1. ## [Solved]Integral with residue, question about poles.

Hello everyone, I am having a little problem with the following integral:

$\displaystyle \int \frac{sinz}{z^3(z^2+1)}$

Where D is the Domain D = {z in the complex field : |z|<2}

Now, my understanding is I have to look where my function goes to zero.
On the denominator I have :

z = j (simple pole)
z = -j (simple pole)
and z = 0 (pole of order 3)

Now, on the numerator, I have
z = 0,pi, 2pi

Since zero appears both at the numerator and the denominator I understand it becomes a pole of 2order, and not a 3 order one.
So my integral becomes:

$\displaystyle 2\pi j(Res(j) + Res(-j) + Res(0))$

Now what happens is, if I calculate the residue in ZERO it's ZERO! So it means that That's a removable singularity! Now, I double checked with wolfram and it doesn't even bother to calculate the residue in zero.
Could anyone tell me why? Please?

sinz/(z^3(z^2+1)) residue - Wolfram|Alpha

Thanks a lot!!

2. ## Re: Integral with residue, question about poles.

Vanishing residue says the coefficient of the -1 term in the series expansion is 0( since it is 1/z^2), which doesn't imply the singularity is removable.

3. ## Re: Integral with residue, question about poles.

sin(z)/z would be a removable singularity. $\displaystyle sin(z)/z^3$ is, as you say, a pole of order 2. Yes, the residue at 0 is 0. That does NOT mean the the singularity is removable.

sin(z) is analytic and has McLaurin series $\displaystyle z- \frac{1}{3!}z^3+ \frac{1}{5!}z^5- \frac{1}{7!}z^7+ \cdot\cdot\cdot$

So the Laurent series for $\displaystyle \frac{sin(z)}{z^3}$ is $\displaystyle \frac{sin(z)}{z^3}= z^{-2}- \frac{1}{3!}+ \frac{1}{5!}z^2- \frac{1}{7!}z^4+ \cdot\cdot\cdot$
z= 0 is not a "removable singularity" because that series has negative powers. It is a pole of order 2 because the largest negative power is -2. The residue at z= 0 is the coefficient of the -1 power which is 0.

4. ## Re: Integral with residue, question about poles.

Alright, I thought that when the residue was 0 it was a removable singularity that's why I had so many doubts! Thanks for clearing that up! .