Why is $=\int\limits_{-\infty}^{\infty}\cos(-tx)dF(x)+i\int\limits_{-\infty}^{\infty}\sin(-tx)dF(x)=\int\limits_{-\infty}^{\infty}\cos(tx)dF(x)-i\int\limits_{-\infty}^{\infty}\sin(tx)dF(x)$?
I know $\sin(-tx)=-\sin(tx)$ because of it´s symmetry. But can I just take the minus out of an integral?
That's integral linearity : $$\int_a^b c\ f(t)dt = c\int_a^b f(t)dt$$ Here with $c = -1$.
For example, $$\int_a^b c\ dt = c\int_a^b dt = c\ (b-a)$$