The Fourier Transform of $u(t)$ is $\frac{1}{jω} + \piδ(ω)$.
However I'm troubled with the following fact:
$\frac{1}{jω} + πδ(ω) = \frac{1+jωπδ(ω)}{jω} = \frac{1+j0πδ(ω)}{jω} = \frac{1}{jω}$
So is the Fourier Transform of $u(t)$ just $\frac{1}{jω}$?
Any help is appreciated.
Let us just look at what is going on with the $\delta(\omega)$ term: $$ \pi\delta(\omega) \stackrel{1}{=} \pi\delta(\omega)\cdot 1 \stackrel{2}{=} \pi\delta(\omega)\cdot(\omega\cdot\operatorname{pv}\frac{1}{\omega}) \stackrel{3}{=} (\pi\delta(\omega)\cdot\omega)\cdot\operatorname{pv}\frac{1}{\omega} \stackrel{4}{=} 0\cdot\operatorname{pv}\frac{1}{\omega} \stackrel{5}{=} 0. $$ (Here, "pv" stands for principal value and is just a notation to specify how integration against $\frac{1}{\omega},$ which has divergent integrals at origin, should be done.)
Step 1: Multiplication with $1$ is well-defined no matter if we consider $1$ to be just a number, a smooth function, or a distribution. Generally multiplication of distributions is not defined, but when the two distributions have disjoint singular support, as is the case here, the product can be defined.
Step 2: It is true that $1=\operatorname{pv}\frac{1}{\omega}\cdot\omega$ in the distribution sense, but now we are getting close to a problem: if we remove the parentheses we have a product involving two distributions that have non-disjoint singular support. Both $\delta(\omega)$ and $\operatorname{pv}\frac{1}{\omega}$ have singular support $\{0\}.$
Step 3: Here is where the problem occurs. Although both sides might be defined, multiplication of distributions obviously is not associative. This is most probably related to what I wrote before about $\delta(\omega)$ and $\operatorname{pv}\frac{1}{\omega}$ having non-disjoint singular support.
Step 4: This step is similar to step 2 and I cannot see any real problem here.
Step 5: Also this step is fine, just as step 1.