# Exp(i infinity) = 0 ?

1. ### kakarukeys

190
: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. Science news on Phys.org
3. ### LeonhardEuler

864
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.

4. ### kakarukeys

190
That's what my math teacher told me,
but my physics teacher told me entirely different thing.

5. ### Hurkyl

16,089
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.

6. ### kakarukeys

190
Do you mean, for computation purpose, we can define the limit to be zero
it can't be proven?

7. ### kakarukeys

190

this is no college level!
Feymann Path Integral!

8. ### inha

576

College=university. Don't get freaked out.

9. ### kakarukeys

190
Feynmann Path Integral is at graduate level :rofl: