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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 problem

Can anyone help me? Thanks!

Last question: ML inequality is the same as Jordan's Lemma?

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?

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