I have a simple Fourier transform problem, originating from mathematical physics (system of linear PDEs), which reduces to taking the integral $$ I(\alpha)\equiv\int_{-\infty}^\infty e^{ikr} \cfrac{\alpha^2 + \beta k^2}{k(k^2+\alpha^2)}\, dk $$ where $\alpha$ and $\beta$ are some free real-valued parameters. I need to compute $I(0)$. It turns out that if I simply set $\alpha=0$ in the integral above, I get an absolutely different answer than if I first compute the integral and set $\alpha\to 0$ in the final expression.

My question is as follows: why do these two procedures lead to different answers? From the physical viewpoint this means that a massless field behaves in a totally different way than a massive with infinitesimal mass, which seems unreasonable.

My attempt is as follows.

I lift the pole at $k=0$ to the upper half-plane: $$ I_\varepsilon(\alpha) \equiv \int_{-\infty}^\infty e^{ikr} \cfrac{\alpha^2 + \beta k^2}{(k-i\varepsilon)(k+i\alpha)(k-i\alpha)}\, dk \equiv \int_{-\infty}^\infty \cfrac{g(k)}{h(k)}\, dk $$ where $$ h(k) = (k-i\varepsilon)(k+i\alpha)(k-i\alpha)=k^3-i\varepsilon k^2+a^2k +i \varepsilon a^2, $$ $$ h'(k)=3k^2-2ik\varepsilon+\alpha^2 $$ I take the integral making use of the Jordan's lemma and Cauchy theorem: I choose a contour in the upper half-plane $\mathbb H$, so that the integral reduces to the sum of residues at $k=i\varepsilon$ and $k=i\alpha$: $$ I_1(\alpha)=2\pi i \lim_{\varepsilon \to 0}\left[\cfrac{g(i\varepsilon)}{h'(i\varepsilon)}+\cfrac{g(i\alpha)}{h'(i\alpha)}\right] $$ $$ =2\pi i\lim_{\varepsilon\to 0}\left[ \cfrac{\alpha^2 + \beta (i\varepsilon )^2}{3(i\varepsilon )^2-2i(i\varepsilon )\varepsilon +\alpha^2}\,e^{-\varepsilon r}+ \cfrac{\alpha^2 + \beta (i\alpha)^2}{3(i\alpha)^2-2i(i\alpha)\varepsilon +\alpha^2}\,e^{-\alpha r} \right] $$ $$ =2\pi i \left[ 1+ \cfrac{1 - \beta}{-3+1} \right]=2\pi i \cfrac{1+\beta}{2}=\pi i(1+\beta). $$ Thus, $I_1(\alpha) = \pi i(1+\beta)$. Clearly then, $\lim_{\alpha\to0}I_1(\alpha) = \pi i(1+\beta)$.

However, if I consider $$ I_2\equiv I(\alpha=0)= \int_{-\infty}^\infty e^{ikr} \cfrac{\beta}{k}\, dk = \lim_{\varepsilon \rightarrow 0}\int_{-\infty}^\infty e^{ikr} \cfrac{\beta}{k-i\varepsilon }\, dk = \lim_{\varepsilon \rightarrow 0}2\pi i \beta e^{-\varepsilon r} = 2\pi i \beta. $$ Hence, $I_2\neq \lim_{\alpha\to0}I_1$!!! Please, give a hint why this sort of thing happens. I clearly understand that in my reasoning there is a flaw -- but it escapes me.

Thank you for any help!

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    $\begingroup$ The integral under consideration diverges at the origin because of $k^{-1}$ in the denominator. Do you mean its principal value? $\endgroup$ – user64494 May 11 at 10:07

Care should be taken because of the pole at $k=0$, let me first take the principal value of the integral. I note that $I(\alpha,-r)=\bar{I}(\alpha,r)$ (complex conjugate), for convenience I will restrict myself to $r>0$.

The principal value integral evaluates to $$I(\alpha,r)=\int_{-\infty}^\infty dk\, e^{ikr} \cfrac{\alpha^2 + \beta k^2}{k(k^2+\alpha^2)}=i\pi+i\pi(\beta-1)e^{-|\alpha|r}.$$ So for $\alpha=0$ the result is $I(0,r)=i\pi\beta$. There is no discontinuity at $\alpha=0$, but there is a discontinuous derivative. The same result would have been obtained if we would have set $\alpha=0$ before carrying out the integral, because the principal value integral $\int dk e^{ikr}k^{-1}=i\pi$ for $r>0$.

Alternatively, you could shift the pole off the real axis, still taking $r>0$ the answer then becomes $$I(\alpha,r)=\lim_{\epsilon\downarrow 0}\int_{-\infty}^\infty dk\, e^{ikr} \cfrac{\alpha^2 + \beta k^2}{(k-i\epsilon)(k^2+\alpha^2)}=2i\pi+i\pi(\beta-1)e^{-|\alpha|r}.$$ So now $I(0,r)=i\pi(\beta+1)$, still continuous and with a discontinuous derivative.

We have recoved the result $I_1$, where the limit $\epsilon\downarrow 0$ is taken before the limit $\alpha\rightarrow 0$. These two limits do not commute, which is why the result $I_2$ in the OP differs from $I_1$.

  • $\begingroup$ Thank you very much for your clarifying comment, @CarloBeenakker. The source of my confusion is that I cannot understand why these limits ($\lim_{\alpha\to 0}$ and $\lim_{\varepsilon\to 0}$) don't commute and what strategy should I follow to obtain a result that does not depend on the order I take those limits. The problem is from physics field theory, so that with $\alpha$ being mass of some field, I need to be able to restore the massless limit by setting $\alpha\to0$, as we restore, for instance, the Coulomb potential $1/r$ from the Yukawa $e^{-mr}/r$ simply by letting $m\rightarrow 0$. $\endgroup$ – jonathan wolf May 10 at 21:39
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    $\begingroup$ Philosophically, you ought to assume that limits don't commute until you can prove that they do, e.g. using uniform convergence. $\endgroup$ – Nate Eldredge May 10 at 22:46
  • $\begingroup$ @NateEldredge or, even dominated/monotone convergence $\endgroup$ – vidyarthi May 14 at 12:03

It is not unreasonable that a massless field behaves in a way that is totally different from a massive one with arbitrarily small mass. Already at an elementary level, you can always perform a Lorentz transformation to the rest frame of a massive excitation; there is no such transformation for a massless one.

Whenever you encounter a mathematical ambiguity in a physics problem, it means that you have not taken into account all the necessary physics information. Physics has to tell you which order of limits is the relevant one. The $\epsilon $ prescriptions you are using usually serve to implement causality in the propagators you are evaluating - that may yield a clue. In mathematical terms: You are solving a PDE - what boundary conditions are you trying to satisfy?

Without knowing the full details of what you're calculating, one possibility is that you are considering the propagation of an actual massive particle, regardless of how small the mass is. In that case, $\epsilon $ has to be kept much smaller than $\alpha $, i.e., the order of limits is opposite to the case of the massless field. Another possibility is that you are treating the propagation of a massless particle, and merely introducing a mass as an infrared regulator at an intermediate stage. That is a rather subtle thing to do! One would then require that final physical results are not altered; e.g., that the additional polarization state induced for a photon is not counted in, say, a partition function.


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