Calc Residue of \frac{z \sin{z}}{\left( z - \pi \right)^3}

[eXorikos]

Homework Statement

Calculate the residu for the singularity of $$\frac{z \sin{z}}{\left( z - \pi \right)^3}$$

Homework Equations

$$R \left( a_0 \right) = \frac{1}{2 \pi i} \oint \frac{z \sin{z}}{\left( z - \pi \right)^3} dz$$

The Attempt at a Solution

$$\pi$$ is an essential singularity so the residue cannot be calculated using the limit.
I have tried calculating the integral around the path from 3-i to 4-i to 4+i to 3+i to 3-i. In principle this should work, but Maple gives me a wrong answer if I try to evaluate.

$$\frac{1}{2 \pi i} \left( \int^4_3 \frac{\left( x - i \right) \sin{x - i}}{\left( x - i - \pi \right)^3} dx + \int^1_{-1} \frac{\left( 4 + iy \right) \sin{4 + iy}}{\left( 4 + iy - \pi \right)^3} dy + \int^3_4 \frac{\left( x + i \right) \sin{x + i}}{\left( x + i - \pi \right)^3} + \int^{-1}_1 \frac{\left( 3 + iy \right) \sin{3 + iy}}{\left( 3 + iy - \pi \right)^3} dy \right)$$

 

[vela]

How did you conclude that $z=\pi$ is an essential singularity?

 

[eXorikos]

I was wrong on that. Calculating the residue using the limit works fine, but my problem stands. The integral should give the same result.

 

[vela]

You're missing a factor of i on the y integrals since dz = i dy.

 

[eXorikos]

It works fine thanks!

Is there a better choice that makes it easier to calculate the integral?

 

[vela]

Perhaps let x run from 0 to 2 pi to take advantage of the periodicity of sin z, but I'd try to avoid the contour integrals in the first place.