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I have an integral of the form

$$\int^\infty_{-\infty}\mathrm d \omega \, \frac{\omega^2}{k^2 + \gamma^2 \omega^2}$$

which diverges. This integral should have a finite value, as it must related to some physical measurement. I am trying to assign a value to the integral, kind of like how one does using regularisation. In a few papers on theoretical physics (Which is the field I am in), I have seen people use the Cauchy principal value in the form

$$-\!\!\!\!\!\!\!\int^\infty_{-\infty}\mathrm dx \, f(x) = \lim_{L \to \infty} \, \frac1{L} \int^L_{-L}\mathrm dx \, f(x)$$

but I am not sure how one deduces that from

http://en.wikipedia.org/wiki/Cauchy_principal_value

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Dividing by L seems to be an invention of your own and makes the result, if it exists, behave very differently than any (reasonably defined) integral on the real line. In any case this is not Cauchy principal value (and the only reasonable value for the integral you are interested in is $+\infty$). –  Did Sep 17 '11 at 16:05
    
I don't know if you access to this article, but prl.aps.org.ez.sun.ac.za/pdf/PRL/v89/i14/e144101 does something to this effect. Sorry, maybe I got confused by terminology, they use "principal value" in quotes, so it seems to be a theoretical physics invention perhaps? In any case, is there a sensible way to assign a value to said integral? $\omega$ is a frequency, so of course it cannot really become infinite, but there needs to be some finite cutoff... –  Karl Sep 17 '11 at 16:12
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Karl, Off-campus access is restricted to Stellenbosch University staff and currently registered students only... Anyway, the limit you want to consider is more a mean value of $f$ than an integral. –  Did Sep 17 '11 at 16:18
    
Sorry, copy/paste the wrong thing: prl.aps.org/pdf/PRL/v89/i14/e144101 . Anyways, a mean value might be what I want. Thanks. I will think a bit more about this. –  Karl Sep 17 '11 at 16:21
    
Karl, flunked again: Authorization Required. Individual Subscribers. Please log in with your APS Journals Account below. –  Did Sep 17 '11 at 16:26
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4 Answers

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Average value is not that it usually meant as a regularized value of (say) series. One of the standard ways of regularization is to construct the analytical continuation. If it diverges at the point of interest, then the constant term of the expansion at this point is taken as the regularized value. In this case it seems to work like follows. For $s>0$ consider convergent integrals $$ \int_{-\infty }^{\infty } \frac{w^2}{\left(s^2 w^2+1\right) \left(k^2+\gamma ^2 w^2\right)} \, dw=\frac{\pi }{\gamma k s^2+\gamma ^2 s}= \frac{\pi }{\gamma ^2 s}-\frac{\pi k}{\gamma ^3}+O(s), $$ $$ \int_{-\infty }^{\infty } \frac{w^2 e^{-s w^2}}{k^2+\gamma ^2 w^2} \, dw= \frac{\sqrt{\pi }}{\gamma ^2 \sqrt{s}}-\frac{\pi k}{\gamma ^3}+O\left(\sqrt{s}\right), $$ $$ \int_{-\infty }^{\infty } \frac{w^2 e^{-s|w|}}{k^2+\gamma ^2 w^2} \, dw= \frac{2}{\gamma ^2 s}-\frac{\pi k}{\gamma ^3}+O(s). $$ Note that the constant term $-\pi k/\gamma ^3$ is the same in all three cases. So it's a candidate for the regularized value.

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It is worth notice that this same value occurs if we just remove the $\frac{2L}{\gamma^2}$ term from the integral. That is, if we subtract off the main term. In other words $$\lim_{L\rightarrow\infty}\left(\int_{-L}^{L}\frac{\omega^{2}}{k^{2}+\gamma^{2}‌​\omega^{2}}d\omega-\frac{2L}{\gamma^2}\right)=\frac{-\pi k}{\gamma^3}.$$ –  Eric Naslund Sep 17 '11 at 17:35
    
@Andrew do you perhaps have any advice where I can read up more about regularization? I come from a theoretical physics background, but not really quantum field theories (even though I make heavy use of path integrals), so I would really like to learn more. EDIT: Also, when moving the integral into the complex plane, I found also $-\pi k/\gamma^3$ + a contour integral that follows a semi circle. That contour integral seemed to diverge so I just gave up using that method. –  Karl Sep 17 '11 at 21:31
    
@Karl There are different ways to regularize series/integrals. For series there is a classic book of Hardy Divergent Series archive.org/details/divergentseries033523mbp. As for using the analytical continuation it is well-known method. In Terry Tao blog terrytao.wordpress.com/2010/04/10/… it is discussed how to obtain such results staying on real line, without use of complex analysis instruments like analytic continuation. I think the same ideas are applicable to integrals. –  Andrew Sep 18 '11 at 11:30
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We can evaluate $$ \int_{-L}^{L}\frac{\omega^{2}}{k^{2}+\gamma^{2}\omega^{2}}d\omega$$ completely. Our integral is $$\frac{1}{k^{2}}\int_{-L}^{L}\frac{\omega^{2}}{1+\frac{\gamma^{2}}{k^{2}}\omega^{2}}d\omega=\frac{k}{\gamma^{3}}\int_{-\frac{\gamma L}{k}}^{\frac{\gamma L}{k}}\frac{u^{2}}{1+u^2}d\omega=\frac{2L}{\gamma^2}-\frac{k}{\gamma^3}\arctan\left(\frac{\gamma}{k}L\right).$$ The last equality follows since the anti derivative of $\frac{u^{2}}{1+u^{2}}$ is $u-\arctan(u)$.

The Average Value: From the above we can evaluate $\frac{1}{L}\lim_{L\rightarrow \infty}\frac{1}{2L} \int_{-L}^L \frac{\omega^{2}}{k^{2}+\gamma^{2}\omega^{2}}d\omega$. In particular we have $$\lim_{L\rightarrow \infty}\frac{1}{2L} \int_{-L}^L \frac{\omega^{2}}{k^{2}+\gamma^{2}\omega^{2}}d\omega =\frac{1}{\gamma^2},$$ which means that the average value of the function on the real line is $\frac{1}{\gamma^2}$. (notice I divided by $2L$ rather than $L$ because the interval is of length $2L$.)

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I would have thought $\lim_{L\rightarrow\infty}\left(u \biggr|_{-L}^L\right) = \lim_{L\rightarrow\infty}\left(2L\right)$. –  Henry Sep 17 '11 at 16:14
    
@Henry, Didier: Thanks. My mistake was the classic $L-(-L)=L-L=0$. It has been corrected. –  Eric Naslund Sep 17 '11 at 16:33
    
@eric-naslund Thanks that seems to be the correct answer for my question. –  Karl Sep 17 '11 at 16:39
    
@Karl: It may be worth noting for future reference that for this case the name is the "Expected Value" rather than the "Cauchy Principle Value." –  Eric Naslund Sep 17 '11 at 16:43
    
@eric-naslund Noted. Physicists have a tendency to abuse mathematics... Either way, I guess this is the "best" (not that there is any way to assign a measure of good/worse...) to solve my problem. Unfortunately I still want the value of the integral and not just the average value of the function I am integrating over. I guess this will be a case of "the important behaviour is captured by the mean value" as one usually encounters in fairly gross approximations... –  Karl Sep 17 '11 at 16:48
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Hint: Assume that $f(x)\to \ell_+$ when $x\to+\infty$ and $f(x)\to \ell_-$ when $x\to-\infty$. Then $$ \lim\limits_{L\to+\infty}\frac1{L}\int\limits_{-L}^Lf(x)\mathrm dx=\ell_++\ell_-. $$ In your case $f(x)=\dfrac{x^2}{k^2+\gamma^2x^2}$ hence $\ell_+=\ell_-=\dfrac1{\gamma^2}$ and the limit is $\dfrac2{\gamma^2}$.

As you can see the result is robust in the sense that it has nothing to do with whether we know exactly a primitive of $f$ or not, nor even with the exact form of $f$.

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I am not sure what $1/L$ is doing in your expression for the Cauchy principal value.

In any case, your integral is as clearly infinite as $\dfrac{\int^\infty_{-\infty} d \omega}{\gamma^2} $ is as the limit of the average value is $\dfrac{1}{\gamma^2}$.

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Sorry, My Mistake! –  Eric Naslund Sep 17 '11 at 16:17
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@Eric: I suspect the question might be asking for the Cauchy principal value of $\int^\infty_{-\infty}\mathrm d \omega \, \frac{\omega}{k^2 + \gamma^2 \omega^2}$ (note my change to the numerator) but that it is not what it says. –  Henry Sep 17 '11 at 16:17
    
I want $\omega^2$ in the numerator..either way, yes the integral diverges...I just want to "fix" it in the theoretical physics sense. (renormalization/regularization). I expect great flak from proper mathematicians... –  Karl Sep 17 '11 at 16:26
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