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I) This is discussed around eq. (23.7.1) on p. 462 in Ref. 1. The task is to perform the path integral $$\tag{1} \int_{BC} [d\phi]e^{\frac{i}{\hbar}S[\phi]} ~=~\sum_{\nu}\int\! du \int_{BC_0} [d\phi_q]e^{\frac{i}{\hbar}S[\phi_{cl}+\phi_{\nu,u}+\phi_q]}$$ over fields $\phi$ with some (possible inhomogeneous) boundary conditions $BC$. This is done by ...

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Due to asymptotic freedom, the perturbation series expansion of QCD breaks down at low energy. The expansion parameter, the coupling constant, becomes too large and therefore we cannot rely on results anymore. The value of the coupling constant approaches the order of $1$ at an energy of several hundred MeV, a scale referred to as $\Lambda_{QCD}$ ...

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First of all, you can simplify this by rewriting it as an indefinite integral of $x = Z/F$ like so: $$\int_\epsilon^\infty \frac{ds}{s^3} \left ( \frac{s/2}{\sinh s/2} \right )^2 e^{-s x} = - \frac{1}{4} \int_\epsilon^\infty ds \int_0^x dx \frac{1}{\sinh^2 s/2} e^{-s x}$$ From here, you have roughly ...

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I don't get what you're asking? What it means that non-perturbative QCD is the exploration of phases of quark matter, including the quark-gluon plasma? Because, at high energies, you're at the so called asymptotic freedom regime. At this regime, your quarks interact weekly and the perturbative calculations are possible. And quark-gluon plasma is ...

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He doesn't really "state" the result without deriving it. He derives it. See e.g. this sentence on page 2: It is further shown that since the D-brane tension arises from the disk, it scales in string units as $g^{−1}$, $g$ being the closed string coupling The tension of D-branes goes like $1/g_s$ because the tension may be calculated from the disk ...

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