Contour Integration for Trigonometric Functions with Removable Singularities

[muzz]

Homework Statement

∫cos(pi*x)/(1-4x^2) dx from -inf to +infI need to solve this by contour integration but I couldn't find the appropriate contour to use.
Any contour suggestions?

Thanks

 

[jackmell]

Homework Statement

∫cos(pi*x)/(1-4x^2) dx from -inf to +inf


I need to solve this by contour integration but I couldn't find the appropriate contour to use.
Any contour suggestions?

Thanks

Need to show some effort first. That's the rules. Why don't you first try picking one, anyone even if it's the wrong one but use it anyway and then proceed to analyze the integral over each leg of the contour and see what you get. That way, you've opened the door and fulfilled the requirements for getting help in here.

 

[HallsofIvy]

A pretty standard contour for a problem like this is along the x-axis from -R to R, then around the semi-circle, in the upper half plane, from R to -R. There is, of course, one pole inside the contour. You will then need to take the limit as R goes to infinity- and, with any luck, will be able to show that the integral on the semi-circular part of the contour goes to 0.

 

[jackmell]

I don't think so Hall. I mean the pole thing. They're none aren't there? But that's not what he has to use. He needs to consider a related integral which has two then use the Residue Theorem.

 

[muzz]

I have tried that ,( semi circular contour) Actually what I found is the following:

There are 2 poles on the real axis, on the contour, (1/2, -1/2) and they do not contribute to the residue.
And integral on the semi circular part goes to 0.


with the substitution of

z= r e^(iθ) as r goes to inf.

Anyway;

There is, of course, one pole inside the contour.

which pole is it ?

 

[jackmell]

I have tried that ,( semi circular contour) Actually what I found is the following:

There are 2 poles on the real axis, on the contour, (1/2, -1/2) and they do not contribute to the residue.

Lemme' ask you this: what is

\lim_{z\to 1/2} \frac{\cos(\pi z)}{1-4z^2}

 

[muzz]

lemme' ask you this: what is

limz→1/2cos(πz)1−4z2

well it is: pi/4

so ?

 

[jackmell]

Then it's not a pole but rather a removable singularity. But getting back to the problem, often when you have an integral in terms of sines and cosines, it's sometimes helpful to consider the expression:

e^{iw}=\cos(w)+i\sin(w)

and then just work on the expression in terms of e^{iw}, then extract the real or imaginary part.

 

[HallsofIvy]

Homework Statement

∫cos(pi*x)/(1-4x^2) dx from -inf to +infI need to solve this by contour integration but I couldn't find the appropriate contour to use.
Any contour suggestions?

Thanks

Ouch! You are right. I'm so used to requiring imaginary numbers in these I just automatically read it as 1+ 4x^2! Thanks for the correction. It might make sense, then, to do effectively what I said but use the imaginary axis rather than the real axis.

 

[muzz]

Then it's not a pole but rather a removable singularity. But getting back to the problem, often when you have an integral in terms of sines and cosines, it's sometimes helpful to consider the expression:

As you suggested I wrote sin as an imaginary part of a exponential and take the imaginary part of the integral in the end and got the right answer. But anyway , without changing sin to exp, I couldn't do it. In fact I did and it came out --> 0. Thanks for help, but still I am still wondering if I can solve it without using exponential form.