Evaluation of this line integral $\int z^2\cdot e^\left(\frac{1}{z}\right)\cdot\sin\left(\frac{1}{z}\right)dz$

Question

where the contour is $$|z|=1$$

if I use residual theorem $$res f(z) = \lim\limits_{z\to 0} z^3\cdot e^\left(\frac{1}{z}\right)\cdot \sin\left(\frac{1}{z}\right)$$

this limit is undefined .However , if i expand the function in its laurent series i can find coefficient of first negative power which is equal to $\left(\frac{2}{3}\right)$ How is this possible $?$

You can always find residues via the Laurent series. However, the limit formula won't work for an essential singularity or any singularity where the Laurent series is infinite in the negative direction. — Ben Grossmann, Mar 26 '16 at 05:10

score 2 · Answer 1 · answered Mar 26 '16 at 05:12

2

The "quick" formulae for residue only works for simple poles and poles "of finite order". They won't work for essential singularities though. However, Laurent series expansion always works.

answered Mar 26 '16 at 05:12

Henricus V.

18,694

Evaluation of this line integral $\int z^2\cdot e^\left(\frac{1}{z}\right)\cdot\sin\left(\frac{1}{z}\right)dz$

1 Answers1