I have the following integral:
$$\dfrac{1}{2\pi} \int_{-\pi}^0 f(\theta)e^{-in\theta} d\theta + \dfrac{1}{2\pi} \int_0^{\pi} f(\theta)e^{-in\theta} d\theta$$
With $f(\theta + \pi) = f(\theta)$
An we make the substitution: $\theta \rightarrow \theta - \pi$
I don't understand why $$\dfrac{1}{2\pi} \int_{-\pi}^0 f(\theta)e^{-in\theta} d\theta + \dfrac{1}{2\pi} \int_0^{\pi} f(\theta)e^{-in\theta} d\theta = \dfrac{1}{2\pi} \int_{0}^{\pi} f(\theta)e^{-in\theta}(-1^n + 1) d\theta$$
Here's what I understand:
I think that they only apply the substitution to the left hand integral of the sum which allows us to get 2 integrals with the same limits. But when I try to do that, I simply get:
$$\dfrac{1}{2\pi}\int_{0}^{\pi} f(\theta)(e^{-in(\theta - \pi} + e^{-in(\theta - \pi} )d\theta$$ and not $$\dfrac{1}{2\pi} \int_{0}^{\pi} f(\theta)e^{-in\theta}(-1^n + 1) d\theta$$
Can someone tell me where I'm wrong or what I'm missing?
They only apply the substitution on the first integral to - as you say - get the limits of the integrals the same. So you get: $$\dfrac{1}{2\pi} \int_{-\pi}^0 f(\theta)e^{-in\theta} \,\mbox{d}\theta + \dfrac{1}{2\pi} \int_0^{\pi} f(\theta)e^{-in\theta} \,\mbox{d}\theta \to \dfrac{1}{2\pi} \int_{\color{red}{0}}^\color{red}{\pi} f(\theta)e^{-in(\color{red}{\theta-\pi})} \,\mbox{d}\theta + \dfrac{1}{2\pi} \int_0^{\pi} f(\theta)e^{-in\theta} \,\mbox{d}\theta$$ Now note that since $e^{i\pi}=-1$: $$e^{-in(\theta-\pi)} = e^{-in\theta}\color{blue}{e^{i\pi n}}=e^{-in\theta}\color{blue}{\left(e^{i\pi}\right)^n}=e^{-in\theta}\color{blue}{\left(-1\right)^n}$$ So you get: $$\dfrac{1}{2\pi} \int_{{0}}^{\pi} f(\theta)e^{-in\theta}\color{blue}{\left(-1\right)^n} \,\mbox{d}\theta + \dfrac{1}{2\pi} \int_0^{\pi} f(\theta)e^{-in\theta} \,\mbox{d}\theta=\int_0^{\pi} f(\theta)e^{-in\theta}\left(1+\left(-1\right)^n \right) \,\mbox{d}\theta$$