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Help with Jordan's Lemma (Collorary)

  1. Jan 22, 2015 #1
    Hello everyone. I'm having some troubles with Jordan's Lemma and I can't find how to fix it, because everyone in internet and in the books I read do not prove this. Now, I'm following Wunsch (Complex Variables with Applications) but it only does this part for a easy problem.

    For example, if I have an improper integral like this: (page 377 & 378 from Wunsch book)
    [tex]\int_{-\infty}^{\infty} \frac{cos(3.x)} {(x-1)^2+1} dx[/tex]
    To solve it, I do this:
    [tex]\int_{-\infty}^{\infty} \frac{exp(i.3.x)} {(x-1)^2+1} dx[/tex]
    Integrating that on the upper half circle, with [tex]R \rightarrow \infty[/tex] and using the Theorem of Residues:
    [tex]\int_{-R}^{R} \frac{exp(i.3.x)} {(x-1)^2+1} dx+\int_{C_1}^{ } \frac{exp(i.3.z)} {(z-1)^2+1} dx=2.\pi.i .\sum\limits_{k}^{ } Res( \frac{exp(i.3.z)} {(z-1)^2+1}, z_k)[/tex]

    Now, I have to prove in a correct and properly way why this occurs:
    [tex]\int_{C_1}^{ } \frac{exp(i.3.z)} {(z-1)^2+1} dx=0[/tex]

    I'm trying to use the ML inequality, like this:
    [tex]| \int_{C_1}^{ } \frac{exp(i.3.z)} {(z-1)^2+1} dx | \leq M.L = M.\pi.R[/tex], because[tex] L=\pi.R[/tex] (it's the length of the semicircunference)
    So, (I don't understand this step)
    [tex] | \frac{exp(i.3.z)} {(z-1)^2+1} | \leq M [/tex]

    Now I have to put something like:
    [tex] | \frac{exp(i.3.z)} {(z-1)^2+1} | \leq "something" [/tex]
    where I say that [tex] M = "something" [/tex]
    and [tex] "something" \rightarrow 0[/tex], so [tex]\int_{C_1}^{ } \frac{exp(i.3.z)} {(z-1)^2+1} dx=0 [/tex]

    But I don't know how to do it in this case. I know I have to use, for example, [tex] z=R.exp(i.\theta) [/tex]

    For example, something I got from:
    [tex]|exp(i.3.z)|[/tex]
    is this:
    [tex]|exp(i.3.z)|=|exp(-3.y)|.|exp(i.3.x)|=|exp(-3.y)|[/tex]
    but I suppose that is not helpful for this problem


    Can anyone help me? Thanks!!!

    Last question: ML inequality is the same as Jordan's Lemma?
     
    Last edited: Jan 22, 2015
  2. jcsd
  3. Jan 22, 2015 #2

    mathman

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    Gold Member

    What is varying to make "something" -> 0?

    As an aside the original integral is easily shown to exist. |integrand| < 1 for all x, while |integrand| < x^2 as |x| becomes infinite.
     
  4. Jan 24, 2015 #3
    Jordan's Lemma is not ML inequality, it is a bit stronger. But for your integral ML inequality is enough. Namely, in the upper half plane ##|e^{i3z}|\le 1##. Therefore for ##|z|=R>1##, ##\operatorname{Im} z\ge 0## one can estimate $$\left|\frac{e^{i3z}}{(z-1)^2+1}\right| \le \frac{1}{|(z-1)^2+1|}\le \frac1{(R-1)^2}$$. Then ML estimate gives that the integral over the semicircle of radius ##R## is estimated by $$\frac{\pi R}{(R-1)^2} \to 0$$ as ##R\to\infty##.

    This works for integrands of form ##e^{iaz} g(z)##, ## a> 0## such that ##\lim_{R\to\infty}RM(R) =0##, where ##M(R)## is the maximum of ##|g(z)|## on the semicircle of radius ##R##.

    Jordan's lemma asserts that the integral over the semicircle tends to ##0## even if we only have ##\lim_{R\to\infty}M(R) =0## and the ML estimate does not help. You can look it up in the Wikipedia
     
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