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So I have function $$f(t) = 5+H(t-2)$$

and from that I can gather that $$f(t) = 6, t[2,\infty]$$ $$f(t) = 0, t[-\infty,2[$$

So I calculated $$\int_{2}^\infty6e^{-i\omega t}$$

and this one does get a "sensible answer", but is much different from what I get when I use Wolfram Alpha's Fourier Transformation widget to check the answer. My answer is $\frac{-6ie^{-2i\omega}}{\omega}$, but Wolfram Alpha spits out answer with delta functions and other scrap.

Wolfram widget

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    What is your "sensible answer?" – 2017-02-21
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    Do you know what the Fourier transform of a step function is? If so, then use time shifting properties. – 2017-02-21
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    My answer was what you'd get from integrating from 2 to inf of $6e^{-i\omega t}$, [like so](https://www.wolframalpha.com/input/?i=from+2+to+inf+integral+of+6e%5E(-iwt)+dt) – 2017-02-21
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    @Grak Your analysis is not correct unless $\omega$ has a negative imaginary part. – 2017-02-21
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    I noticed what I was thinking wrong with the function, but how could I work out the heaviside function turn into something more sensible. I'm thinking that I will either want to move the function on time-level so that it jumps at y-axis (why?) or that it will be at 0 on x-axis when heaviside equals 0 (how?). – 2017-02-21

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Unfortunately you can't give sense to the Fourier transform of $f(t) = 5 + H(t-2)$ unless you interpret it in the sense of (Schwartz) distributions.

It might be useful for an engineer to write, for example, $$ \int_{-\infty}^{\infty} 5\cdot e^{-i t \omega } \,dt = 10 \pi \delta \tag{1} $$ but since the integral on the left-hand side of $(1)$ does not converge, this has no mathematical sense. However if we see the constant function $t\mapsto5$ as an element of the space $\mathcal{S}'(\Bbb{R})$ of tempered distributions, then it is true that $$ \mathcal{F}\{5\}(\omega) = 10\pi\delta \tag{2} $$ Here $(2)$ has a precise meaning, but one must know some distribution theory in order to understand it.

Now, one learns that

$ \phantom{-}\phantom{-}\phantom{-}\phantom{-}\phantom{-}\phantom{-}\phantom{-}\phantom{-}\phantom{-}\phantom{-}\phantom{-}\phantom{-}\phantom{-}\phantom{-}\phantom{-}\phantom{-}\phantom{-}\displaystyle\mathcal{F}\{H\}(\omega) = \pi \delta -i$ $\displaystyle\operatorname{p.v.}\left(\frac{1}{\omega}\right)$

We can now find the Fourier transform of $f$ by making use (as hinted in the comments) of the following time shifting property.

Proposition: Let $u\in \mathcal{S}'(\mathbb{R})$ and $a\in\Bbb{R}$. Then $$ \mathcal{F}\{\tau_a u\}(\omega) = e^{-i a \omega}\mathcal{F}\{u\}(\omega) $$ Here $\tau_au := u \circ f$ where $f(t) := t-a$.

We obtain \begin{align} \mathcal{F}\{5+H(t-2)\} &= \mathcal{F}\{5\} + \mathcal{F}\{H(t-2)\}\\ &=10\pi\delta+\pi e^{-2i\omega}\delta-i e^{-2i\omega} \operatorname{p.v.}\left(\frac{1}{\omega}\right) \\ &=10\pi\delta+\pi\delta-i e^{-2i\omega} \operatorname{p.v.}\left(\frac{1}{\omega}\right) \end{align}

Note. Be careful interpreting results from Wolfram Alpha (which are wrong sometimes!).

The Fourier transform documentation says (just after equation $(16)$) that Wolfram Alpha adopted the convention $$ \mathcal{F}\{f\} := \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} f(t) e^{i\omega t} \,dt $$

The answer obtained here is for $$ \mathcal{F}\{f\} := \int_{-\infty}^{\infty} f(t) e^{-i\omega t} \,dt $$

The "scrap" that Wolfram Alpha "spits out" is correct if you interpret the $\displaystyle\frac{1}{\omega}$ in $\displaystyle\frac{i e^{2 i \omega}}{\sqrt{2\pi}\omega}$ as the distribution $\displaystyle\operatorname{p.v.}\left(\frac{1}{\omega}\right)$.