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There are many badly defined integrals in physics. I want to discuss one of them which I see very often. $$\int_0^\infty \mathrm{d}x\,e^{i p x}$$ I have seen this integral in many physical problems. Many people seem to think it is a well defined integral, and is calculated as follows:

We will use regularization (we introduce a small real parameter $\varepsilon$ and after calculation set $\varepsilon = 0$.

$$I_0=\int_0^\infty \mathrm{d}x\,e^{i p x}e^{ -\varepsilon x}=\frac{1}{\varepsilon-i p}=\frac{i}{p}$$

But I can obtain an arbitrary value for this integral! I will use regularization too, but I will use another parametrization:

$$I(\alpha)=\int_0^\infty \mathrm{d}x\,e^{i p x}=\int_0^\infty dx \left(1+\alpha\frac{\varepsilon \sin px}{p}\right)e^{i p x}e^{ -\varepsilon x}$$ where $\varepsilon$ is a regularization parameter and $\alpha$ is an arbitrary value using $\int_0^\infty \mathrm{d}x\,\sin{(\alpha x)} e^{ -\beta x}=\frac{\alpha}{\alpha^2+\beta^2}$

After a not-so-difficult calculation I obtain that $I(\alpha)=\frac{i}{p}\left(1+\frac{\alpha}{2}\right)$.

This integral I have often seen in intermediate calculation. But usually people do not take into account of this problem, and just use $I_0$. I don't understand why?

I know only one example when I can explain why we should use $I_0$. In field theory when we calculate $U(-\infty,0)$, where $U$ is an evolution operator, It is proportional to $\int^0_{-\infty} \mathrm{d}t\,e^{ -iE t}$. It is necessary for the Weizsaecker-Williams approximation in QED, or the DGLAP equation in QCD, because in axiomatic QFT we set $T\to \infty(1-i\varepsilon)$.

My question is: Why, in calculation of the integral $\int_0^\infty \mathrm{d}x\,e^{i p x}$, do people use $I_0$? Why people use $e^{ -\varepsilon x}$ regularization function? In my point of view this regularization no better and no worse than another.

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    $\begingroup$ You have discovered the difference between physics and mathematics. $I_0$ just works better. $\endgroup$ – Wolphram jonny Nov 22 '14 at 5:33
  • $\begingroup$ Many smart people have worked on trying to formalise the regularisation technique self consistently. Eventually though, to make sense of anything in physics you have to compare with experiment. Calculation of the Casimir force by using Zeta functions is the famous example here. One must then choose a specific parameterisation to obtain an answer that agrees. As for your $I_{0}$ specifically, I could not say. I have never been really satisfied with the 'divergence removing' techniques much either. $\endgroup$ – Arthur Suvorov Nov 22 '14 at 5:42
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    $\begingroup$ Amazingly, it doesn't matter how you regularize, as long as you do so consistently. The finite part of the regularisation of the divergent series/integral will always agree, see this post from Terry Tao. $\endgroup$ – ACuriousMind Nov 22 '14 at 14:12
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    $\begingroup$ I hope this adds another little clarification. Step function: $\mathbb{1}_{[0,+\infty)}(x)\in \mathscr{S}'(\mathbb{R})$, that gives $1$ if $x\in[0,+\infty)$, $0$ otherwise. Fourier transform maps $\mathscr{S}'$ into itself. The Fourier transform of the step function above uniquely defines your integral, as a distribution on $\mathscr{S'}$; see here for an explicit form. The ambiguities mathematically rise from the fact that "exchanging the limit with the integral" is not permitted in this case. $\endgroup$ – yuggib Nov 22 '14 at 19:29
  • $\begingroup$ Crossposted to math.stackexchange.com/q/1507474/11127 $\endgroup$ – Qmechanic Nov 1 '15 at 20:28
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When you introduce an auxiliary variable, such as a regularization parameter, at the end of the calculation you have to take the limit that sets the expression back to the original one. If you introduce multiple auxiliary variables, you have to do this for all of them. Otherwise you're just doing a different integral. In this case specifically,

$$\begin{align} I(\alpha) &=\int_0^\infty \mathrm{d}x\,e^{i p x} \\ &=\lim_{\alpha\to 0}\lim_{\varepsilon\to 0}\int_0^\infty dx \left(1+\alpha\frac{\varepsilon \sin px}{p}\right)e^{i p x}e^{ -\varepsilon x} \\ &= \lim_{\alpha\to 0} I(\alpha) \\ &= I(0) = I_0 \end{align}$$

Without the $\lim_{\alpha\to 0}$, you're not calculating $\int e^{ipx}$, you're calculating something else.

Sure, if you put the $\varepsilon$ limit inside the integral then it looks like the value of $\alpha$ doesn't matter, but you can't actually make that conclusion, because limits don't commute with integrals in general. And clearly, in this case, the value of $\alpha$ does matter.

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    $\begingroup$ $\alpha$ is not a regularization parameter. It is arbitrary number.I mean that $\left(1+\alpha\frac{\varepsilon \sin px}{p}\right)e^{ -\varepsilon x}$ is a regularization function for arbitrary $\alpha$ $\endgroup$ – Peter Nov 22 '14 at 6:50
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    $\begingroup$ Yes, but the point of my answer is that $\bigl(1 + \alpha\frac{\varepsilon\sin px}{p}\bigr)e^{ipx}e^{-\epsilon x}$ is not the same as $e^{ipx}e^{-\epsilon x}$. It's a different integrand and integrating it will give you a different result. This is true whether you call $\alpha$ a regularization parameter or not. $\endgroup$ – David Z Nov 22 '14 at 6:53
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    $\begingroup$ @DavidZ I think Peter's point is that both integrands are equal to $e^{ipx}$ when you let $\varepsilon \rightarrow 0$. $\endgroup$ – childofsaturn Nov 22 '14 at 7:16
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    $\begingroup$ @Ahsan but my point is precisely that the OP has not demonstrated that there are many different non-equivalent choices. Only the choice where $\alpha=0$ corresponds to the original integral. $\endgroup$ – David Z Nov 23 '14 at 5:56
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    $\begingroup$ No. In mathematical point of view both regularization function are normal. This integral is bad define, because the answer dependence on regularization function $\endgroup$ – Peter Nov 23 '14 at 6:28
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Something that fixes $\frac{i}{p}$ uniquely is that, independently of regulator, it is the constant term of the asymptotic expansion of $$\int_0^b \mathrm{e}^{ipx} \,\mathrm{d}x = \frac{i - \mathrm{e}^{ipb}}{p}$$

Moreover, consider applying your regulator with $\alpha \neq 0$ to a case where the integral does converge, such as $p=i$. Does it still yield the correct answer? Naively, one might think the second term in the integrand doesn't matter because $\varepsilon \rightarrow 0$ so $\frac{\varepsilon \sin p x}{p} \rightarrow 0$. The same would be true in this convergent scenario. But it clearly does matter because $\alpha \neq 0$ yields the wrong answer. Thus you must let $\alpha \rightarrow 0$ at the end, or your method will be inconsistent with convergent integrals.

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