It is always true that $ \int\limits_{-t_0}^{t_0} f(t)\,e^{-iwt}\,dt = 2\,\int\limits_{0}^{t_0} f(t)\,e^{-iwt}\,dt$?? I think that is wrong but I can´t find my mistake.
First, let $f(t)$ and $F(w)$ been a symmetric-time-limited function and its symmetric-time-limited Fourier transform defined as: $$f(t) = x(t)\cdot \left(\theta(t+t_0)-\theta(t-t_0) \right)$$ for a time $t_0 >0$, $\theta(t)$ the standard unitary step function, and $x(t)$ a arbitrary continuous and differentiable function (so, it could be even, odd, or whatever - not necessarily symmetric), so its Fourier transform is given by the following definition (if you choose other definition is going to be at least a difference of a constant, so feel free to use which is more comfortable for you): $$F(w) = \int\limits_{-t_0}^{t_0} f(t)\,e^{-iwt}\,dt $$ $$f(t) = \frac{1}{2\pi}\int\limits_{-\infty}^{\infty} F(w)\,e^{iwt}\,dw $$ To avoiding any problem at its compact-support boundaries $\partial t =\{-t_0,\,t_0\}$ assume that $x(-t_0) = x(t_0) = 0$ (without lost of generality I think, since that problem is easy avoidable using the transform $\mathring{\mathbb{F}}\{\cdot\}$ I am using into this another question, so I am indistinguishable using a transform for which its Fourier spectra is "well behaved").
I am splitting the transform in two, so I will list step by step what I am doing, to see if someone can spot where I am making my mistake: $$ F(w) = \int\limits_{-t_0}^{t_0} f(t)\,e^{-iwt}\,dt = \int\limits_{-t_0}^{0} f(t)\,e^{-iwt}\,dt+\int\limits_{0}^{t_0} f(t)\,e^{-iwt}\,dt$$ and let name the positive-time part as $$G(w) = \int\limits_{0}^{t_0} f(t)\,e^{-iwt}\,dt$$ so now, working only in the negative-time part: $$I(w) = \int\limits_{-t_0}^{0} f(t)\,e^{-iwt}\,dt$$ by using the following change of variable: $$\begin{array}{c} u \cong -t \Rightarrow du = -dt \\ t=0 \Rightarrow u=0 \\ t=-t_0 \Rightarrow u=t_0 \end{array}$$ I will have that $$ I(w) = \int\limits_{t_0}^{0} f(-u)\,e^{-iw(-u)}\,(-du)= - \int\limits_{t_0}^{0} f(-u)\,e^{iwu}\,du = \int\limits_{0}^{t_0} f(-u)\,e^{iwu}\,du$$ now, using another change of variable $w = - z$, I will have that $$I(w) = I(-z) = \int\limits_{0}^{t_0} f(-u)\,e^{-izu}\,du$$ which is the same form of the transform $G(w)$ but applied to $f(-t)$. Now, using the time-scaling property of the Fourier transform: $$\text{If}\,f(t) \overset{\mathbb{F}\{\cdot\}}{\rightarrow} F(w),\,\text{then}\,f(-t) \overset{\mathbb{F}\{\cdot\}}{\rightarrow} F(-w)$$ I will have that $$I(w) = I(-z) = \int\limits_{0}^{t_0} f(-u)\,e^{-izu}\,du \overset{\text{time-scaling prop.}}{=} G(-z) = G(w)$$ So $$F(w) = 2\,G(w)$$ As I am asking at the beginning, but I believe is wrong... Hope you can spot the mistake.
