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Q&A

Comments on With what classes of functions does the equality $\int_0^\infty f(x)dx=\int_0^\infty\mathcal{L}_t[t f(t)](x) dx$ lead to paradoxes?

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With what classes of functions does the equality $\int_0^\infty f(x)dx=\int_0^\infty\mathcal{L}_tt f(t) dx$ lead to paradoxes?

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The following operators keep the area under the convergent integrals unchanged:

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

But with divergent integrals applying the Laplace-based transform $\mathcal{L}_t[t f(t)](x)$ (I would denote it $\mathcal{T}f(x)$) to a function more than once may lead to a strictly greater or smaller function, which I refer to as an unacceptable paradox.

One such example is the function $f(x)=\frac1{\sqrt{x}}$. Applying the transform $\mathcal{T}$ to it twice, one arrives at function $\frac{\pi }{2 \sqrt{x}}$, which is strictly greater than the original function, and the problematic step seems to be the first one. It seems, one class of the problematic functions is $f(x)=\frac1{x^p}$, where $ 0< p< 1$.

So, I wonder, what is the whole description of the class of the functions that lead to such paradoxes. I outline the two paradoxes I am seeking to avoid:

  1. Functions such that $f(x)>\mathcal{T^n}[f(x)]$ or $f(x)<\mathcal{T^n}[f(x)]$ for all $x>0$, that is the function becomes strictly smaller or greater after the transform.

  2. Functions such that the integral $\int_0^\infty (f(x)-\mathcal{T^n}[f(x)])dx$ is finite but nonzero.

Is it possible to clearly describe the class of functions that lead to such paradoxes so to avoid them?

History

3 comment threads

x-post https://math.stackexchange.com/questions/4768302/with-what-classes-of-functions-the-equality-i... (1 comment)
Why "paradox"? (6 comments)
"applying $\mathcal{T}$ more than once"—what if we apply $\mathcal{T}$ only once? (5 comments)
Why "paradox"?
clemens‭ wrote 4 months ago

Why do you call instances where $\mathcal{T}^2(f) > f$ "paradoxical"?

Anixx‭ wrote 4 months ago · edited 4 months ago

The transform is intended to be an equivalent class for divergent integrals. If the function becomes strictly smaller or greater, it contradicts the expectation.

clemens‭ wrote 4 months ago

"is intended"—by whom? What do divergent integrals have to do with it?

Anixx‭ wrote 4 months ago

The transform works well as an equivalent class in most cases (even keeping the finite regularized part) but fails in others, this question is intended to find out in what cases it is not applicable.

tommi‭ wrote 4 months ago

Maybe there is some theorem that specifies the property you have and gives a sufficient condition for it to hold. Referring to such a theorem in the question might make it easier to answer, if such a theorem exists.

Anixx‭ wrote 4 months ago

I am unaware of any theorem on this.