So far I have tried the following:
$$\begin{align}
\mathscr{F}(f)&=\mathscr{F}\{1-\frac{|\tau|}{2T}\}\\
&=\int_{-\infty}^{+\infty}(1-\frac{|\tau|}{2T})e^{-i\omega\tau}d\tau\\
&=\int_{-\infty}^{0}(1+\frac{\tau}{2T})e^{-i\omega\tau}d\tau+\int_{0}^{+\infty}(1-\frac{\tau}{2T})e^{-i\omega\tau}d\tau\\
\end{align}$$
We use the method change of variables for the first integral $\tau=-\phi$ , $d\tau=-d\phi$ , $\tau=0\Rightarrow \phi=0$ , $\tau=-\infty\Rightarrow \phi=+\infty$ so the first integral will be:
$\int_{-\infty}^{0}(1+\frac{\tau}{2T})e^{-i\omega\tau}d\tau=\int_{+\infty}^{0}(1-\frac{\phi}{2T})e^{i\omega\phi}(-d\phi)=\int_{0}^{+\infty}(1-\frac{\phi}{2T})e^{i\omega\phi}d\phi=\int_{0}^{+\infty}(1-\frac{\tau}{2T})e^{i\omega\tau}d\tau$
So we will have
$$\begin{align}
\mathscr{F}(f)&=\int_{0}^{+\infty}(1-\frac{\tau}{2T})e^{i\omega\tau}d\tau+\int_{0}^{+\infty}(1-\frac{\tau}{2T})e^{-i\omega\tau}d\tau\\
&=\int_0^{+\infty}(1-\frac{\tau}{2T})(e^{i\omega\tau}+e^{-i\omega\tau})d\tau
\end{align}$$
Now from integration by parts method we have:
$$\begin{array}{c|c}
\text{Derivation}& \text{Integration} \\
\hline
\oplus \quad 1-\frac{\tau}{2T} & e^{i\omega\tau}+e^{-i\omega\tau} \\
\ominus\quad -\frac{1}{2T} & \frac{1}{i\omega}(e^{i\omega\tau}-e^{-i\omega\tau})\\
\oplus \quad0 & \frac{-1}{\omega^2}(e^{i\omega\tau}+e^{-i\omega\tau})
\end{array}$$
$$\mathscr{F}(f)=[(1-\frac{\tau}{2T})\frac{1}{i\omega}(e^{i\omega\tau}-e^{-i\omega\tau})-\frac{1}{2T\omega^2}(e^{i\omega\tau}+e^{-i\omega\tau})]_0^{+\infty}$$
But I don't seem to get to the right answer, which is: $2T[\frac{\sin(\omega T)}{\omega T}]^2$
P.S. please don't recommend me to use propositions of fourier transform like linearity and etc. Also I don't want to use known fourier transdorms. I want to see what's wrong with my integration and my limit calculation in the infity an etc.
I think you really want
$$\begin{align}\int_{-\infty}^{\infty} d\tau \left (1-\frac{|\tau|}{2 T}\right ) \theta(2 T-|\tau|) e^{i \omega \tau} &= \int_{-2 T}^{2 T} d\tau \left (1-\frac{|\tau|}{2 T}\right ) e^{i \omega \tau}\\ &= \frac{2 \sin{2 \omega T}}{\omega} - \frac{1}{T} \int_0^{2 T} d\tau \, \tau \, \cos{\omega \tau}\\ &= \frac{2 \sin{2 \omega T}}{\omega} - \frac{1}{T}\frac{\partial}{\partial \omega} \int_0^{2 T} d\tau \, \sin{\omega \tau} \\ &=\frac{2 \sin{2 \omega T}}{\omega} - \frac{2}{T}\frac{\partial}{\partial \omega} \frac{\sin^2{\omega T}}{\omega} \\ &= 2 T \frac{\sin^2{\omega T}}{(\omega T)^2}\end{align}$$
as you wanted to show.