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4

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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Leaving out numerical factors, we have that $$\mathrm{d}j_A = \mathrm{Tr}(F \wedge F)$$ This already shows that we are dealing with a topological quantity, since the RHS is the second Chern character of the gauge field (or rather, the principal bundle associated to it). Now, there is also the (3D) Chern-Simons form  \omega = \mathrm{Tr}(F \wedge A - ...

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If you assume separability of the wave function, i.e., $\psi(\mathbf x)=u(x)v(y)w(z)$, you can solve the individual components separately: \begin{align} -\frac{\hbar^2}{2\mu}\frac{d^2u(x)}{dx^2}+V_1(x)u(x)&=E_1u(x)\\ -\frac{\hbar^2}{2\mu}\frac{d^2v(y)}{dy^2}+V_2(y)v(y)&=E_2v(y)\tag{1}\\ ...

2

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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No. Feynman diagrams are made by summing over the perturbative contributions of quantum amplitudes. They cannot hold non-perturbative information.

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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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