A Polylogarithms integrals in Nastase QFT book

[Pnin]

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 do not see at all how that's possible.

Unfortunately, the book (first edition) seems to have some fair amount of typos, as far as I can tell. The exponent (n-1) above -1 in equation (1) should be (n-2), as on the polylogarithm wiki page noted under integral representation, no. 5.

Has someone an idea what the author does to get from equation (3) to (4)?

 

[mathman]

You should move the question to calculus forum.

 

[Pnin]

You should move the question to calculus forum.

Good idea. How do I do that? Or could some admin move my question? thank you

 

[mfb]

I moved it.

Did you try integration by parts? Take the integral of 1, take the derivative of the ln^k(t).

 

[Pnin]

I assume repeated (twice) integration by parts. That makes me believe the integral in (3) can be morphed into the integral in (1). With surface terms disappearing somehow. But where does the sum sigma come from?

 

[mfb]

Let's look at $$\int_0^1 \frac{z\ln^k(t)}{(1-tz)^2} dt$$
For k=0, we get $$\int_0^1 \frac{z}{(1-tz)^2} dt = \frac{-z}{z-1}$$
Subtracting that from 1/(1-z) we get 1, as expected.

For k>0 I confirmed the first cases with WolframAlpha (e.g. k=3). I started integration by parts but I'm not sure if the diverging first term will cancel later:

Let k>0. As the integrand diverges for t->0, set the lower bound to $\epsilon$ for now. Define $f'(t)=\frac{z}{(1-tz)^2}$, $g(t)=\ln^k(t)$. Then $$\int_\epsilon^1 f'(t)g(t) dt = [f(t)g(t)]_\epsilon^1-\int_\epsilon^1 f(t)g'(t) dt \\
= \left[\frac{z}{1-tz}\ln^k(t)\right]_\epsilon^1 - \int_\epsilon^1 \frac{k}{t} \frac{z}{1-tz}\ln^{k-1}(t) \\
= \dots$$

Edit: Forgot a 1/t term.

 

[Pnin]

Thanks, mfb. Much appreciated.

 

[Pnin]

Unfortunately, still a bit unclear.

If f'(t) = z/(1-zt)^2, shouldn't it be f(t) = z⋅(z/(1-zt)?

Also, what next? I need to integrate one more time by parts. But I have a new 1/t term. I want that term in my final result (4), not now.

Integrating by parts without the 1/t would give me k(k-1)ln^(k-2)(t)⋅ln(1-tz). And since k(k-1) =k!/(k-2)!, that starts to look good. Still, I need that 1/t term, but in the final equation, not in the step before..

And what about those z‘s. They appear only in ln(1-tz) in equation (1) and (4).