Cauchy integral, why is it equal zero?

Question

I have the following Cauchy integral evaluated only in $\mid\zeta\mid\leq 1$,

$$\int_\gamma t\cdot\frac{t+\zeta}{t-\zeta}\frac{dt}{t}$$

which has one (?) pole at $\zeta$.

$\gamma$ is circle of unit radius in the plane of complex variables $\zeta$

$\zeta$ is arbitrary points in the plane of complex variables

$t$ are points located in the unit circle $\gamma$

Thus, the solution can be evaluated by,

$$\int_\gamma t\cdot\frac{t+\zeta}{t-\zeta}\frac{dt}{t}=\int_\gamma \frac{t}{t}\cdot\frac{t+\zeta}{t-\zeta} dt=2\pi i(\zeta+\zeta)=4\pi i\zeta$$

Now, suppose I want to evaluate the following integral:

$$\int_\gamma \frac{1}{t}\cdot\frac{t+\zeta}{t-\zeta}\frac{dt}{t}$$

is it correct to say that the integral also has one (?) pole at $\zeta$? and hence,

$$\int_\gamma \frac{1}{t}\cdot\frac{t+\zeta}{t-\zeta}\frac{dt}{t}=\int_\gamma \frac{1}{t^2}\cdot\frac{t+\zeta}{t-\zeta}dt = 2\pi i\left(\frac{\zeta+\zeta}{\zeta^2}\right)=2\pi i\left(\frac{2}{\zeta}\right)=\frac{4\pi i}{\zeta}$$

The reason I am asking is that the solution says the result of the last integral equals zero instead of $\frac{4\pi i}{\zeta}$.

Which one is correct?

Can anyone show me the correct solution please?

Answer 1

Switching the $t$ out for $\frac{1}{t}$ changed the integral quite a bit

$$\oint_\gamma\frac{1}{t^2}\frac{t+\zeta}{t-\zeta}dt=2\pi i\sum_{j=1}^n\operatorname{Res}(f(t),t_j)$$

Now there's a pole at $t=\zeta$ (order $1$) and $t=0$ (order $2$), so

$$\operatorname{Res}(f,0)=\lim_{t\to0}\frac{d}{dt}\left(t^2\frac{1}{t^2}\frac{t+\zeta}{t-\zeta}\right)=\lim_{t\to0}\frac{d}{dt}\left(\frac{t+\zeta}{t-\zeta}\right)= \lim_{t\to0}\left(\frac{-2\zeta}{(t-\zeta)^2}\right)=-\frac{2}{\zeta}$$

Assuming that $\gamma$ encloses both $t=0$ and $t=\zeta$ then,

$$\sum_{j=1}^n\operatorname{Res}(f(t),t_j)=\frac{2}{\zeta}-\frac{2}{\zeta}=0$$

And the integral is $0$

Answer 2

By the residue theorem the value of $$ \oint \frac{z+a}{z-a}\,dz =2a\oint\frac{dz}{z-a}$$ depends on $a$ falling inside or outside the (simple) integration contour. If $a$ lies on the integration contour such integral is non-convergent in the usual sense, but its principal value is just the average of the values computed by considering it inside/outside the given contour.

Answer 3

Note that for $|z| > |\zeta|$ we have ${1 \over z -\zeta} = { 1 \over z} (1+ {\zeta \over z} + {\zeta^2 \over z^2} + \cdots)$.

Then the coefficient of ${ 1\over z}$ in the Laurent expansion of $z \mapsto {z+\zeta \over z -\zeta}$ can be read off as $ 2 \zeta$ (one from the $z$ in $z+\zeta$ and one from the $\zeta$ in $z+\zeta$).

Similarly, for $z \mapsto {1 \over z^2}{z+\zeta \over z -\zeta}$, we can see that the Laurent series expansion has only terms of ${1 \over z^2}$ and higher, hence the integral is zero.