I have this integral: $\int_z^1 dz_1\frac{z}{z_1(z_1 - z)} \Bigg(\ln z_1 \ln(1 - z_1) - \ln z \ln(1-z)\Bigg)$.

If I try to solve it in Mathematica it doesn't give any result, though it can solve the indefinite version of it. If I take then the limit of that result for $z_1\rightarrow z$ and $z_1\rightarrow 1$ to have the answer for the definite integral there are some infinities in separate terms, but in the whole expression they cancel. So terms like this for example: $-\ln 0 \ln z + \ln 0 \ln z$ which is obvious the infinities cancel (as they should since this integral describes physical quantity). So far I have been dealing with this problem by hand and cancel these apparent infinities term by term.
My question is: Is there a way to tell Mathematica to manipulate these terms and cancel them itself in the result?
I have tried taking the limit, but it just gives "Indeterminate" every time. I would really appreciate some help.


1 Answer 1


There seems to be no problems with MA 11.3. There are no divergencies for real values of z. One needs to wait around 40s.

 Integrate[z/(z1(z1-z)) (Log[z1]Log[1-z1]-Log[z]Log[1-z]),{z1,z,1},Assumptions->0<z<1]//Timing

Out[1]= {41.7505,-(1/6) Log[1-z] (Log[1-z]^2+3 Log[1-z] Log[z]+3 Log[z]^2
                  +6 PolyLog[2,z])+PolyLog[3,z/(-1+z)]}

It should be noted that for $0<z<1$ the integrand is real, continuous and free of singularities in the interval $z\le z_1 \le 1$. In fact $z_1=z,1$ are removable singularities. Therefore, PrincipalValue->True is not needed.

  • 3
    $\begingroup$ Very interesting, Mathematica v12 doesn't evaluate. Option PrincipalValue->True is needed. $\endgroup$ Commented Aug 17, 2020 at 11:01
  • $\begingroup$ @UlrichNeumann I think that is actually the desired behavior. As the OP states, you can do it for "arbitrary" boundaries and then take the limit. $\endgroup$
    – Natas
    Commented Aug 17, 2020 at 12:30
  • $\begingroup$ It evaluates on a Mac with either v12.1.0 or 12.1.1 without needing the option PrincipalValue->True $\endgroup$
    – Bob Hanlon
    Commented Aug 17, 2020 at 13:39

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