Integration by parts (convolution)

Question

I am trying to evaluate the integral

$$F(\nu) = \frac{jT}{2 \pi} \int^{\infty}_{- \infty} \delta ' (\nu ') \ \text{sinc} (T (\nu-\nu')) e^{-j 2 \pi (\nu-\nu') (T/2)} \ d \nu'$$

Where $\delta$ represents the Dirac delta function.

Attempt

I tried using integration by parts but the answers I got did not look right.I think it is better to use the $\text{sinc}$ function as u (since it is easier to differentiate than to integrate).

I also tried it with the substitution $\delta ' (\nu ')d \nu' = d\delta(\nu ')$, so that the integral becomes

$$F(\nu ) = \frac{jT}{2 \pi} \int^{\infty}_{- \infty} \ \text{sinc} (T (\nu-\nu')) e^{-j 2 \pi (\nu-\nu') (T/2)} \ d\delta(\nu ')$$

I am stuck here. Any help would be greatly appreciated.

P.S. This integral is the convolution of the Fourier transform of two functions.

Answer 1

by using sampling property( we know Dirac's delta function is the unity of convolution in other word convolution of everything with delta is itself ) of Dirac delta function we get: $$\int_{-\infty}^{\infty}\delta (t-t_0)f(t)dt=\int_{-\infty}^{\infty}\delta (t)f(t+t_0)dt=f(t_0)$$ of equivalently: $$\delta(t) * f(t)=f(t)*\delta(t)=f(t)$$ where $*$ stands for convolution. we can simplify the integral as follow: $$F(\nu) = \frac{jT}{2 \pi} \int^{\infty}_{- \infty} \delta ' (\nu ') \text{sinc} (T (\nu-\nu')) e^{-j 2 \pi (\nu-\nu') (T/2)} \ d \nu'=\frac{jT}{2 \pi}\text{sinc} (T \nu) e^{\frac{-j 2 \pi (\nu)T} {2}}$$