I am doing a fourier series example, and the solution contains of a somewhat surprising claim.
Let $f(t)$ be a real-valued, even function, that is, $f(t) = f(-t)$.
Then
$$\int\limits_{-\infty}^\infty f(t) (\cos(\omega t) + i\sin(\omega t)) \mathrm{dt} = \int\limits_{-\infty}^\infty f(t) \cos(\omega t) \mathrm{dt} $$
Because "even $\times$ odd = 0, even $\times$ even is not necessarily zero"
Is there a way to justify the above explanation?
If $f$ is even and $g$ is odd, then $h := f \cdot g$ is also odd, so its integral over an interval $[-T,T]$ is zero. $$\int_{-T}^Th(t) \mathop{dt} = \int_0^Th(t) \mathop{dt} + \int_{-T}^0 h(t) \mathop{dt} = \int_0^T h(t) \mathop{dt}- \int_0^T h(t) \mathop{dt} = 0 $$ Here, $g(t) = i \sin (\omega t)$.
If $f$ is even and $g$ is even, then $f \cdot g$ is also even, and all we can say is $$\int_{-T}^T h(t) \mathop{dt} = 2 \int_0^T h(t) \mathop{dt}.$$ Here, $g(t) = \cos(\omega t)$.
In response to your comment: in general ($h$ doesn't have to be odd or even), we have $$\int_{-T}^0 h(t) \mathop{dt} = -\int_T^0 h(-u) \mathop{du} = \int_0^T h(-u) \mathop{du}.$$ When $h$ is odd, then this becomes $- \int_0^T h(u) \mathop{du}$. When $h$ is even, this becomes $\int_0^T h(u) \mathop{du}$. I used these facts in my above computations.
I think you can consider integral on $(-\infty,0)$ and $[0,+\infty)$
