How to integrate on a complex interval?

Question

In textbook,it gives me a theorem:$w(t) \in C^{1}([a,b]),then \int^{b}_{a}w^{'}(t)dt=w(b)-w(a)$,How do I deal with the case where a,or b is a not real number?For example:$\int^{ \frac{\pi}{4}i}_{0}e^{it}dt=?$，It is different from definition,I can't integrate the real part and the imaginary part separately.I dont know how to handle $[0,\frac{\pi}{4}i] $ and integrand is a real-valued function.

Answer 1

In general there is no well-defined integral between two complex numbers.

But there is a theory of path integration: if $\gamma : [a,b] \to \mathbb C$ is a path in the complex plane, and if $f : \mathbb C \to \mathbb C$ is a continuous function then one can define the path integral $$\int_a^b f(\gamma(t)) \, dt $$ One of the things you learn in a complex analysis course is that if $f$ is a holomorphic function on the whole plane then $f$ has a complex antiderivative $F : \mathbb C \to \mathbb C$, i.e. $F'(z)=f(z)$ for all $z \in \mathbb C$, and furthermore there is a version of the fundamental theorem of calculus: $$\int_a^b f(\gamma(t)) \, dt = F(\gamma(b)) - F(\gamma(a)) $$ By a somewhat severe abuse of notation, perhaps one might write the first integral in the format $$\int_{f(a)}^{f(b)} f(\gamma(t)) \, dt $$ which sorta-kinda matches what you have written in your post, although I would not advise that in general.

Anyway, you could work this out for the case $$\gamma : [0,\pi/4] \to \mathbb C, \qquad \gamma(t) = i \, t $$ $$\gamma(0) = 0 \qquad \gamma(1) = \frac{\pi}{4} \, i $$ $$f(z) = e^{z}, \qquad f(\gamma(t)) = e^{it} $$ Using antiderivative $F(z) = e^{z}$ (and continuing the abuse of notation) you get $$\int_0^{\frac{\pi}{4} i} e^{it} \, dt = F(\pi / 4) - F(0) = e^{\pi/4} - 1 $$