I have already asked a similar question, albeit far more extensive, but it was criticized and closed for being too extensive and promotional. So, here is a greatly truncated and focused version.

Since there are Laplace-based transforms that keep the area under the integral of the function intact, it is reasonable to generalize this to assume that the divergent integrals are kept equal as well under the transform and its inverse:

$$\int_0^\infty f(x)dx=\int_0^\infty\mathcal{L}_t[t f(t)](x)dx=\int_0^\infty\frac1x\mathcal{L}^{-1}_t[ f(t)](x)dx$$

This gives several equivalence classes of divergent integrals that we under the above assumption consider the same. Their regularized values are also preserved. So, it is possible to treat them as various integral representations of infinite "constants".

So here is a short list of the classes, and I wonder, where they can be encountered in areas of math and/or physics:

  • $\int_0^{\infty } \, dx=\int_0^{\infty } \frac{1}{x^2} \, dx$

  • $\int_{1/2}^\infty dx=\int_0^\infty \frac{e^{-\frac{x}{2}} (x+2)}{2 x^2}dx=\int_0^2\frac1{x^2} dx$

  • $\int_{-1/2}^\infty dx=\int_0^\infty \frac{4-e^{-\frac{x}{2}} (x+2)}{2 x^2} dx$

  • $\int_0^1 \frac1x dx-\gamma=\int_1^\infty \frac1xdx=\int_0^\infty\frac{e^{-x}}{x}dx=\int_0^\infty\frac{dx}{x+1}=\int_0^\infty\frac{e^x x \text{Ei}(-x)+1}{x}dx=\int_0^\infty\frac{x-\ln x-1}{(x-1)^2}dx$

  • $\int_0^1 \frac1x dx=\int_0^\infty\frac{1-e^{-x}}{x}dx=\int_0^\infty\frac{1}{x^2+x}dx=-\int_0^\infty e^x \text{Ei}(-x)dx=\int_0^\infty\frac{x\ln x-x+1}{(x-1)^2 x}dx$

  • $\int_1^\infty \sqrt{x^2-1}dx=\int_0^\infty \frac{K_2(x)}{x}dx=\int_0^\infty \left(x-\frac{1}{2 x}\right) dx =\int_0^\infty \left(\frac{2}{x^3}-\frac{1}{2 x}\right)dx$

Are there other notable divergent integrals that appear persistently?

  • 2
    $\begingroup$ the most prominent "divergent integral" I know of is $\int_0^\infty e^{ikx}dx$. $\endgroup$ Jun 2, 2021 at 8:36
  • $\begingroup$ @CarloBeenakker yes, but that integral is not divergent to infinity, it is Cesaro-summable. I missed in the question to note that I meant only integrals divergent to infinity. For my intents and purposes the Cesaro(or Abel)-summable integrals are similar to convergent integrals, being equal to their finite regularized values. $\endgroup$
    – Anixx
    Jun 2, 2021 at 8:42

1 Answer 1


Perhaps the most "famous" divergent integral in physics is the Casimir vacuum energy of the electromagnetic field, $$\int_0^\infty x^2\sqrt{1+x^2}\,dx"="-\frac{1}{10},$$ and the related class of integrals $$\int_0^\infty x^p\,dx"="\frac{(-1)^{p+1}}{(p+1)(p+2)},\;\;p=0,1,2,\ldots,$$ see Calculation of the Vacuum Energy Density using Zeta Function Regularization.

  • $\begingroup$ Why is the = double quoted ? $\endgroup$
    – Soleil
    Jun 2, 2021 at 16:24
  • 1
    $\begingroup$ "=" because I am too timid to write $\int_0^\infty x\,dx=1/6$. $\endgroup$ Jun 2, 2021 at 17:05
  • $\begingroup$ Hmm, this is not in line with the regularization methods I know, on my clock, an integral of monomial from 0 to infinity regularizes to 0. $\endgroup$
    – Anixx
    Jun 4, 2021 at 4:58
  • 1
    $\begingroup$ More details to your interesting answer are welcome. I am sure you, as a physicist can do it. Otherwise from the mathematical point of view your answer is along the medieval proof of the existence of God: 0=(1-1)+(1-1)+...=1+(-1+1)+(-1+1)+...=1. Something was created from nothing. God exists!!! $\endgroup$ Jun 5, 2021 at 3:49
  • $\begingroup$ I have the following formula: $\int_0^\infty x^n dx=\frac{\left(\tau +\frac{1}{2}\right)^{n+2}-\left(\tau -\frac{1}{2}\right)^{n+2}}{(n+1)(n+2)}=\frac{\omega _+^{n+2}-\omega _-^{n+2}}{(n+1)(n+2)}$ which regularizes to $\frac{B_{n+2}(1)-B_{n+2}(0)}{(n+1)(n+2)}$, which is zero. Integrals of monomials always regularize rto zero as far as I know. $\endgroup$
    – Anixx
    Jun 6, 2021 at 10:29

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