I have the following integral $$I = \int_{\gamma_{1}} \frac{e^{z^2}}{(z-1)^2}dz,$$ where $$\gamma_{1} : [0,2Pi] \rightarrow \mathbb{C}, \\ \quad t \quad \mapsto \quad 2e^{i t} .$$
I want to show that the value is $$I = \int_{\gamma_{2}} \frac{e^{z^2}}{(z-1)^2}dz = (2 \pi i) (1!) (e^{1^2}),$$
which is calculated easy by Cauchy's Integral Formula for discs, knowing that $$\gamma_{2} : [0,2Pi] \rightarrow \mathbb{C}, \\ \quad t \quad \mapsto \quad 1+e^{i t} .$$
My attempts are the following:
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Let $\gamma = \gamma_{1} – \gamma_{2}$ and using Cauchy's theorem, the integral of my initial fuction $e^{z^2}/(z-1)^2$ is zero along this path and I obtain the equality I was looking for follows.
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Using the Generalized Cauchy's theorem, consider the region $\Omega_2 = \mathbb{D}(0,3) \setminus {1}$ and conclude the equality since my function is analytic in $\Omega_2$ (except for the removable singularity at zero) the path $\gamma = \gamma_1 – \gamma_2$ is homologous to zero with respect to $\Omega_2$.
Here my problems:
I know that I can't apply directly the Cauchy's theorem for discs since my function is not analytic in, for example, the disc $\mathbb{D}(0,3)$. The problems is the point $z=1$. I don't know if the fact that $n(\gamma,1) = 0$ (index) allows us to forget about the point $z=1$, but something sais me that this fact is relevant.
In the second attempt, I think I'm concluding correctly the equality between the integrals in the region $\Omega_2$, but it's still true (the equality) if I extend the region to $\Omega_1= \Omega_2 \cup {1}$?. For this problem I think, that intuitively, the integral only depends in the values of the function along the path (and z = 1 is irrelevant for our study), but it's still a bit confusing to me the role that the topology (z = 1 is relevant topologically) of the set plays in integration theory of complex functions (beyond the connectivity, in the general sense).
It's desirable to obtain a solution to the first attempt, because the second attempt is out of the content of my course (but it's more convincing for me). Despite, some words about the second question will be welcome (and needed since I have no idea about the role of the underlying sets in all this theory).
Thanks to everyone!
Best Answer
You can apply Cauchy's Integral formula to your integrals. To evaluate the first integral you can just use Cauchy's Generalised Integral formula which states:
We can just rearrange this formula to get:
$$\int_C \frac{f(z)}{(z-s)^{n+1}}dz = \frac{2\pi i}{n!}f^{(n)}(s) $$
In the case of your first integral $\int_{\gamma_{1}} \frac{e^{z^2}}{(z-1)^2}dz$. The curve your integrating around is the positively oriented closed circle centred at the origin of radius 2. Since this contour contains the point $z=1$ (the singularity of your integrand) you can just directly apply Cauchy's Integral Formula:
$$ \int_{\gamma_{1}} \frac{e^{z^2}}{(z-1)^2}dz = \frac{2\pi i}{1!} f^{(1)}(1)$$
where $f(z)=e^{z^2}$.
Then $f^{(1)}(z)=2ze^{z^2}$ and hence,
$$ \int_{\gamma_{1}} \frac{e^{z^2}}{(z-1)^2}dz = \frac{2\pi i}{1!} 2e^1=4\pi i e$$
The value for your integral around the contour $\gamma_2$ there is incorrect. The integrand has only one singularity at $z=1$ which is not contained in or on the closed contour $\gamma_2$ you have described there. Since the integrand contains no singularities in $\gamma_2$ it must be analytic on the region bounded by $\gamma_2$ and hence $\int_{\gamma_{2}} \frac{e^{z^2}}{(z-1)^2}dz =0$ using Cauchy's Integral Theorem.