I have recently run into a number of divergent oscillating integrals in various contexts. Thus, I have been led to desire general methods for assigning values to divergent oscillating integrals. All of the integrals I am interested in have the following form $$ \int_0^\infty f(x) \sin(x) dx \text{ or } \int_0^\infty f(x) \cos(x) dx $$ Where $f(x)$ is usually an eventually monotonically increasing function.


In the case where $f(x)$ grows like a polynomial I believe there are various approaches that all provide the same value. For instance, consider
$$ \begin{split} \int_0^\infty \sin(x) dx &= \frac{1}{2}\int_0^\infty \left(\sin(x) + \sin(x)\right) dx \\ &= \frac{1}{2}\int_0^{\pi} \sin(x)dx + \frac{1}{2}\int_0^\infty \left(\sin(x) + \sin(x + \pi)\right) dx = 1. \end{split}$$ Alternatively, in an analgous manner to applying a smooth cutoff function to a divergent series, we can also apply a smooth cutoff function to get $$ \lim_{\varepsilon\to 0} \int_0^\infty \sin(x) (1-\varepsilon)^{1+x} dx = 1$$ An analogue to Cesaro summation for integrals also provides the same values for this integral.

We can apply these method to higher powers of $x$ as well. Doing this generates the following values $$ \int_0^\infty x^n \sin(x) = \cos\left( \frac{n \pi}{2}\right) \Gamma(n+1), \int_0^\infty x^n \cos(x) = -\sin\left( \frac{n \pi}{2}\right) \Gamma(n+1)$$

Thus, if $f(x)$ has a power series presentation, we can write $$\int_0^\infty f(x) \sin(x) = \int_0^\infty \sum_{n=0}^\infty \frac{f^{n}(0)}{n!} x^n \sin(x) = \sum_{n=0}^\infty (-1)^n f^{2n}(0)$$ This formula is a good start, but the series typically doesn't converge.

Is there a general way to assign a value to divergent oscillating integrals, especially those where $f(x)$ grows at an exponential rate or faster? Since it seems unlikely that there might be a way to assign a value to a general function-- are there intereting/broad categories of functions which can be assigned values?


2 Answers 2


There is indeed such a method which is elementary and allows one to rigorously give a numerical value to such integrals, in particular, the value $1$ to $\int_0^\infty \cos x\, dx$. This uses two facts:

  1. Every continuous function has a primitive. In your case this is $\cos x$.

  2. About 60 years ago, a notion of the limit of a distribution at a point (in your case at $0$ and $\infty$) was developed in an elementary context (i.e., at thelevel of a good calculus or real single variable course).

Combining these concepts in the usual manner as in defining improper Riemann integrals, one obtains a notion of definite integrals of distributions which contains the classical ones and also ones of the type mentioned in your query.

The details can be found in many places—one of the most easily accessible ones (also in the mathematical sense) is the lecture notes on the theory of distributions by J. Sebastião e Silva which are available (in english) at the site


Under the section


you will find

Vol. 3 Theory of Distributions,

which gives an introduction to distributions at sophomore level (without using functional analysis).

The relevant definitions and concepts for your query can be found in

Chapter 4–Limits and Integrals of Distributions.

  • $\begingroup$ Is there a form of the theory in which $\int_0^\infty \cos(x) = 0$? In all of the contexts for which I have looked at, the integral should be zero (for instance, Cesaro means, shifting, and smoothing all give zero). $\endgroup$ Nov 4, 2022 at 22:28
  • 1
    $\begingroup$ This is indeed the case—the reason is that the primitive of cosine vanishes at $0$ and has (distributional) limit $0$ at infinity. I suggest you take a look at the reference I suggested where such simple examples are computed explicitly. $\endgroup$
    – order
    Nov 6, 2022 at 8:48

If you do not want to work within the framework of distribution theory, an alternative reading might be "Oscillatory Integrals" by Ioannis Parissis.

  • $\begingroup$ Thank you I will take a look at this $\endgroup$ Nov 4, 2022 at 22:29

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