logarithmic Derivative of B*)
LLD[s_]:=-Total[Log[PT]*DirichletCharacter[4,2,PT]/(PT^s-DirichletCharacter[4,2,PT])] (*Logarithmic Derivative of L(s,(Subscript[χ, 4]))*)
MP[s_]:=A'[s]B[s]+BLD[s]*M[s] (*This is M'(s)*)

A0[s_]:=((1-5^(-s+1))/(s-1))

F[s_]:=Zeta[s-1]*M[s-1]*(((s-1)Zeta[s])^(1/2))Gamma[s]

F0[s_]:=A0[s]*F[s]

FP[s_]:=Zeta'[s-1]*M[s-1]*(((s-1)Zeta[s])^(1/2))Gamma[s]+Zeta[s-1]*MP[s-1]*(((s-1)Zeta[s])^(1/2))Gamma[s]+Zeta[s-1]*M[s-1]*(((Zeta[s]+(-1+s) Zeta'[s])/(2 (Sqrt[(-1+s) Zeta[s]]))))Gamma[s]+Zeta[s-1]*M[s-1]*(((s-1)Zeta[s])^(1/2))Gamma'[s] (*FP(s) = F'(s), I just computed the derivative by hand and defined FP that way*)

int0[H_]:=(1/(π*k^2))NIntegrate[(FP[s]*H^(s-1)+F[s]*H^(s-1)*(Log[H]-(A0[s]/2)))/((1-s)^(1/2)),{s,0.5+10^(-10),1}] (*This is integral for v=0 on the interval (-1/2+(10^(-10)),1). Below is the integral on a larger interval.*)

int0[10^8]

NumberForm[10^8/5+Out[47],16]

goodint0[H_]:=(1/(π*k^2))NIntegrate[(FP[s]*H^(s-1)+F[s]*H^(s-1)*(Log[H]-(A0[s]/2)))/((1-s)^(1/2)),{s,0.5+10^(-30),1},PrecisionGoal->20,WorkingPrecision->40,MaxRecursion->100] (*This is integral for v=0 on the interval (-1/2+(10^(-30)),1). Below is the integral on a larger interval.*)

goodint0[10^8] (*If we compare goodint0[10^8] with int0[10^8] we see they at least agree in the first 5 digits.*)

Lcc4[chi_]:=N[Sum[DirichletCharacter[5,chi,n]*DirichletCharacter[4,2,n]/n,{n,2*10^5}]] (*This is L(1,χ.(Subscript[χ, 4]))*)

Lcc42=Lcc4[2]
Lcc43=Lcc4[3]
Lcc44=Lcc4[4]

C5[chi_]:=(N[Times@@(1-DirichletCharacter[5,chi,P]^2/P^2)]^(-1/2))*((N[Sum[DirichletCharacter[5,chi,n]*DirichletCharacter[4,2,n]/n,{n,2*10^5}]])^(-1/2))*(DirichletL[5,chi,0])*(DirichletL[5,chi,1]^(1/2))*((1-DirichletCharacter[5,chi,2]/2)^(-1/2))*(1-DirichletCharacter[5,chi,2]+DirichletCharacter[5,chi,4]) (*This is (Subscript[C, q,χ])*)

C52=C5[2]

C53=C5[3]

C54=C5[4]

constinint[v_]:=(1/(8*k^2))(1/DirichletCharacter[5,2,v]*C52+1/DirichletCharacter[5,3,v]*C53+1/DirichletCharacter[5,4,v]*C54) (*This is the constant term which goes in front of the integral when v not equal 0*)

CI1=constinint[1]
CI2=constinint[2]
CI3=constinint[3]
CI4=constinint[4]

intv[H_]:=(1/(8*π*k^2))NIntegrate[(F0[s]*H^(s-1))/((1-s)^(1/2)),{s,0.5+10^(-30),1},PrecisionGoal->20,WorkingPrecision->40,MaxRecursion->100] (*This is integral for v not equal 0 on the interval (-1/2+(10^(-30)),1).*)

intv[10^6]

sq1h=CI1+Out[53] (*integral + constant, v=1, H=10^6*)
sq2h=CI2+Out[53] (*integral + constant, v=2, H=10^6*)
sq3h=CI3+Out[53](*integral + constant, v=3, H=10^6*)
sq4h=CI4+Out[53](*integral + constant, v=4, H=10^6*)
(*Comaparing these 4 values with those of tables 10-14 in the paper, the fit is pretty 3 or 4 digits after decimal, which agrees with RH ET*)

intv[-1/Log[1-k/Sqrt[Log[10^12]]]] (*integral for nonzero v at H=6.365...*)

sq1h6=CI1+Out[69] (*integral + constant, v=1, H=6.365...*)
sq2h6=CI2+Out[69](*integral + constant, v=2 H=6.365...*)
sq3h6=CI3+Out[69](*integral + constant, v=3 H=6.365...*)
sq4h6=CI4+Out[69](*integral + constant, v=4 H=6.365...*)
(*These are the values in Table 7 when v not 0*)

-1/Log[1-k/Sqrt[Log[10^12]]] (*H=6.365...*)

goodint0[-1/Log[1-k/Sqrt[Log[10^12]]]] (*Integral v=0 at H=6.365..., this is the value in table 7 when v=0*)