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!!