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Consider the integral:

$$\int_0^\infty\sin(x)dx.\tag1$$

It's clearly divergent, but if we regularize it as

$$\int_0^\infty\sin(x)e^{-x/a}dx=\frac{a^2}{a^2+1},\tag2$$

we can take the limit of $a\to\infty$ and get some value (here $1$).

This reminds me of an approach similar to the following: a function, e.g. $x\mapsto e^{1/x}$, isn't defined at $x=0$. But if we take the limit $x\to0^-$, we get some value (here $0$). But if we take the limit from another side, we won't get the same value (actually we'll get either infinity (for $x\to0^+$), or no limit at all).

It seems that doing regularization $(2)$ and taking the limit $a\to\infty$ is very similar to approaching the integral's value from some "side" of function space on which it is defined. So I assume that it can, as the function discussed above, have different limits when approached from other "sides", or not have any. Is it true? What would be an example of such approach directions (i.e. regularizing functions $\xi(a)$), which would give some finite value different from $1$?

Ruslan
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  • Your integral is not "clearly divergent", it is uncertain without indication what it means and how it should be understood. Its value $I$ oscillates as $1-\cos(L)$, $0\le I\le 1$, where $L$ is the upper limit of the integration interval. – Vladimir Kalitvianski Mar 07 '17 at 13:08
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    @VladimirKalitvianski that's what it means to be divergent: the limit as $L\to\infty$ does not exist. In physics it's normal to assume "by default" that some regularization must exist and work, but mathematically, the improper integral $(1)$ diverges. – Ruslan Mar 07 '17 at 13:49
  • My answer has been deleted, but the currently accepted answer is not regularization. Whatever method you use (for instance, Cesaro or Laplace transform) the result is 1. – Anixx Sep 05 '23 at 06:35
  • @Anixx You need to elaborate on your criteria of what is regularization and what isn't. Otherwise your comment adds zero value. – Ruslan Sep 05 '23 at 06:47
  • As you can see, the average value of the integral $I(u)=\int_0^u\sin(x)dx$ as $u$ goes to infinity is $1$. This is called Cesaro regularization, and the most basic regularization technique. All other techniques should agree with Cesaro regularization (generalize it), otherwise they are invalid (one can call "regularization" whatever they want but that would not be so in essense). So in sum: Cesaro is the most basic regularization technique. It is out of the question, because it is basically the average value. – Anixx Sep 05 '23 at 07:16
  • @Anixx "All other techniques should agree with Cesaro regularization (generalize it)" — why, under what (widely used) definition? – Ruslan Sep 05 '23 at 08:51
  • Well, this is a general principle, if you extend some notion you generalize it to newer area with a method that coincides with old method on old area. Like, rules for complex numbers coincide with rules for real numbers on the set of real numbers. – Anixx Sep 05 '23 at 09:12
  • The very idea of regularization comes from the infamous Grandi series $1 − 1 + 1 − 1 + 1 − 1 + 1 − 1 + ...$ and the idea that this series should be assigned the average value, $1/2$. – Anixx Sep 05 '23 at 10:28

2 Answers2

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Here's a regulator which yields a different value for the regularized integral:

$$\xi(x,\alpha)=\exp\left(-\frac x\alpha\right)\frac{\alpha+\sin(x)+1}\alpha.\tag1$$

With it we have

$$\int_0^\infty \sin(x)\xi(x,\alpha)\,dx=\frac{6\alpha^4+4\alpha^3+3\alpha^2+\alpha}{4\alpha^4+5\alpha^2+1}\tag2$$

and

$$\lim_{\alpha\to\infty}\int_0^\infty \sin(x)\xi(x,\alpha)\,dx=\frac32\ne1.\tag3$$

Replacing $\sin(x)+1$ in definition $(1)$ with $(\sin(x)+1)\beta$, we can assign an arbitrary regularized value to the integral: after such replacement the limit $(3)$ will be $1+\beta/2$.

Ruslan
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    Here you, roughly speaking, add to your integrand $\sin(x)$ an addendum $[\sin^2(x)+\sin(x)]\frac{\beta}{\alpha}$ within the interval $0<x<\alpha$, i.e., a non-zero contribution. No wonder it depends on $\beta$. – Vladimir Kalitvianski Mar 07 '17 at 12:04
  • @VladimirKalitvianski right, I specifically tried to find a regulator which would not let the integral diverge to infinity, still not making the "anomal" contribution negligible — and yet maintaining the property of $\lim_{\alpha\to\infty}\xi(x,\alpha)=1$ (so that the integrand would converge to $\sin(x)$). It's of course an ad hoc regulator, but the question was asking for any. – Ruslan Mar 07 '17 at 13:52
  • If we were discussing the integrand itself, then your "regulator" would be a regulator in the limit $\alpha\to\infty$. But we are discussing the integral itself, so your "regulator" $\xi$ is not truly a regulator since it still brings a contribution to the integral value in this limit. In other words, this regulator is not sufficient yet to yield a certainty: it contains factually a similar integrand and this causes some sort of $L$-dependence of $I$. – Vladimir Kalitvianski Mar 07 '17 at 15:24
  • @VladimirKalitvianski Why would $L$-dependence be a problem? The integral $(2)$ as an improper integral always converges for $0<\alpha<\infty$, so $\xi$ is a regulator for the given integral $(1)$ in the OP. – Ruslan Mar 07 '17 at 17:44
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    Mathematically, I've done nothing wrong here: I've chosen a path in the function space, which does lead to a different limit. That's all the OP was asking for. And it's completely OK that to achieve this I used the knowledge of the original integrand. I could in fact even use a square or triangular wave instead of sinusoid and achieve a similar effect. – Ruslan Mar 07 '17 at 17:51
  • That is what I said: its value depends on how you define it. If you want to add some addendum, nobody can prevent you from doing it. You can add an addendum like $\beta/L$ within the interval $0<x<L$ to the integrand, it itself vanishes when $L\to\infty$, but the integral gets and addendum $\beta$. As I previously said, it is not a regulator, don't fool yourself. – Vladimir Kalitvianski Mar 07 '17 at 18:35
  • @VladimirKalitvianski so you suggest to use $\xi(x,L)=\exp(-x/L)+\theta(L-x)\beta/L$? This is actually even nicer option than my own answer. I don't see any reason why it wouldn't be a regulator: it does approach $1$ as $L\to\infty$, and it does make the integral converge. And it can be made analytic if you don't like Heaviside theta — just replace it with something like $\operatorname{erf}$. – Ruslan Mar 07 '17 at 20:14
  • Oh wait... This example $\xi$ actually gives the regulated value of $1$. I have to divide Heaviside theta by $\sin(x)$ to get $1+\beta$. Not quite nice really, since $\xi$ will not converge pointwise to $1$ (only if we define is values at $\pi n$ as $1$). – Ruslan Mar 07 '17 at 20:25
  • Yes, I meant an addendum to the integrand, not a factor like $\xi$. Your factor (in the answer) brings an addendum of a similar sort: $\exp(-x/\alpha)$ is some (rough) approximation to the Heaviside function standing at an addendum giving a growing as $L$ integral. $\alpha\approx L$ in the denominator makes the contribution finite, but arbitrary. – Vladimir Kalitvianski Mar 08 '17 at 06:15
  • This is not regularization. – Anixx Jan 24 '22 at 19:49
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The most basic method of regularization is Cesaro regularization, which is basically, finding the average value at infinity. More advanced methods should generalize it to greater set of integrals, so to be meaningful.

The average value of $\int_0^u \sin (x) \, dx$ as $u$ goes to infinity is

$\underset{v\to \infty }{\text{lim}}\frac{1}{v}\int _0^v\int _0^u\sin (x)dxdu=1$

All more advanced methods of regularization should coincide with Cesaro regularization where it is applicable to make any sense.

The following Mathematica code calculates regularized values of integrals, including far more complicated ones using 3 different methods: 1) representing the integral as Riemann sum and then regularizing the sum using Dirichlet method, 2) using Laplace transform, 3) using Fourier transform.

RegIntegral1[Integrate[e_, {t_, 0, Infinity}]] := 
   Limit[(\[FormalK] Sum[e /. t -> \[FormalK] t, {t, 1, Infinity}, 
          Regularization -> "Dirichlet"] + -\[FormalK] Sum[
          e /. t -> -\[FormalK] t, {t, 1, Infinity}, 
          Regularization -> "Dirichlet"])/2 // Simplify, \[FormalK] ->
      0]; // FullSimplify
RegIntegral2[Integrate[e_, {t_, 0, Infinity}]] := 
 Limit[Evaluate[Evaluate[LaplaceTransform[ e, t, \[FormalK]]]] + 
      Evaluate[LaplaceTransform[ e, t, -\[FormalK]]], \[FormalK] -> 
      0]/2 // Re // FullSimplify 
RegIntegral3[Integrate[e_, {t_, 0, Infinity}]] := 
 Limit[Evaluate[
     I/2 FourierTransform[(e /. t -> I t) Sign[t], t, \[FormalK], 
        FourierParameters -> {1, -1}] + (I/
          2 FourierTransform[(e /. t -> I t) Sign[t], t, \[FormalK], 
          FourierParameters -> {1, -1}] /. \[FormalK] -> \
-\[FormalK])]/2, \[FormalK] -> 0] // FullSimplify

Use:

RegIntegral1[Integrate[x^2 Cosh[x]^4 Sinh[3 x], {x, 0, Infinity}]]
RegIntegral2[Integrate[x^2 Cosh[x]^4 Sinh[3 x], {x, 0, Infinity}]]
RegIntegral3[Integrate[x^2 Cosh[x]^4 Sinh[3 x], {x, 0, Infinity}]]

Output:

-157106/385875
-157106/385875
-157106/385875

You can see that for sine function under integral all methods will return 1.

Anixx
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