# Exp(i infinity) = 0 ?

1. Sep 9, 2005

### kakarukeys

:grumpy: :yuck:
Hello there,

please tell me whether $$e^{iS} = 0$$ when $$S = +\infty$$

S is the action of a particle $$= \int L dt$$
e.g. when the particle goes to infinity and comes back, kinetic energy blows up,
then S blows up.

2. Sep 9, 2005

### LeonhardEuler

It does'nt:
$$e^{ix}=\cos{x}+i\sin{x}$$
$$\lim_{x\rightarrow \infty}e^{ix}=\lim_{x\rightarrow \infty}\cos{x}+i\lim_{x\rightarrow \infty}\sin{x}$$
But niether the real nor imaginary parts of the limit exist; as x approaches infinity they oscilate between -1 and 1.

3. Sep 9, 2005

### kakarukeys

That's what my math teacher told me,
but my physics teacher told me entirely different thing.

4. Sep 9, 2005

### Hurkyl

Staff Emeritus
I would suspect both your math teacher and physics teacher are right!

The map x → e^(ix) is, indeed, discontinuous at +∞, and thus one cannot continuously extend this map to have a value at +∞. (continuous extension is the process that is generally used to justify statements like arctan +∞ = π / 2 or x/x = 1).

However...

As is often the case, there is probably some related concept that is practical to use here, and whether he knows it or not, it's what your physics professor meant.

5. Sep 9, 2005

### kakarukeys

Do you mean, for computation purpose, we can define the limit to be zero
it can't be proven?

6. Sep 12, 2005

### kakarukeys

this is no college level!
Feymann Path Integral!

7. Sep 13, 2005

### inha

College=university. Don't get freaked out.

8. Sep 13, 2005

### kakarukeys

Feynmann Path Integral is at graduate level :rofl: