All,
If an integral is of the form
$$L(\xi)=\int_C \frac{g(t)}{t-\xi}dt$$
and the function $g(t)$ is analytic inside the unit circle $C$, then we know that, according to the residue theorem
$$L(\xi)=\int_C \frac{g(t)}{t-\xi}dt=2\pi i \cdot \sum residue$$
Now, let consider this function $g(t)$
$$g(t) = t^{k-1}(t+\xi)$$
where $k \geq 1$, then $g(t)$ is indeed analytic the unit circle $C$, then
$$L(\xi)=\int_C t^{k-1}\frac{t+\xi}{t-\xi}dt=2\pi i(\xi^{k-1})(\xi+\xi)=4\pi i \xi$$
My question is now as follow:
Suppose that we have another function $h(t)$ as such
$$h(t) = \frac{1}{g(t)} = \frac{1}{t^{k-1}(t+\xi)}$$
can we get the value of $L(\xi)$ from $h(t)$ as
$$L(\xi)=\int_C \frac{1}{t^{k-1}}\frac{t+\xi}{t-\xi}dt=\frac{4\pi i}{\xi^k}$$
is it correct to put it that way?
Since $h(t) = 1/g(t)$, what can we say about the analyticality of $h(t)$ with respect to our understanding of $g(t)$?
Thank you. Any comments would be appreciated. I am not a mathematician so I am open to any comments.
No you cannot do this. $\int_C\frac 1 {t^{k-1}} \frac {t+\xi} {t-\xi} dt=\frac {4\pi i} {\xi^{k-2}}$ for $k >0$. You can get this by noting that the poles are $0$ and $\xi$ and the residue at $0$ is $0$.