Inverse Fourier transform Of a triangular impulse I have to find the expression of this graphic and after find the inverse Fourier transform of it. First of all I found that the expression of the graphic is $$ X(f) = \frac{1}{2} tri (\frac{f+f_0}{B}) - \frac{1}{2} tri(\frac{f-f_0}{B})$$. Now to find inverse Fourier transform , my book give me the advice to multiply numerator and denominator for i. But $$ i=e^{i \frac{\pi}{2}} $$ and $$ -i=e^{-i \frac{\pi}{2}} $$. Now I know that the Fourier transform of a triangular impulse is $$ (sinc(f)^{2}) $$ and that $$ \frac{d}{dt} tri(t) = rect ( t + \frac{1}{2}) - rect ( t - \frac{1}{2}) $$ but I don’t know how to apply correctly integration property of my x(t). Thank you so much
It's nice to see alternative paths, like this one: First of all, properties:


Frequency shifting: $x(t)e^{j2\pi f_0 t} \longleftrightarrow X(f-f_0)$ 
Duality: if $x(t) \longleftrightarrow X(f)$ then $X(t) \longleftrightarrow x(-f)$


You have correctly expressed the depicted transform: $$X(f) = \frac{1}{2}\mathrm{tri}\Big(\frac{f+f_0}{B}\Big) - \frac{1}{2}\mathrm{tri}\Big(\frac{f-f_0}{B}\Big)$$

As you said, $\mathrm{tri}\Big(\frac{t}{B}\Big) \longleftrightarrow B\mathrm{sinc}^2(fB)$. From the duality property, you have $$B\mathrm{sinc}^2(tB) \longleftrightarrow \mathrm{tri}\Big(\frac{-f}{B}\Big)$$ but since $\mathrm{tri(\cdot )}$ is an even function, you can write $$B\mathrm{sinc}^2(tB) \longleftrightarrow \mathrm{tri}\Big(\frac{f}{B}\Big)$$

Now you have your FT-pair you need. If you apply the frqeuency shifting property on $\mathrm{tri}\Big(\frac{f\pm f_0}{B}\Big)$, you can easily get... what? :-) You can continue from this point.