Let $$f(t)=\cases{\frac{1}{(1+t)^3}&t>0\\0&t<0}$$ Does:
a.$\hat{f}$ is differentiable?
b.$\hat{f}\in L^1(\mathbb{R})$?
c.$\hat{f}\in L^2(\mathbb{R})$?
Seems like we need to calculate the transform $\hat{f}$. We know that it's the second derivative of $F(t)=\frac 1 {1+t}$ so we need only to calculate the $\mathscr{F}(F)(\omega)$ but now I got a divergent integral since $$\int_{-\infty}^\infty\frac{\exp(-i\omega t)}{1+t}dt=\exp(i\omega)\int_{-\infty}^\infty\frac{\exp(-i\omega u)}{u}du$$ which is divergent. How can I find explicitly the transform? Can I use some other theorems to get it?
You can answer the questions without computing $\hat f$. I give you hints that will allow you to find the answer.
a)$x\,f(x)$ is integrable.
b) $f$ is not continuous.
c) Is $f\in L^2(\mathbb{R})$?
By the way, $$ \hat f(\omega)=\int_0^\infty\frac{e^{-i\omega}}{(1+t)^3}\,dt. $$