Talk:Jordan's lemma

Untitled
Isn't there a restriction on the power of Jordan's Lemma? Something like, g(r) must go to 0 as fast or faster than 1/R? There's a slightly weaker theorem like 1/R^2, and then Jordan's covers all your bases. Can't find it on the web, and I don't see it here...

You don't need it 203.200.95.130 20:20, 26 August 2007 (UTC)

There needs to be some restriction, right? i.e. $$g(z)=e^{-2 i a}$$ is clearly not going to work... although any polynomial with finite terms should work, I imagine. Wackywendell (talk) 00:16, 1 November 2011 (UTC)

> The proof given here would indicate not also wolfram Mathworld thinks not http://mathworld.wolfram.com/JordansLemma.html Jekowl (talk) 21:06, 4 December 2015 (UTC)

Error near top?
$$g(z)=\frac{1}{1+z^2}$$ is a valid function for Jordan's Lemma, but $$g(z)=\frac{1}{z}$$ is not.

Surely that's only for the case $$a=0$$?

I'm not sure that we should even include this case, as the references don't seem to include it as part of Jordan's lemma.

--Ryanl (talk) 10:47, 14 April 2008 (UTC)

final limit in proof
Shouldn't this limit equal pi/a rather than 0? —Preceding unsigned comment added by 131.111.8.97 (talk) 12:01, 29 May 2009 (UTC)

Thanks for this.131.111.216.15 (talk) 20:02, 31 May 2009 (UTC) (At Cambridge also? Mathmo?)

Holomorphicity/Continuity required?
The requirement that f and g should be smooth on the entire half-plane is not really necessary and often not given in practice, especially if Jordan's lemma is used with the residue theorem. See, e.g., http://mathworld.wolfram.com/JordansLemma.html It is also not quite clear where it is used in the proof listed here. —Preceding unsigned comment added by 218.82.2.216 (talk) 03:14, 8 October 2009 (UTC)

Assessment comment
Substituted at 02:15, 5 May 2016 (UTC)