Explaining integrals equality

Question

$$\int_{-2\pi}^0f(y)e^{iny} dy = \int_0^{2\pi}f(y)e^{-i(-n)y} dy $$

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?

Answer 1

Assuming that $f$ is $2\pi$-periodic and $n\in\mathbb{Z}$, this holds after the change of variables $y\mapsto y-2\pi$: $$ \begin{align} \int_{-2\pi}^0f(y)e^{iny}\mathrm{d}y &=\int_0^{2\pi}f(y-2\pi)e^{in(y-2\pi)}\mathrm{d}y\\ &=\int_0^{2\pi}f(y)e^{iny}\mathrm{d}y \end{align} $$

Answer 2

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

$\int_0^{2\pi}f(y)e^{-i(-n)y} dy = \int_0^{2\pi}f(y)e^{iny} dy = \int_{-2\pi}^0f(y)e^{iny} dy$