I'm interested in the (obviously divergent) integral

$$ \int_{-\infty}^\infty dx e^{-x f}\ ,$$

where $f$ is real. Is there any way to meaningfully assign a value to this integral? I was thinking of the following procedure:

$$\lim_{\epsilon\rightarrow 0}\int_{-\infty}^\infty dx e^{-x f- \epsilon x^2} = \lim_{\epsilon\rightarrow 0} \sqrt{\frac{\pi}{\epsilon}} e^{\frac{f^2}{4 \epsilon}} = \lim_{\epsilon\rightarrow 0} \sqrt{\frac{\pi}{\epsilon}} e^{-\frac{(if)^2}{4 \epsilon}}\ .$$

The last formula can be recognized as a limit representation of Dirac delta function. Thus the result is $\delta(if)$.

Does this make any sense? If it does, how do I interpret delta function over imaginary argument?

EDIT: Perhaps I should have explained why I'm thinking about this. The reason comes from physics and calculation of partition functions. Suppose we have a partition function involving some field $\psi$, i.e.:

$Z = \int \mathcal{D} \psi e^{-S[\psi]}$

Now suppose we have some restriction for the values of $\psi$, for example we require that $F[\Psi] = c$, where $c$ is a real number. Now we can introduce this restriction in two ways. First is to add a delta function inside the integral, i.e.:

$\delta (F[\psi] -c) = \int dx e^{- i x (F[\psi]-c)}$,

and the partition function becomes

$Z = \int \mathcal{D} \int dx \psi e^{-S[\psi] - ix(F[\psi]-c)}$

Now suppose we want to evaluate this partition function within the saddle point approximation. This amounts to just minimizing the action plus the term that comes from the restriction, i.e:

$\frac{\delta }{\delta \psi} [S[\psi] - ix(F[\psi]-c)] = 0$ .

On the other hand, since this is a minimization problem, we could introduce the restriction as Lagrange multiplier. In this case, we would be minimizing a slightly different expression:

$\frac{\delta }{\delta \psi} [S[\psi] - x(F[\psi]-c)] = 0$ ,

where $x$ is, in this case, the lagrange multiplier. The difference between the two is in the imaginary unit in front of the restriction condition. Now there seems to be some deep connection between the two methods (delta function inside the integral and Lagrange multipliers). In fact, the delta function method would amount exactly to the Lagrange multiplier method if we could interpret the divergent integral I wrote above as the Dirac delta function.

  • 2
    $\begingroup$ Doesn't make sense ;-) $\endgroup$ – Fabian Jun 24 '14 at 10:13
  • $\begingroup$ I added some explanation about why I'm considering this problem. $\endgroup$ – Echows Jun 24 '14 at 10:36
  • $\begingroup$ Yes, with the substitution $\,f=i\,t\,$ your initial (divergent!) integral may indeed be considered as the Fourier transform of $1$ as seen here and here with the result $2\pi\delta(i\,f)$ (in 'distribution theory' at least). $\endgroup$ – Raymond Manzoni Jun 24 '14 at 11:47
  • $\begingroup$ And how should I interpret $\delta(if)$? For example, suppose I want to calculate integral $\int_{-\infty}^\infty dx g(x) \delta(ix)$. I would say that the result is $-i g(0)$. Is this correct? $\endgroup$ – Echows Jun 24 '14 at 11:55
  • $\begingroup$ @Echows: the substitution $t=ix$ will usually help (even if an ambiguity remains on the sign of the result...). But your integral will be from $-i\infty$ to $+\i\infty$ and thus different. The Fourier transform of the Heaviside function $H(x)$ amay be of interest too from scattering theory. See here and this chapter. The prescription with the $\pm i\epsilon$ may help) and this file. $\endgroup$ – Raymond Manzoni Jun 24 '14 at 12:48

I like the stile. Ever since I learned about limits I found that there should be some magic way to assign a value to those limits. One way I find is to work in $\mathbb{R}[\epsilon]\equiv \frac {\mathbb{R}[x]}{ < x^2 > } $ for limits converging to zero. For divergent limits there is something in your delta idea.


Lets divide the integral in two (I suppose $f$ is a constant).

$$\operatorname{st} \int_0^\infty e^{-xf}=\frac{e^{f-xf}}{\left(e^f-1\right) }$$

$$\operatorname{st} \int_{-\infty}^0 e^{-xf}=\operatorname{st} \int_0^\infty e^{xf}=\frac{e^{xf}}{\left(e^f-1\right) }$$


$$\operatorname{st} \int_{-\infty}^{+\infty} e^{-xf}=\frac{e^{-xf } \left(e^{2 xf}-e^f\right)}{e^f-1}$$


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