>Is this always true that Re[$\int$ $w(t)dt$ ]$=$ $\int$ $Re[w(t)]$ $dt$.

Question

Let $w$ be a complex valued function over an interval $ a \leq t \leq b$.That is $w : [a,b] \to \mathbb C$

Is this always true that \begin{align} \text{Re}\left[\int_a^b w(t)\, dt \right] &= \int_a^b \text{Re}[w(t)]\, dt, \end{align} where $\text{Re}[w(t)]$ means real part of $w(t)$?

Answer 1

yes this is always true, same goes for the imaginary part. This is because the only operation on the imaginary part or the real part once it is converted to the polar form to simplify integration is a rotation in the complex
plane.The real and imaginary cannot interact with each other otherwise hence the integral $\int_a^b cos(P(x)dx)=Re(\int_a^b exp(P(x))dx , \int_a^b sin(P(x))dx) = Im(\int_a^b exp(P(x))dx)$.

Answer 2

We know that

$w(t) = u(t) + iv(t)$ [Since $w$ is a complex valued function] thus

$\int_a^b w(t)\, dt$ =$\int_a^b u(t)\, dt$ $+$ $i$$\int_a^b v(t)\, dt$

And hence

\begin{align} \text{Re}\left[\int_a^b w(t)\, dt \right] &= \int_a^b \text{Re}[w(t)]\, dt, \end{align}

Thank you Torsten Schoeneberg