This is another question about the problem of collecting of the logariths.
Suppose the function
F[s2_, Ss_, t1_, m_, m2_] := Evaluate[-6 ((s2^6 + (Ss - 7 t1) s2^5 + (-4 Ss^2 + 13 t1 Ss + 16 t1^2) s2^4 + 2 (Ss^3 - 13 t1 Ss^2 - 10 t1^2 Ss - 8 t1^3) s2^3 + t1 (18 Ss^3 + 6 t1 Ss^2 - 4 t1^2 Ss + 7 t1^3) s2^2 + t1^2 (18 Ss^3 + 22 t1 Ss^2 + 11 t1^2 Ss - t1^3) s2 + Ss t1^3 (2 Ss^2 + 2 t1 Ss - t1^2)) m^6 - Ss (s2 - t1) ((11 t1 - 2 Ss) s2^4 + 2 (Ss^2 - 16 t1 Ss - 7 t1^2) s2^3 + 6 (4 Ss^2 t1 - t1^3) s2^2 + 2 t1^2 (15 Ss^2 + 16 t1 Ss + 5 t1^2) s2 + t1^3 (4 Ss^2 + 2 t1 Ss - t1^2)) m^4 + Ss^2 (s2 - t1)^2 (s2^4 - (Ss + 9 t1) s2^3 + 3 (3 Ss - t1) t1 s2^2 + t1^2 (15 Ss + 11 t1) s2 + Ss t1^3) m^2 + Ss^3 (s2 - t1)^3 t1 (s2^2 - 4 t1 s2 + t1^2)) Log[((-m^2 + m2^2 + s2 - Sqrt[m^4 - 2 (m2^2 + s2) m^2 + (m2^2 - s2)^2]) (s2 - t1))/(2 s2)] s2^3 + 6 ((s2^6 + (Ss - 7 t1) s2^5 + (-4 Ss^2 + 13 t1 Ss + 16 t1^2) s2^4 + 2 (Ss^3 - 13 t1 Ss^2 - 10 t1^2 Ss - 8 t1^3) s2^3 + t1 (18 Ss^3 + 6 t1 Ss^2 - 4 t1^2 Ss + 7 t1^3) s2^2 + t1^2 (18 Ss^3 + 22 t1 Ss^2 + 11 t1^2 Ss - t1^3) s2 + Ss t1^3 (2 Ss^2 + 2 t1 Ss - t1^2)) m^6 - Ss (s2 - t1) ((11 t1 - 2 Ss) s2^4 + 2 (Ss^2 - 16 t1 Ss - 7 t1^2) s2^3 + 6 (4 Ss^2 t1 - t1^3) s2^2 + 2 t1^2 (15 Ss^2 + 16 t1 Ss + 5 t1^2) s2 + t1^3 (4 Ss^2 + 2 t1 Ss - t1^2)) m^4 + Ss^2 (s2 - t1)^2 (s2^4 - (Ss + 9 t1) s2^3 + 3 (3 Ss - t1) t1 s2^2 + t1^2 (15 Ss + 11 t1) s2 + Ss t1^3) m^2 + Ss^3 (s2 - t1)^3 t1 (s2^2 - 4 t1 s2 + t1^2)) Log[((-m^2 + m2^2 + s2 + Sqrt[m^4 - 2 (m2^2 + s2) m^2 + (m2^2 - s2)^2]) (s2 - t1))/(2 s2)] s2^3] I need to collect the logarithms with the same prefactors to the form Log[.../...]. If I call already statically defined function F inside the construction
Simplify[F[s2_, Ss_, t1_, m_, m2_]]/. -Log[x_] + Log[y_] -> Log[y/x] the logarithms become to be collected. But if the function F is defined dynamically (in my case - because of an integration, see the precise example below), the construction doesn't work. I.e., for
F[] := Evaluate[Simplify[Integrate[]]/. -Log[x_] + Log[y_] -> Log[y/x]] F[] it doesn't work.
Only the construction FullSimplify works, but it requires so many time that I doesn't have the wish to wait. The Simplify construction, on the contrary, works very fast.
So, does anyone know some alternative ways to force Mathematica to collect the logarithms, especially working for dynamically defined functions?
I'm using Mathematica 11.
Precise example when Simplify doesn't work, but FullSimplify does
Just to illustrate the problem completely, I add the integral:
f[Ss_,s2_,t1_,t2_,m_,m2_] =1/(32 \[Pi]^3 Ss^3 (s2-t1) (m^2-s2+t1-t2)) (-(((s2^2 t2-s2 t2 (Ss+t1)+Ss t1 (t1-s2)-Ss t1 t2) (s2^2-6 s2 (Ss-t1+t2)+Ss^2+2 Ss (t2-t1)-(t1-t2)^2))/(s2-t1)^2)+(s2-Ss) (s2^2+2 s2 (Ss-2 t1+2 t2)+Ss^2+(t1-t2)^2)-(2 (2 s2-2 Ss+t1-t2) (-2 Ss t1 t2 (s2-t1) (2 s2^2-2 s2 (Ss+t1)-Ss t1)+t2^2 (s2^4-2 s2^3 (Ss+t1)+s2^2 (Ss-t1)^2+4 s2 Ss t1 (Ss+t1)+Ss^2 t1^2)+Ss^2 t1^2 (s2-t1)^2))/(s2-t1)^4-(2 (s2^2 t2-s2 t2 (Ss+t1)+Ss t1 (t1-s2)-Ss t1 t2) (2 Ss t1 t2 (s2-t1) (-4 s2^2+4 s2 (Ss+t1)+Ss t1)+t2^2 (s2^4-2 s2^3 (Ss+t1)+s2^2 (Ss^2-6 Ss t1+t1^2)+8 s2 Ss t1 (Ss+t1)+Ss^2 t1^2)+Ss^2 t1^2 (s2-t1)^2))/(s2-t1)^6) F[Ss_,s2_,t1_,m_,m2_]=Assuming[s2> (m+m2)^2 && s2-t1 > 0 && m > 0 && m2 > 0,Simplify[Integrate[f[Ss,s2,t1,t2,m,m2],{t2,t2lower,t2upper}] ]]/.-Log[x_]+Log[y_]:> Log[y/x] Here, the code defining the limits of integration is
lambda[a_, b_, c_] := a^2 + b^2 + c^2 - 2*(a*b + a*c + b*c) t2lower = Simplify[m^2 - 1/(2*s2)*(s2 - t1)*(s2 + m^2 - m2^2) - 1/(2*s2)*(s2 - t1)*Sqrt[lambda[s2, m^2, m2^2]]] t2upper = Simplify[m^2 - 1/(2*s2)*(s2 - t1)*(s2 + m^2 - m2^2) + 1/(2*s2)*(s2 - t1)*Sqrt[lambda[s2, m^2, m2^2]]] For this construction, the logs collecting including Simplify doesn't work. Changing it to FullSimplify works, but it takes a plenty of time. I want to avoid this.
FullSimplifyworks butSimplifydoes not? The origin of the expression should not matter, only its content. $\endgroup$Evaluateand defineFwithSet, i.e.F[ . . . ] = defand (2) you should be usingRuleDelayedwhen you have named patterns on the left-hand-side, unless you specifically know otherwise, i.e.-Log[x_] + Log[y_] :> Log[y/x]$\endgroup$