Is the integral of the Fourier transform of derivative a function equal to the Fourier transform of the function?
For an example, I have a function $\theta(x)=-2\tan^{-1}(\frac{x}{c})$.
I find computing Fourier transform of this function quite difficult. So, I was planning to take the derivative of the function. $$\theta'(x)=-\frac{2}{c \left(\frac{x^2}{c^2}+1\right)}$$ I can easily compute the Fourier transform of this function $\theta'(p)=\sqrt{2 \pi } \left(-e^{-c | p| }\right)$. Integrating on both sides: $\theta(p)=\frac{| p| \left(\sqrt{2 \pi } e^{-c | p| }\right)}{c p}$
Is this the correct Fourier transform of the function $\theta(x)$?
Not true. The FT of $f'$ is $-it \hat {f} (t)$. So to find $\hat {f} (t)$ you have to just divide $\hat {f'}(t)$ by $-it$ instead of integrating it.