How to know if an integral is well defined regardless of path taken.

Question

I can calculate

\begin{equation*} \int_0^i ze^{z^2} dz=\frac{1}{2e}-\frac12, \end{equation*}

but why can I calculate this irrelevant to the path taken? Is this since it is analytic everywhere - if so, how would I go about verifying this? I can't see how to apply the Cauchy Riemann equations here since I don't know how I would break this into the sum of a real and complex component.

Answer 1

Let $f : \mathbb{C} \to \mathbb{C}$ be any complex-valued function over $\mathbb{C}$. let $u, v$ be its real and imaginary part:

$$f(z) = u(z) + i v(z)$$

Let $p, q \in \mathbb{C}$ be any two distinct points and $\gamma : [0,1] \to \mathbb{C}$ be any path in $\mathbb{C}$ joining $p$ to $q$. i.e

$$\gamma(0) = p,\; \gamma(1) = q$$ If one inspect the integral of $f$ over $\gamma$ carefully, one find:

$$\begin{cases} \Re\left[ \int_{\gamma} f dz \right] &= \int_{\gamma} u dx - v dy\\ \,\,\Im\left[ \int_{\gamma} f dz \right] &= \int_{\gamma} v dx + u dy\\ \end{cases}\tag{*1}$$ If we want the integral $\int_\gamma f dz$ independent of choice of $\gamma$ and depends only on the end points $p, q$, the two line integrals on RHS of $(*1)$ need to independent of choice of $\gamma$ too.

Using the Green's theorem, this will happen if $$ \begin{cases} \frac{\partial(-v)}{\partial x} - \frac{\partial u}{\partial y} &= 0\\ \frac{\partial u }{\partial x} - \frac{\partial v}{\partial y} &= 0\\ \end{cases} \quad\iff\quad \begin{cases} \frac{\partial v}{\partial x} + \frac{\partial u}{\partial y} &= 0\\ \frac{\partial u}{\partial x} - \frac{\partial v}{\partial y} &= 0\\ \end{cases} $$ But the two equations on the right is nothing but the Cauchy-Riemann equations.

So if $f$ is analytic (eg. your $z e^{z^2}$), then by definition it satisfies the Cauchy-Riemann equations. You can then use Green's theorem to deduce the integral $\int_\gamma f dz$ is independent of path $\gamma$. This is the core of the Cauchy's integral theorem.

Actually, this path independence can be used as an alternate definition of analyticity. Quoting wiki, Morera's theorm states that a continuous, complex-valued function ƒ defined on a connected open set $D \subset \mathbb{C}$ that satisfies $$\oint_\gamma f(z)\,dz = 0$$ for every closed piecewise $C^1$ curve $\gamma$ in $D$ must be holomorphic on $D$.

I hope this will clarify the relations among the path independence of $\int_\gamma f dz$, Cauchy Riemann equations, Green's theorem and Cauchy's integral theorem.

Answer 2

The integrand is a product of a composition of entire functions. Hence it is analytic everywhere, and in particular analytic on the simply connected domain $\mathbb{C}$. So by Cauchy's integral theorem, for every path $\gamma_1$ in $\mathbb{C}$ from $0$ to $i$, you can create a closed path by completing the path (using any other path $\gamma_2$ from $i$ to $0$), and the integral of that closed path is zero, so what does that say about the integrals along the two paths?

Answer 3

Going to the Cauchy-Riemann equations is not a good way of showing that $z\mapsto ze^{z^2}$ is differentiable everywhere, just like reasoning explicitly about $\lim_{h\to 0}\frac{f(x+h)+f(x)}{h}$ wouldn't be a good way to investigate whether the real function $x\mapsto xe^{x^2}$ is differentiable everywhere.

Instead note that the symbolic differentiation rules you learned in ordinary real calculus still work in the complex case: The product and sum of differentiable functions are differentiable, the composition of differentiable functions is differentiable, and so forth -- with the expected derivatives! So $$ \frac{d}{dz} ze^{z^2} = e^{z^2} + z(\frac d{dz}e^{z^2}) = e^{z^2} + z\cdot 2z \cdot e^{z^2} $$ by the product and rule and then the chain rule. Since this computation works for every $z$, the function $ze^{z^2}$ is differentiable everywhere in the complex plane, and thus analytic, so Cauchy's integral theorem applies to it.