Imaginary Part of the Free Energy - Sohotski Plemenj theorem I have posted this question already on Math Stack Exchange and I hope not to annoy the community if I post it here again, looking maybe for a better suited audience.
I need to understand how the following Limit is calculated (equation (9) in the paper; also this).
$$    \lim_{\epsilon \to +0}Z(-P+i\epsilon) =  \lim_{\epsilon \to +0}  \int_{0}^{\infty}   \mathrm{d}x \, \int_{0}^{\infty} \mathrm{d}y \exp \{ -\alpha y -\beta x^2 - y(-P+i \epsilon)x    \}$$
The author says use is to be made of the Sokhotski–Plemelj theorem
$$ \lim_{\epsilon \to +0} \int \mathrm{d}x f(x)/(x+i \epsilon) = P.V. \int \mathrm{d}x f(x)/x -i \pi f(0) $$
but how to get there?
My best attempt is to integrate on the $y$ variable, obtaining
$$  Z(-P)   \lim_{\epsilon \to +0}  \int_{0}^{\infty}   \mathrm{d}x \,  \exp \{ -\beta x^2 \} \left [- \frac{1}{-\alpha-(-P+i \epsilon)x}    \right]$$
[EDIT: This integration step completely neglects the fact that the integral does not converge for the real part of P less than zero...But then, the author states the function $Z(P)$ is holomorphic in the entire complex plane except a branch cut from negative infinity to zero. This is maybe where I go wrong. Is the function $$Y(P) = \int_{0}^{\infty} e^{-Px}\mathrm dx $$ defined everywhere in the complex plane other than the negative semi-axis?]
To use the Theorem I need the denominator to contain a factor $x+i \epsilon$, which is not as the variable of Integration is multiplied bz $\epsilon$, so I am unable to proceed. I am even more puzzled because if I set $ \epsilon$ to Zero, the integral I wrote coincides with the P.V. integral reported in the paper as the P.V. contribution in the Sohotski-Plemelj formula (last line equation (9). The term corresponding to $i\pi f(0)$ contains $P$ as argument of an exponential, which also I completely fail to understand.
Any help please, even only a faint hint, would be most appreciated.
[EDIT] I think it is fair to report the original derivation I am trying to understand, equation (9) of the above cited paper. My question is how this derivation works: how can the exponential with positive argument be integrated (as occurs when P is positive, which is physically the case of interest), and how exactly use is made of the Sohotski-Plemenj theorem. Also, how can the author claim $Z(P)$ is holomorphic in the entire complex plane except a branch cut from negative infinity to zero, when the real part of the argument of the exponential is positive (as -P is preceded by the minus sign).
The original equation contains a number of constants: for clarity's sake and following advice, I have set them to 1, whenever this is possible. 
\begin{align}Z(-P) &= \lim_{\epsilon \to 0+} Z(-P+i\epsilon) \\ &= \int_{0}^{\infty}\mathrm{d}l \int_{0}^{\infty}\mathrm{d}v \exp\left[-(2\ l + \frac{1 }{4(1-\sigma^2)}v^2+\frac{l (-P+i\epsilon)}{2}v)\right] \\&=  \,\, P.V.\int_{0}^{\infty}\mathrm{d}v\frac{\exp\left[-( v^2)/4(1-\sigma^2)\right]}{4 -  P v} - \frac{2i}{ P}\exp\left[\frac{-4  }{ P^2(1-\sigma^2)}\right]\end{align}
 A: You can restate the Sokhotski–Plemelj theorem as
$$
\int_a^b\frac{f(x)}{x-y-i\varepsilon \color{red}{g(x)}} \mathrm dx=i\pi f(y)+\mathcal P\int_a^b\frac{f(x)}{x-y}\mathrm dx \tag{1}
$$
for any monotonic non-zero function $g(x)$ and any $y\in(a,b)$. In your case, you would take $g(x)=x$ and $y=\alpha$, but any other (well-behaved) function $g(x)$ will do as well.
For example, take
$$
\begin{aligned}
f(x)&\equiv\mathrm e^{-x^2}\sin(x-1)\Gamma(x+3)\\
a&\equiv-\gamma_\mathrm{E}=-0.577...\\
b&\equiv\phi=1.618...\\
g(x)&\equiv\log(2+x^2)
\end{aligned}
$$
If we define
$$
\xi(y)\equiv\left|\int_a^b\frac{f(x)}{x-y-i\varepsilon g(x)} \mathrm dx-i\pi f(y)-\mathcal P\int_a^b\frac{f(x)}{x-y}\mathrm dx\right|^2
$$
we should observe that $\xi(y)\equiv 0$. Indeed, if we evaluate the integrals numerically using Mathematica, we get
$\hspace{200pt}$
where the error is dominated by the finite value of $\varepsilon=0.005$.
To your actual question:
If I understood your question, the main issue is about the following integral:
$$
f(A)\equiv\int_0^\infty \mathrm dy\ \mathrm e^{Ay}=\frac{-1}{A} \qquad \text{Re}[A]<0
$$
where $A$ is any complex number with negative real part. The integral only converges for $\text{Re}[A]<0$, but we may insist that it is valid for any $A$, and assign $1/A$ as its value. I believe this is what the author means by analytically continuing the integral. Mathematically speaking, we can say that $1/A$ is the analitical continuation $f(A)$ outside of the convergence radius, in the same way we analitically continue the sum
$$
\sum_{i=0}^\infty x^i=\frac{1}{1-x}
$$
The actual sum is only convergent for $|x|<1$, but we may assign the value $1/(1-x)$ to it even if $|x|>1$. The philosophy is: evaluate any integral/sum in the region where it converges, and if the result makes sense in a larger region, then call that result the analytical continuation of the integral/sum (of course analytical continuation is more subtle and complex than just this).
Anyway, if we let $A=-\alpha+x(P-i\varepsilon)$, then we have
$$
\int_0^\infty\mathrm dy \ \mathrm e^{-\alpha y+xy(P-i\varepsilon)}=\frac{1}{-\alpha+x(P-i\varepsilon)}
$$
even if $-\alpha+x(P-i\varepsilon)>0$. Next, multiply by $\mathrm e^{-\beta x^2}$ on the left, and integrat over $x\in(0,\infty)$:
$$
\int_0^\infty\mathrm dx \int_0^\infty\mathrm dy \ \mathrm e^{-\beta x^2-\alpha y+xy(P-i\varepsilon)}=\int_0^\infty\mathrm dx\ \mathrm e^{-\beta x^2}\frac{1}{-\alpha+x(P-i\varepsilon)}
$$
The r.h.s. can be evaluated with the Sokhotski–Plemelj theorem, where I'll take $g(x)=x$ and $y=\alpha$ (in the notation of $(1)$):
$$
\int_0^\infty\mathrm dx\ \mathrm e^{-\beta x^2}\frac{1}{-\alpha+x(P-i\varepsilon)}=\frac{1}{P}\left[i\pi\mathrm e^{-\beta\alpha^2/P^2}+\mathcal P\int_0^\infty \frac{\mathrm e^{-\beta x^2/P^2}}{x-\alpha}\right]
$$
