Integral with complex function

 

Def 1

Let $f : \mathbb{R} \to \mathbb{C}$ by

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

In this case, if both $u$ and $v$ are integrable, then $f$ is said to be integrable.

The same applies to differentiation. We define

$\displaystyle \int f \; \mathrm{d}t = \int u \; \mathrm{d}t + i \int v \; \mathrm{d}t$

$\displaystyle f' = u' + i v'$

 

 

 

Prop 1

Let $f : \mathbb{R} \to \mathbb{C}$ . For $\alpha \in \mathbb{C}$, 

$\displaystyle \int \alpha f \; \mathrm{d}t = \alpha \int f \; \mathrm{d}t$

 

  Let $\alpha = a + ib$  and  $f(t) = u(t) + iv(t)$ .

 

  Since $\alpha f = (au - bv) + i (av + bu)$,

  $\displaystyle \int \alpha f(t) = \int (au - bv) \; \mathrm{d}t + i \int (av + bu) \; \mathrm{d}t$

                    $\displaystyle = (a \int u \; \mathrm{d}t - b \int v \; \mathrm{d}t) + i (a \int v \; \mathrm{d}t + b \int u \; \mathrm{d}t)$

                    $\displaystyle = (a + ib)(\int u \; \mathrm{d}t + i \int v \; \mathrm{d}t)$

                    $\displaystyle = \alpha \int f \; \mathrm{d}t$

 

 

 

Prop 2 

Let $f : \mathbb{R} \to \mathbb{C}$

$\displaystyle \left|\int_{a}^{b} f \; \mathrm{d}t \right| \leq \int_{a}^{b} \left|f \right| \; \mathrm{d}t$

 

  Let $\displaystyle \int_{a}^{b} f \; \mathrm{d}t = re^{i \theta}$

 

  By Prop 1,

  $\displaystyle \int_{a}^{b} e^{-i \theta} f \; \mathrm{d}t = r \in \mathbb{R}$

 

  Let $\mathrm{Re} \; (e^{-i \theta}f(t)) = u(t)$

 

  Since $u \leq \left|e^{-i \theta}f \right| = \left|f \right|$,

  $\displaystyle \left|\int_{a}^{b} f \; \mathrm{d}t \right| = r = \int_{a}^{b} e^{-i \theta} f \; \mathrm{d}t = \int_{a}^{b} u(t) \; \mathrm{d}t \leq \int_{a}^{b} \left|f \right| \; \mathrm{d}t$

 

 

 

Prop 3

Let $f : \mathbb{R} \to \mathbb{C}$

$\displaystyle \int_{b}^{a} f \; \mathrm{d}t = - \int_{a}^{b} f  \; \mathrm{d}t$

 

  Let $f(t) = u(t) + iv(t)$

 

  Therefore

  $\displaystyle \int_{b}^{a} f \; \mathrm{d}t = \int_{b}^{a} u \; \mathrm{d}t + i \int_{b}^{a} v \; \mathrm{d}t$
                    $= - \displaystyle \int_{a}^{b} u \; \mathrm{d}t - i \int_{a}^{b} v \; \mathrm{d}t$
                    $= - \displaystyle \int_{a}^{b} f  \; \mathrm{d}t$