# A surface integral over infinite space

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dyn
Hi.
If a function f is normalizable ,ie f→0 as | x | → infinity or r→ infinity then I presume the following surface integral f dS over infinite space is zero ?
But I thought about this again and it seems like a case of zero x infinity. The function is zero at the infinite surface but the area of the infinite surface is infinite. Is the surface integral zero ?
Thanks

Gold Member
Maybe I'm misunderstanding your question, but define $f(x)$ for $x\in\mathbb{R}^2$ to be $1$ if $x$ is in the unit square $[0,1]\times [0,1]$ and zero everywhere else. This is "normalizable" according to your definition. Let $S$ be the plane $\mathbb{R}^2$. Then $\int_S f dS=1$.

Edit: I might not understand what you mean by infinite surface.

Mentor
2022 Award
Hi.
If a function f is normalizable ,ie f→0 as | x | → infinity or r→ infinity then I presume the following surface integral f dS over infinite space is zero ?
But I thought about this again and it seems like a case of zero x infinity. The function is zero at the infinite surface but the area of the infinite surface is infinite. Is the surface integral zero ?
Thanks
Can you explain this question? Should it be a question about physical methods, or what does normalizing here mean? Maybe you can give an example for function and surface?

dyn
The normalisable function is a normalisable wavefunction from quantum mechanics

Mentor
2022 Award
The normalisable function is a normalisable wavefunction from quantum mechanics
Then I assume you should better post this there. AFAIK is normalisability a technique used by physicists to deal with otherwise infinite quantities.

Your question is hard to answer mathematically if you don't say what you mean, i.e. in terms of equations, not words.

eys_physics
Normalizable, in Quantum Mechanics, means that in the 1-dimensional case the integral
N^2=$$\int_{a}^{b} |f(x)|^2dx$$ ,
is well-defined and ##N^2<\infty##. The limits ##a## and ##b## can here either be finite or infinite.
Similarly, in the 2-dimensional case:
$$N^2=\int dS |f(\vec{x})|^2$$

The condition that ##f(x)\rightarrow## when ##|x|\rightarrow \infty## is not enough.