Help with Jordan's Lemma (Collorary)

  • Thread starter juan.
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In summary, Jordan's Lemma is a stronger version of the ML inequality, and for the given integral, we can use the ML inequality to show that the integral over the semicircle tends to zero as the radius of the semicircle approaches infinity. This is because the integrand can be bounded by a function that tends to zero as the radius increases, making the integral itself tend to zero. Jordan's Lemma is useful in cases where the function being integrated cannot be bounded in this way, but the sine function in this case allows us to use the ML inequality.
  • #1
juan.
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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 problemCan anyone help me? Thanks!

Last question: ML inequality is the same as Jordan's Lemma?
 
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  • #2
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.
 
  • #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
 

1. What is Jordan's Lemma and why is it important?

Jordan's Lemma is a mathematical theorem used in complex analysis. It states that for a complex-valued function f(z) and a closed contour C, if the absolute value of f(z) is bounded and f(z) has no poles on or within C, then the integral of f(z) along C approaches zero as the radius of the contour increases to infinity. This lemma is important because it allows us to evaluate complex integrals by finding the residues of the poles inside the contour.

2. Can Jordan's Lemma be applied to any complex-valued function and contour?

No, Jordan's Lemma only applies to functions that are bounded and have no poles on or within the contour. If these conditions are not met, the lemma cannot be used to evaluate the integral.

3. How is Jordan's Lemma related to the Cauchy Residue Theorem?

Jordan's Lemma is a corollary of the Cauchy Residue Theorem. The Cauchy Residue Theorem states that for a function f(z) with a finite number of isolated singularities inside a closed contour C, the integral of f(z) along C is equal to 2πi times the sum of the residues of f(z) at the singularities. Jordan's Lemma is a special case of this theorem when there are no singularities inside the contour.

4. Can Jordan's Lemma be used to evaluate integrals with infinite bounds?

Yes, Jordan's Lemma can be used to evaluate integrals with infinite bounds as long as the function is bounded and has no poles on or within the contour. In these cases, the lemma can be used to show that the integral approaches zero as the radius of the contour increases to infinity.

5. Are there any limitations to using Jordan's Lemma?

Yes, there are limitations to using Jordan's Lemma. It can only be applied to integrals with finite bounds, and the function must be bounded and have no poles on or within the contour. Additionally, the contour must be simple, meaning it does not intersect itself. If any of these conditions are not met, the lemma cannot be used to evaluate the integral.

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