Why $\int_{0}^{\pi} \tan(\theta +ia) \text{d}\theta = \frac{1}{2} \int_{0}^{2\pi} \tan(\theta + ia) \text{d}\theta$

Question

I need to prove that equality in order to solve the integral using the remainder theorem, but I don't really know how to do it. Any help is appreciated.

Answer 1

$\begin{align}&\int_0^{2\pi}\tan(\theta+i\alpha) d\theta=\int_0^{\pi}\tan(\theta+i\alpha)d\theta+\int_{\pi}^{2\pi}\tan(\theta+i\alpha)d\theta \end{align}$

In the second integral change $\theta \to \theta+\pi$. Thus the limits in the second integral become $0$ to $\pi$

So we have,

$\begin{align}&\int_0^{2\pi}\tan(\theta+i\alpha) d\theta=\int_0^{\pi}\tan(\theta+i\alpha)d\theta+\int_{0}^{\pi}\tan(\pi+\theta+i\alpha)d\theta \end{align}$

Using $\tan(\pi+x) = \tan(x)$,

$\begin{align}&\int_0^{2\pi}\tan(\theta+i\alpha) d\theta=\int_0^{\pi}\tan(\theta+i\alpha)d\theta+\int_{0}^{\pi}\tan(\theta+i\alpha)d\theta = 2\int_0^{\pi}\tan(\theta+i\alpha)d\theta\end{align}$

or

$\begin{align} \int_0^{\pi}\tan(\theta+i\alpha)d\theta = \frac12\int_0^{2\pi}\tan(\theta+i\alpha) d\theta\end{align}$