How Do You Evaluate sin(x)/x^4 Over the Unit Circle Using Complex Analysis?

[Physgeek64]

Homework Statement

evaluate sinx/x^4 over the unit circle

Homework Equations

Cauchys Residue theorem
$sinz=1/(2i)(z+1/z)$

The Attempt at a Solution

So we have a branch point at z=0 but its of order 4 so I can't see any direct way of using Cauchys residue theorem. I've tried changing the sin expression to as above but simply end up with poles of order 3 and 5, which again doesn't help me.

So I tried defining a new contour essentially around the unit circle, but also enclosing the branch point by traveling back along the real axis at some small value of y. By Cauchy's residue theorem the integral along this combined contour is zero since no poles are enclosed. Redefining $z=e^(i*theta)$ for the integral along the outer circle, $z=x+ie_0$ along the path from $x to e_0$, $z=e_0e^(i*theta)$ around the small inner circle, and $z=x+ie_0$ for the line running just above the real axis from $e_0 to x$. But I'm not sure how to proceed from here, nor how to cary out any of the integrals.

Many thanks :)

 

[stevendaryl]

Homework Statement

evaluate sinx/x^4 over the unit circle

I think you're making this a lot more complicated than it needs to be. You know the power series for sine:

$sin(x) = x - \frac{x^3}{6} + \frac{x^5}{120} + ...$ (the general term is $(-1)^n \frac{x^{2n+1}}{(2n+1)!}$)

So $\frac{sin(x)}{x^4} = \frac{1}{x^3} - \frac{1}{6x} + \frac{x}{120} + ...$ (the general term is $(-1)^n \frac{x^{2n-3}}{(2n+1)!}$)

You can just integrate each term separately around the unit circle.

Your expression for $sin(x)$ has something wrong with it. It should be:

$sin(x) = \frac{e^{ix} - e^{-ix}}{2i}$

 

[Physgeek64]

I think you're making this a lot more complicated than it needs to be. You know the power series for sine:

$sin(x) = x - \frac{x^3}{6} + \frac{x^5}{120} + ...$ (the general term is $(-1)^n \frac{x^{2n+1}}{(2n+1)!}$)

So $\frac{sin(x)}{x^4} = \frac{1}{x^3} - \frac{1}{6x} + \frac{x}{120} + ...$ (the general term is $(-1)^n \frac{x^{2n-3}}{(2n+1)!}$)

You can just integrate each term separately around the unit circle.

Your expression for $sin(x)$ has something wrong with it. It should be:

$sin(x) = \frac{e^{ix} - e^{-ix}}{2i}$

Oh okay, so is it okay to use the expansion to expose the single pole?

Thank you for the reply :)

Oops, my mistake, Thank you

 

[Physgeek64]

I think you're making this a lot more complicated than it needs to be. You know the power series for sine:

$sin(x) = x - \frac{x^3}{6} + \frac{x^5}{120} + ...$ (the general term is $(-1)^n \frac{x^{2n+1}}{(2n+1)!}$)

So $\frac{sin(x)}{x^4} = \frac{1}{x^3} - \frac{1}{6x} + \frac{x}{120} + ...$ (the general term is $(-1)^n \frac{x^{2n-3}}{(2n+1)!}$)

You can just integrate each term separately around the unit circle.

Your expression for $sin(x)$ has something wrong with it. It should be:

$sin(x) = \frac{e^{ix} - e^{-ix}}{2i}$

Just to check I got $- \frac{pi*i}{3}$