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Please consider the following curve integral: $$I:=\int_{\partial B_1(2i)}\frac{e^{z^2}}{2i-z}dz$$ where $$B_r(z_0):=\left\{z\in\mathbb{C}:|z-z_0|<r\right\}$$

Let $\gamma :[a,b]\to\Omega$ denote a piecewise continuously differentiable path in $\Omega$, $\gamma^*:=\gamma([a,b])$ denote its trace, and $f:\gamma^*\to\mathbb{C}$ be continuously $\Rightarrow$ $$\int_\gamma f(z)dz:=\int_a^b\gamma '(t)f(\gamma (t))\;dt$$ is called curve integral of $f$ along $\gamma$.

Using the definition above, we can calculate $I$. However, I'm sure there is an easier way to calculate the integral. I thought about Cauchy's integral theorem or Cauchy's integral formula . I'm new to them and still unsure how they can be applied here.

0xbadf00d
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To use Cauchy's integral formula, in the notation of the wikipedia link you provided, set $f(z)=-e^{z^2}$ and $a=2i$. This gives you $I=-\left.2\pi ie^{z^2}\right|_{z=2i}=-2\pi e^{-4}i$.

You mentioned Cauchy's Integral Theorem, but it requires the integrand to be holomorphic. It isn't, so it doesn't contradict the above.

As an added bonus, one can even infer that $z\mapsto -\dfrac{e^{z^2}}{z-2i}$ doesn't have an antiderivative in $\mathbb C$ (or in any neighborhood of $2i$) for if it did, since the path is closed, the integral would be $0$.

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Git Gud
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  • If $$f(z):=\frac{e^{z^2}}{2i-z}$$ can't we restrict $f$'s domain to $B_1(2i)$? Then, since $B_1(2i)$ is star-shaped, Cauchy's integral theorem would be applicable. – 0xbadf00d Jun 02 '14 at 12:26
  • @oxbadfood You would need $f$ holomorphic in $B_1(2i)$, it isn't even defined there. – Git Gud Jun 02 '14 at 12:27
  • You're absolutely right. As stated in the comments above, I thought about $B_2(0)$ while writing $B_1(2i)$. Thanks a lot. – 0xbadf00d Jun 02 '14 at 12:32
  • @oxbadfood No problem. – Git Gud Jun 02 '14 at 12:33
  • Cauchy's integral theorem doesn't require the domain to be simply connected. It requires that the cycle over which one integrates is null-homologous (the winding number around any point not in the domain is $0$). – Daniel Fischer Jun 02 '14 at 12:41
  • @DanielFischer I confess that that generalization is beyond me. The versions the OP provided/knows aren't that general. In any case I edited out that part of my answer. Thanks. – Git Gud Jun 02 '14 at 13:02
  • @GitGud You've stated that Cauchy's integral theorem requires the integrand to be holomorphic. But as far as I can tell Cauchy's integral formula does require this too. From the wikipedia article: The formula can be applied if $f : U\to\mathbb{C}$ is holomorphic, $D$ is a closed disk which is completely contained in $U$ and $\gamma$ is the circle forming the boundary of $D$. So, if we need $f$ to be holomorphic in $U$ and $D\subset U$, then $f$ must be holomorphic in $D$. What am I missing? – 0xbadf00d Jun 02 '14 at 13:15
  • Illustration: Consider an annulus, $a < \lvert z\rvert < b$. For $a < \rho < b$, let $\kappa_\rho(t) = \rho\cdot e^{it}$ the circle of radius $\rho$. Then $\kappa_R - \kappa_r$ is a null-homologous cycle in the annulus, and Cauchy's integral theorem says $$\int_{\kappa_R - \kappa_r} f(z),dz = 0$$ for all $f$ holomorphic in the annulus. In other words, the integral of $f$ over a circle doesn't depend on the radius. If you only have the integral theorem for simply connected domains, that result is harder to come by (in such a geometrically simple situation not much harder, admittedly). – Daniel Fischer Jun 02 '14 at 13:16
  • @0xbadf00d Cauchy's integral formula deals with the integral $$\int_\gamma \frac{f(\zeta)}{\zeta-z},d\zeta$$ where $f$ is holomorphic in the region enclosed by $\gamma$. The integral theorem deals with $$\int_\gamma f(\zeta),d\zeta.$$ – Daniel Fischer Jun 02 '14 at 13:17
  • @DanielFischer I've just found that out. At first glance, I didn't noticed that the $f$ in the formula is different from the $f$ in the theorem. But thanks anyway. – 0xbadf00d Jun 02 '14 at 13:26
  • @DanielFischer Thank you. I will read your comment later. – Git Gud Jun 02 '14 at 13:39