Undergrad Can Functions Have Complex Poles Beyond Infinitesimal Limits?

[dRic2]

Suppose I have a function
$$f(x) = \lim_{\eta \rightarrow 0} \int_{-\infty}^{\infty} d \zeta \frac {g(\zeta)}{x - \zeta + i \eta}$$
and suppose $g(\zeta)$ is a continuous (maybe even differentiable) function. Can $f(x)$ have complex poles of the form $a + ib$ with $b$ not an infinitesimal ?

Would a similar result hold if, instead of an integral, I have a summation
$$f(x) = \lim_{\eta \rightarrow 0} \sum_{i}^{\infty} \frac {g_i}{x - \zeta_i + i \eta}$$
?

I'm sorry if I'm not writing any ideas, but I don't have any. It has been quite a while since my last analysis exam and I don't really known where to even start. Btw this is non an exercise, it's just something I'm wondering about.

Thanks
Ric

 

[mfb]

Assuming that integral is over the real axis I don't see how you could get a pole. You would need the integral to change rapidly with arbitrarily small changes of x, but the denominator has an imaginary part that doesn't go away if x is not real.

For the sum, I assume $\zeta_i$ is real, in that case I expect the same result.

 

[dRic2]

You would need the integral to change rapidly with arbitrarily small changes of x, but the denominator has an imaginary part that doesn't go away if x is not real.

Sorry I don't understand this sentence. Can you explain a little more? Thanks for the answer

 

[mfb]

It's one of the properties of pole. More formally, for all $\epsilon,\delta>0$ you can find z and z' within $\epsilon$ of the pole such that $|f(z)-f(z')|>\delta$. But what could lead to such a rapid change in the integrals if the denominator has an imaginary part that doesn't disappear?

 

[dRic2]

I don't see how the condition you just mentioned implies that integrand should oscillate vigorously. $\frac 1 {(x -1)^2}$ has a pole in 1 but doesn't oscillate. A part from this I get your argument and I agree. And what if the numerator is some strange "function" maybe something with a dirac's delta? The only request I have is that the numerator doesn't carry a pole by itself. I would still be tempted to say that only real poles survive

 

[wrobel]

If $g\in L^1(\mathbb{R})$ then you have two holomorphic functions: in $\{\mathrm{Im}\,x>0\}$ and in $\{\mathrm{Im}\,x<0\}$

 

[mfb]

$\frac 1 {(x -1)^2}$ has a pole in 1 but doesn't oscillate.

I'm not talking about oscillations. Approach 1 from the real axis and it goes to infinity, approach it along the imaginary axis and it goes to minus infinity.

And what if the numerator is some strange "function" maybe something with a dirac's delta?

That's not a function. For "functions" I don't know.

 

[hutchphd]

Are you by any chance studying the Mittag-Leffler Theorem? This is a way of doing sums by putting in poles using meromorphic functions. Sort of similar to your question.

 

[dRic2]

Thanks for all the replays. I'm sorry if my question looks weird but I started thinking about this as a consequence of thinking about other stuff, but the complete question would be to long to post here. Anyway, I think I solved this particular problem.

Are you by any chance studying the Mittag-Leffler Theorem?

no sorry. I started thinking about this after reasoning about some properties of the green function in many body theory. Might be linked though...

 

[hutchphd]

OK When you realize it is all wizardry then you are halfway there. Sure wish I knew it routinely.

 

[dRic2]

OK When you realize it is all wizardry then you are halfway there. Sure wish I knew it routinely.

I usually trust those wizards who made the machinery, but I also like to know what I trust them about :D