From a mathematical perspective, ostensibly dodgy distributional issues often arise because the familiar formula for the Fourier transform is not the whole story.
If a function (lets say of one variable) $f\in L^1(\mathbb R)$, then its Fourier transform $\hat f\in L^1(\mathbb R)$, and is given by
$$ \hat f(k) = \int\mathrm dx \ e^{ikx} f(x)$$
If $f\in L^2(\mathbb R)$ but $f\notin L^1(\mathbb R)$, then this integral does not converge. In such cases, the Fourier transform is obtained as follows: Let $f_n\rightarrow f$ be a sequence of functions converging to $f$, with $f_n\in L^1(\mathbb R)\cap L^2(\mathbb R)$. Then the Fourier transform of $f$ is given by
$$\hat f(k) = \lim_{n\rightarrow \infty} \hat{f_n}(k) = \lim_{n\rightarrow \infty} \int \mathrm dx \ e^{ikx} f_n(x)$$
This justifies the use of a regulator to compute e.g. the Fourier transform of the Coulomb potential in $\mathbb R^3$. The Fourier transform can be shown to be a continuous map, which means that any such sequence will produce the same result.
We can generalize even further to the case where $f$ is a tempered distribution. More specifically, $D_f$ is a map from the Schwartz space to $\mathbb R$ (or $\mathbb C$) which we write as
$$D_f[\varphi] = \int \mathrm dx \ f(x) \varphi(x)$$
We then define the Fourier transform of this distribution via
$$\hat{D_f}[\varphi] = D_f[\hat \varphi]= \int \mathrm dx \int \mathrm dk \ f(x) \varphi(k)e^{ikx} \sim D_{\hat f}[\varphi]$$
where $\hat f(k) \equiv \int \mathrm dx f(x) e^{ikx}$. Note that this expression is generically not well-defined on its own! We have assumed that we could switch the order of integration $x\leftrightarrow k$, which is typically not justified. In that sense, this $\hat f(k)$ is to be generally understood as a formal object which only truly makes sense when integrated against a Schwartz function (though of course, it may or may not end up being a perfectly well-defined function in its own right).
A standard example of the third definition comes when Fourier transforming the Heaviside distribution. We define
$$\Theta[f] := \int_0^\infty \mathrm dx\ f(x) = \int_{-\infty}^\infty \mathrm dx\ \theta(x) f(x)$$
The Fourier transform is then
$$\hat{\Theta}[f] = \Theta[\hat f]= \int_0^\infty \mathrm dk \int_{-\infty}^\infty \mathrm dx \ e^{ikx} f(x) $$
We can massage this into a more familiar form by writing
$$\hat{\Theta}[f] = \lim_{a\rightarrow\infty} \int_0^a \mathrm dk \int_{-\infty}^\infty \mathrm dx e^{ikx} f(x)=\lim_{a\rightarrow \infty} \int_{-\infty}^\infty\mathrm dx \left(\frac{e^{iax}-1}{ix}\right) f(x)$$
where we've used the fact that we can switch the order of integration inside the limit, where both integrals are finite. It is a standard exercise in contour integration to demonstrate that the right-hand side is equal to
$$\lim_{a\rightarrow \infty} \pi f(0) + \mathrm{P.V.}\int \left(\frac{e^{iax}-1}{ix}\right)f(x)$$
In the limit as $a\rightarrow \infty$, the oscillatory contribution vanishes and we are left with
$$\hat{\Theta}[f] = \int_{-\infty}^\infty\mathrm dx \left[\pi \delta(x) + i \mathrm{p.v.}\left(\frac{1}{x}\right) \right]f(x)$$
which leads us to identify the kernel of the integral as the Fourier transform of the Heaviside step function (note that different conventions yield different signs and factors of $2\pi$, as per usual with Fourier transforms).
With that out of the way, in your reference (Breuer, p. 140), the appearance of the P.V. distribution comes when defining the (one-sided) Fourier transform
$$\Gamma_{ij}(\omega)=\int_0^\infty \mathrm ds \ e^{i\omega s}\langle E_i(t) E_j(t-s)\rangle = \int_{-\infty}^\infty \mathrm ds \ e^{i\omega s} \theta(s)\langle E_i(t)E_j(t-s)\rangle$$
This is not the Fourier transform of $\theta(s)$ (as above), but it is the Fourier transform of $\theta(s)$ multiplied by the quantity $\left<E_i(t)E_j(s-t)\right>$. If the latter quantity does not decay sufficiently quickly as $s\rightarrow \infty$, then this will be a Fourier transform in the distributional sense, and we should not be overly surprised to see a P.V. show up.
Indeed this is what happens. In the subsequent example in the text, we see that the average value becomes
$$\left<E_i(t)E_j(s-t)\right>\propto \delta_{ij} \int_0^\infty \mathrm d\omega_k \omega_k^3\left[\left(1+N(\omega_k)\right)e^{-i\omega_k s}+N(\omega_k) e^{i\omega_k s}\right]$$
which is not well-defined as a function of $s$, as that integral does not converge. The divergent part requires renormalization in order to interpret properly, which the text delays until Chapter 12; the remainder is oscillatory in $s$, and the Fourier transform is to be understood in the distributional sense. Making use of the result I derived above allows us to perform the appropriate transform, yielding the desired answer.
A Note on Distributions
As mentioned, a tempered distribution is a map which eats Schwartz functions and spits out numbers. There are many ways in which we could define such a map.
- Given a polynomially-bounded function $f$, we could define a distribution $D_f:\varphi \mapsto \int_{-\infty}^\infty \mathrm dx \ f(x)\varphi(x)$ which eats a Schwartz function $\varphi$ and spits out the number obtained by integrating $\varphi$ against $f$. If we do this, we call $f$ the kernel of the distribution $D_f$. In an abuse of terminology, we often refer to $f$ itself as the distribution, but the distinction between a distribution and its kernel is conceptually important.
- We could define a distribution $\Theta:\varphi \mapsto \int_0^\infty \mathrm dx\ \varphi(x)$. On its face, this distribution is not defined via a kernel; however, we could cast it in the same form as the previous example by defining
$$\theta(x)=\begin{cases}0 & x<0 \\ 1 & x\geq 0\end{cases}$$
and then writing $\Theta: \varphi \mapsto \int_{-\infty}^\infty \mathrm dx\ \theta(x)\varphi(x)$.
- We could define a distribution $\Delta_a:\varphi \mapsto \varphi(a)$ which simply evaluates the input function at $a$. This distribution cannot be put into the same form as the above two examples because it can be shown that there is no function which could serve as the kernel of this distribution. However, we perform a slight of hand; we write down the formally meaningless set of symbols $\int_{-\infty}^\infty \mathrm dx \ \delta(x-a)\varphi(x)$ and then define this to be equal to $\varphi(a)$. We therefore write
$$\Delta_a:\varphi \mapsto \int_{-\infty}^\infty \mathrm dx \ \delta(x-a)\varphi(x) \equiv \varphi(a)$$
in an effort to continue with the same kind of notation. In other words, when you see $\delta(x-a)$ in an expression we are to understand that we are working with an object which must be integrated with respect to $x$, and when we do so, we simply evaluate whatever is next to it at the point $x=a$.
- Our final example is the strangest of all. We define a distribution which eats a function $\varphi$ and spits out
$$\lim_{\epsilon\rightarrow 0} \left[\int_{-\infty}^{-\epsilon} \mathrm dx\ \frac{\varphi(x)}{x} + \int_\epsilon^\infty \mathrm dx\ \frac{\varphi(x)}{x}\right]$$
which can be shown to be well-defined for all Schwartz functions $\varphi$. Just as the $\Delta_a$ distribution, this distribution cannot be written as the integral of a function against $\varphi$ - it involves a limiting procedure, after all. However, we engage in the same slight of hand as before. We define the set of symbols
$$\int_{-\infty}^\infty \mathrm dx\ \left[\mathrm{p.v.}\left(\frac{1}{x}\right)\right]\varphi(x) \equiv \lim_{\epsilon\rightarrow 0} \left[\int_{-\infty}^{-\epsilon} \mathrm dx\ \frac{\varphi(x)}{x} + \int_\epsilon^\infty \mathrm dx\ \frac{\varphi(x)}{x}\right]$$
where we understand $\mathrm{p.v.}\left(\frac{1}{x}\right)$ to be shorthand for a set of instructions which are to be executed when integrated - just like we understand $\delta(x-a)$ to be shorthand for the instructions "evaluate at $x=a$".
Now, when we extend Fourier transforms to distributions, it is important to understand that it is the distributions which are being transformed, not the kernels. If the distribution $D_f$ is defined by a distribution kernel $f\in L^1(\mathbb R)$, then its Fourier transform $\hat{D_f}$ is defined by $\hat f$; this is perhaps not surprising.
On the other hand, if $f(x)=1$ then the Fourier transform of $D_f$ is not defined by a kernel at all - it is (up to a factor of $2\pi$) the $\Delta_0$ distribution. In an abuse of terminology, we say that the Fourier transform of $f(x)=1$ is $2\pi\delta(x)$.
If $f(x)=\theta(x)$, the problem is even worse. The Fourier transform of $\Theta$ is a perfectly well-defined distribution, but we really have to tie ourselves in knots to define it in terms of a kernel. Doing so yields the result that the Fourier transform of $\theta(x)$ is $\pi\delta(x) + i\mathrm{p.v.}\left(\frac{1}{x}\right)$, which involves the principal value distribution "kernel" defined above.
