4
$\begingroup$

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}

$\endgroup$
  • $\begingroup$ But $i\varepsilon=ia\varepsilon$ for any $a>0$. $\endgroup$ – AccidentalFourierTransform Mar 20 '16 at 13:05
  • 1
    $\begingroup$ For $a$ constant I might understand...but $x$ is the integration variable..I am probably missing your point entirely... $\endgroup$ – Smerdjakov Mar 20 '16 at 13:23
  • $\begingroup$ I think your not missing my point at all: for every $x\in(0,\infty)$ you can use $i\varepsilon=ix\varepsilon$ point-wise. $\endgroup$ – AccidentalFourierTransform Mar 20 '16 at 13:35
  • 1
    $\begingroup$ According to what I understand of what you say, one could reformulate the Sohotski-Plemenj theorem as $$ \lim_{\epsilon \to +0} \int \mathrm{d}x f(x)/(x+i \epsilon x) = P.V. \int \mathrm{d}x f(x)/x -i \pi f(0) $$. I trying to convince myself...In the meantime, thanks a lot.. $\endgroup$ – Smerdjakov Mar 20 '16 at 16:15
  • 1
    $\begingroup$ Crossposted from math.stackexchange.com/q/1701964/11127 $\endgroup$ – Qmechanic Mar 20 '16 at 18:55
3
+100
$\begingroup$

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

$\hspace{200pt}$enter image description here

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 $(1)$): $$ \int_0^\infty\mathrm dx\ \mathrm e^{-\beta x^2}\frac{1}{-\alpha+x(P-i\varepsilon)}=\frac{1}{P}\left[i\pi\mathrm e^{-\beta\alpha^2/P^2}+\mathcal P\int_0^\infty \frac{\mathrm e^{-\beta x^2/P^2}}{x-\alpha}\right] $$

$\endgroup$
  • $\begingroup$ That is very helpful...it made me realise that I have a new problem, i.e. how to perform an integral over the positive real semiaxis of an exponential function with an argument with positive real part..I have edited the post to highlight this fact..thanks a lot.. $\endgroup$ – Smerdjakov Mar 20 '16 at 21:04
  • $\begingroup$ I'm glad I could help :) anyway, note that $\int_0^\infty \mathrm dx\ \mathrm e^{-ax}$ for complex $a$ is only defined if $\text{Re}(a)>0$. Otherwise the integral is divergent. (The imaginary part is irrelevant for convergence) $\endgroup$ – AccidentalFourierTransform Mar 20 '16 at 21:28
  • $\begingroup$ Thanks for confirming this. Iam struggling then to understand how can then the function $Z(P)$, equation (8) in the paper (reported for everybody's convenience at the end of my original post, be holomorphic in the entire complex plane except a branch point from $(-\infty,0]$ as stated in the paper, if it is not even defined when P has a negative real part..thanks again $\endgroup$ – Smerdjakov Mar 20 '16 at 21:37
  • 1
    $\begingroup$ It makes sense now. Thanks a lot, truly much appreciated. $\endgroup$ – Smerdjakov Mar 22 '16 at 21:54
  • 1
    $\begingroup$ @AccidentalFourierTransform Ya that is what I meant. Nice answer. $\endgroup$ – Brian Moths Mar 22 '16 at 22:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.