Integral

asi123 · #1 $Old$ January 1st, 2009, 04:07 AM

Hey guys.
Any idea about how to solve this integral?

10x.

Mathstud28 · #2 $Old$ January 1st, 2009, 04:35 AM

Quote:

Originally Posted by asi123 $View Post$

Hey guys.
Any idea about how to solve this integral?

10x.

Hello, this can be solved analytically letting $\theta=2\arctan(\phi)$

NOTE Take what I am about to say with extreme caution...I am a novice at this and just learning so wait for confirmation!

I think this integral can also be solved using Residues by considering the function

$f(z)=\frac{i}{(z-a)(az-1)}$ and taking

$\int_{|z|=1}f(z)~dz$

asi123 · #3 $Old$ January 1st, 2009, 04:46 AM

Well, I think you are right with the second idea because we exactly learning how to work with Cauchy's integral formula so it's got to be it

My question due is how can you break it into $f(z)=\frac{i}{(z-a)(az-1)}$ ?

Thanks a lot.

mr fantastic · #4 $Old$ January 1st, 2009, 04:48 AM

Quote:

Originally Posted by Mathstud28 $View Post$

[snip]
NOTE Take what I am about to say with extreme caution...I am a novice at this and just learning so wait for confirmation!

I think this integral can also be solved using Residues by considering the function

$f(z)=\frac{i}{(z-a)(az-1)}$ and taking

$\int_{|z|=1}f(z)~dz$

The integral can be solved this way. Details available upon request.

mr fantastic · #5 $Old$ January 1st, 2009, 04:53 AM

Quote:

Originally Posted by asi123 $View Post$

Well, I think you are right with the second idea because we exactly learning how to work with Cauchy's integral formula so it's got to be it

My question due is how can you break it into $f(z)=\frac{i}{(z-a)(az-1)}$ ?

Thanks a lot.

Substitute $z = e^{it}$ :

1. $dt = \frac{dz}{iz} = \frac{-i \, dz}{z}$ .

2. $\cos t = \frac{e^{it} + e^{-it}}{2} = \frac{z + \frac{1}{z}}{2} = \frac{z^2 + 1}{2z}$ .

Simplify. The integrand becomes $\frac{i}{az^2 - z - a^2 z + a}$ which factorises in the way Mathstud gave.

asi123 · #6 $Old$ January 1st, 2009, 04:59 AM

Thanks a lot guys.
I have a question due, is this substitute $z = e^{it}$ usually works?

mr fantastic · #7 $Old$ January 1st, 2009, 05:07 AM

Quote:

Originally Posted by asi123 $View Post$

Thanks a lot guys.
I have a question due, is this substitute $z = e^{it}$ usually works?

Integration around the contour $|z| = 1$ , that is, making the substitution $z = e^{it}$ , is typically used to evaluate a real integral of the form $\int_0^{2 \pi} F(\cos t, \sin t) \, dt$ .

asi123 · #8 $Old$ January 1st, 2009, 05:15 AM

I'm having a problem with the boundaries due.
I'm getting an integral from 1 to 1, is this normal?

mr fantastic · #9 $Old$ January 1st, 2009, 05:24 AM

Quote:

Originally Posted by asi123 $View Post$

I'm having a problem with the boundaries due.
I'm getting an integral from 1 to 1, is this normal?

You're integrating around the unit circle. You need to consider singularites of f(z) that lie inside this circle and apply Cauchy's Integral Formula.

asi123 · #10 $Old$ January 2nd, 2009, 12:43 AM

As I was saying, I'm completely new with this Cauchy's integral formula so just to make sure I uploaded my solution to this problem and marked the place where I got to yesterday thanks to you.
Can someone please give me the OK?

Again, thanks a lot.

mr fantastic · #11 $Old$ January 2nd, 2009, 02:44 AM

Quote:

Originally Posted by asi123 $View Post$

As I was saying, I'm completely new with this Cauchy's integral formula so just to make sure I uploaded my solution to this problem and marked the place where I got to yesterday thanks to you.
Can someone please give me the OK?

Again, thanks a lot.

And you can check it by giving $a$ an appropriate value and then comparing your answer with what a CAS gives you.

You should now consider the case |a| > 1 (check your answer using the above mentioned way of checking).