# Integration by parts (convolution)

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.

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