Indented contour integration + applying jordans lemma

[MM360]

Homework Statement

Hi I am new on here so I am just writing out the question.

Real integral ie from - infinity to + infinity of [cos(x) -1]/(x^2)

2. Homework Equations + solution attempt
I know you have to use the exp cosx idenity and then split the integral.
also it has a removable singularity at z =0 and also the contour has to be indented
I have the ans for this question but my main problem is how this happens

integral of [exp(iz) - 1]/2(z^2) = 0

but the integral of

[exp(-iz)-1]/2(z^2) = pi ( once residue thoerm is applied)

Also could anyone please explain how to apply Jordans lemma to these kind of questions

I have exams soon so any help would be much appreciated

 

[xiaoB]

I don't know this for you is useful or not?

Try using this formula to match it.

Res(f,a)=1/(m-1)!lim_(z→a)〖d^(m-1)/〖dz〗^(m-1) 〗 [(z-a)^mf(z)]

m is the pole of the order;a is pole.

After that, give your answer as:

integral of f(z)=2(pi)i(∑ⁿ_(j=1)〖Res(f;zj)〗)

 

[MM360]

thanks for the reply - but I am fine with calculating the residue - its the reasoning of why the first integral i mentioned goes to 0

its easier to expand the top in its Taylor expansion and divide to get a residue of i/2pi - this is not a problem only the reasoning

 

[MM360]

I used cosx = 1/2[exp(ix) + exp(-ix)]

x and z are just dummy variables so that is correct - even the solution has the same thing

its just the indenting contour reasoning I am stuck on

 

[xiaoB]

If not going wrong:

no ! the first integral where got 0 ?

Are you trouble at e⁰ is equal to 1 .

If not, i am sorry about that because this sem my exam also going with this topic.:shy:

 

[snipez90]

By the triangle inequality we have
\left|\frac{exp(iz) - 1}{2z^2}\right | \leq \frac{1}{|z|^2}
whence the conclusion follows by a simple estimate on the first integral.

 

[hunt_mat]

The idea in that the singularity is on the contour, and so to counter that you make a small indentation in the form of a semi-circle of radius \varepsilon so that the singularity is outside the contour. You just examine what this gives, it looks to be quite an easy calculation in your case.

This sort of problem has popped up in my research, which is why I know about it. Complex analysis is very useful for waves.

 

[MM360]

Thanks everyone for the help/replies

I have found the reason why and my lecturer has confirmed it

so i understand it now