Explaining integrals equality $$\int_{-2\pi}^0f(y)e^{iny} dy = \int_0^{2\pi}f(y)e^{-i(-n)y} dy $$
Can you please explain why is this equality true? 
I know that $\int_a^b f= - \int_b^a f$ but how is this applied here?
 A: Assuming that $f$ is $2\pi$-periodic and $n\in\mathbb{Z}$, this holds after the change of variables $y\mapsto y-2\pi$:
$$
\begin{align}
\int_{-2\pi}^0f(y)e^{iny}\mathrm{d}y
&=\int_0^{2\pi}f(y-2\pi)e^{in(y-2\pi)}\mathrm{d}y\\
&=\int_0^{2\pi}f(y)e^{iny}\mathrm{d}y
\end{align}
$$
A: Starting with the right hand side
$\int_0^{2\pi}f(y)e^{-i(-n)y} dy = \int_0^{2\pi}f(y)e^{iny} dy$
f(x) being $2\pi$ periodic means that $f(x) = f(x+2\pi)$ for all values of x
also note that $e^{ix}$ is $2\pi$ periodic since $e^{ix} = cos(x) + i sin(x)$ from eulers formula (and sin and cos themselves are $2\pi$ periodic)
Note that the product of two functions ($f$ and $g$) that are $2\pi$ periodic produces another $2\pi$ periodic function ($h$). 
ie: $f(x) g(x) = h(x)$
$f(x) g(x) = f(x+2\pi) g(x+2\pi) = h(x+2\pi)$
Therefore $h(x) = h(x+2\pi)$ for all values of x
so $f(y)e^{iny}$ is a $2\pi$ periodic function
so shifting the domain of the integral by $2\pi$ wont change the result
$\int_0^{2\pi}f(y)e^{-i(-n)y} dy = \int_0^{2\pi}f(y)e^{iny} dy = \int_{-2\pi}^0f(y)e^{iny} dy$
