How to compute $\int_c \frac{ \cos(z)}{ z^2 (z-1)}$ on the circle |z|=3/2 without the residue theorem

Question

How does one integrate

$\int_C \frac{ \cos(z)}{z^2 (z-1)}$ on the circle C: |z|=3/2?

We cannot use partial fractions and the function is not analytic at z=0 and z=1 which are both in C.

Thanks

Answer 1

By cauchy integral formula We know that $$f^n(a) =\frac{n!}{2i\pi}\int_c\frac{f(z)}{(z-a)^{n+1}}$$ and

$$\frac{\cos z}{z^2(z-1)} =\frac{\cos z}{z-1}-\frac{\cos z}{z}-\frac{\cos z}{z^2}$$ then applies Cauchy integral formula, to get $$\int_c \frac{ \cos(z)}{ z^2 (z-1)}=\int_c \frac{\cos z}{z-1}-\int_c\frac{\cos z}{z}-\int_c\frac{\cos z}{z^2} = \color{red}{2i\pi(\cos 1- \cos 0 - (\cos z)'|_{z=0})}\\ = \color{red}{2i\pi(\cos 1-1)}$$

Answer 2

Hint: Note that the poles $z=0$ (of order $2$) and $z=1$ (a simple pole) are inside circle $|z|=\dfrac32$ and since $\cos z$ is analytic, you can use residue theorem. Also with Cauchy integral formula and drawing a suit contour around $z=0$ and $z=1$ (for example $C_1$ and $C_2$) you may write $$\int_{|z|=\frac32}\dfrac{\cos z}{z^2(z-1)}dz=\int_{C_1}\dfrac{\frac{\cos z}{z^2}}{(z-1)}dz+\int_{C_2}\dfrac{\frac{\cos z}{z-1}}{z^2}dz$$

Answer 3

Use power (Laurent) series for the pole at zero:

$$\frac{\cos z}{z^2(z-1)}=-\frac1{1-z}\frac1{z^2}\left(1-\frac{z^2}2+\frac{z^4}{24}-\ldots\right)=$$

$$-\frac1{z^2}(1+z+z^2+\ldots)\left(1-\frac{z^2}2+\frac{z^4}{24}-\ldots\right)=$$

$$=-\frac1{z^2}\left(1+z+\frac{z^2}2\ldots\right)=\ldots-\frac1z+\ldots\implies Res(f)_{z=0}=-1$$

For the residue at $\;z=1\;$ much easier than using Laurent series is to use limits:

$$Res(f)_{z=1}=\lim_{z\to1}(z-1)f(z)=\lim_{z\to1}\frac{\cos z}{z^2}=\cos 1\implies$$

finally

$$\oint_{|z|=\frac32}\frac{\cos z}{z^2(z-1)}dz=2\pi i(-1+\cos 1)$$