Question: how to go from equation (3) to equation (4)?
This is from Horatio Nastase "Intro to Quantum Field Theory" book (Cambridge University Press, 2019) , chapter 59. The reader is supposed to massage equation (3) into equation (4) with the help of the given polylogarithm formulas (1) and (2). I suppose I must integrate by parts twice. That gives a k(k-1) term, which equals k!/(k-2)!. Also, for k=0 the surface term from the first integration gives -z/(z-1), which substracted from 1/(1-z) gives the 1 in (4). But what about the 1/t term? And what about the z? I need the 1/t in (4) and only z in log(1-zt) in (4). Do I have to use equation (2) somewhere?
Let's drop the limits for a bit ... we shall put them in at the end.
Integrate by parts \begin{eqnarray*} \int \frac{z}{(1-zt)^2} ( \ln(t))^k dt = \frac{( \ln(t))^k}{(1-zt)} -k \int \frac{ (\ln(t))^{k-1}}{t(1-zt)} dt. \end{eqnarray*} Partial fractions \begin{eqnarray*} \frac{ 1}{t(1-zt)} = \frac{1}{t} +\frac{z}{1-zt}. \end{eqnarray*} So we have two integrals to consider \begin{eqnarray*} \int \frac{ ( \ln(t))^{k-1}}{t} dt = \frac{( \ln(t))^k}{k}. \end{eqnarray*} And the other integral \begin{eqnarray*} \int ( \ln(t))^{k-1} \frac{z }{1-tz} dt = ( \ln(t))^{k-1} \ln(1-zt)- (k-1) \int ( \ln(t))^{k-2} \ln(1-zt) dt. \end{eqnarray*} Putting these together & including the limits, gives \begin{eqnarray*} \int_{0}^{1} \frac{z}{(1-zt)^2} ( \ln(t))^k dt = \left[ \frac{( \ln(t))^k}{(1-zt)} -( \ln(t))^k-k ( \ln(t))^{k-1} \ln(1-zt) \right]_0^1 \\ +k(k-1) \int ( \ln(t))^{k-2} \ln(1-zt) dt. \end{eqnarray*} There is some subtle cancellations with the lower limit, but the whole square bracket evaluates to zero & we are done.