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I have to calculate this integral, and I also have the exact result ($- \pi$) but honestly I really have no idea how to calculate it. Can someone please explain me this problem?

Thank you :( enter image description here enter image description here

EDIT: $ \hat g(w) = \int_{-\infty}^{\infty} g(x) \cdot e^{-iwx} dx $

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    It would be useful to include your definition of $\hat{g}(\omega)$ because the normalization constants differ from context to context. – 2017-01-25
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    Additionally, are you aware of results linking $$\int_{-\infty}^{+\infty}\hat{g}(\omega)\,f(\omega)\,d\omega\quad\text{and}\quad \int_{-\infty}^{+\infty}g(t)\,\hat{f}(t)\,dt $$ ? – 2017-01-25
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    I added the $ \hat g(w) $ . And yes, I am. – 2017-01-25
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    Well, in such a case you just have to compute the Fourier transform of $\omega \sin(\omega/2)$ (that is a distribution, a linear combination of two $\delta'$) and check the problem is actually asking you to compute the derivative of $g(x)$ at $x=\frac{1}{2}$ and nothing more. – 2017-01-25
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    In the end I used Eulero for the sin and added a i/i in order to anti-transform g'. Thank you so much – 2017-01-27

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In the end I solved the problem: $$ \frac{1}{2i} \int_{-\infty}^{+\infty} \omega \hat g(\omega) (e^{i\omega\frac{1}{2}} - e^{i\omega-\frac{1}{2}}) dw $$ $$ = \frac{1}{2i} [ \frac{1}{i} \int_{-\infty}^{+\infty}i\omega \hat g(\omega) e^{i\omega\frac{1}{2}} dw - \frac{1}{i} \int_{-\infty}^{+\infty}i\omega \hat g(\omega) e^{i\omega\frac{-1}{2}} dw \ ] $$ $$ = \frac{1}{2i} \frac{2\pi}{i} ( g'(\frac12 ) - g'(-\frac12 ) ) = -\pi $$