Integrating a complex $e$-function

Question

Could someone explain in slow steps how to integrate the complex function: $$ \int_0^{2\pi} e^{i(\pi-x)2\alpha} dx $$ where $\alpha \in \mathbb{R} \backslash \mathbb{Z}$ is just some constant. I'm unfamiliar with complex integration and I keep getting weird outcomes so there's not much use in showing my attempts so far.

Answer 1

If we have a function $f:[0,2\pi]\to\mathbb{C}$ then we can write it in the form $f=u+iv$ where $u,v:[0,2\pi]\to\mathbb{R}$. Then by definition $f$ is integrable if both $u$ and $v$ are integrable and if they are then $\int_0^{2\pi} f=\int_0^{2\pi}u+i\int_0^{2\pi}v$.

Alright, so your function is $f(x)=e^{i(\pi-x)2\alpha}$ where $\alpha$ is a real number. How can we write it in the form $u+iv$? Of course using Euler's formula $e^{i\theta}=\cos\theta+i\sin\theta$. So we have $f(x)=\cos((\pi-x)2\alpha)+i\sin((\pi-x)2\alpha)$. In other words $u(x)=\cos((\pi-x)2\alpha)$ and $v(x)=\sin((\pi-x)2\alpha)$. These are real functions and I suppose you know how to integrate them. So now just use the definition to solve the complex integral.

Now, do we have to use the definition of the integral every time? The answer is no. Do it once and convince yourself that if you just integrate this exponential function like you integrate real exponential functions (treating $i$ as a scalar) then you will get the same result.