Writing integral in terms of distributions

Question

EDIT (now asking how to write $F$ as distributions, instead of writing the integral in terms of distributions):

Let $F$ be the distribution defined by its action on a test function $\phi$ as

\begin{equation*} F(\phi)=\int_{\pi}^{2\pi}x\phi(x)dx. \end{equation*}

How would you write $F$ in terms of the delta distribution, heaviside distribution, and a regular distribution $R$ defined by its action on a test function $\phi$ as

\begin{equation*} R(\phi)=\int_{-\infty}^{\infty}g(x)\phi(x)dx \end{equation*}

for a continuous function $g$?

Edit: Q1)b) in this link https://www.maths.ox.ac.uk/system/files/legacy/3422/B5a_13.pdf

Answer 1

Suspect one could not find continuous $g(x)$. I state that $g(x) = x\cdot H(x-\pi)\cdot H(2\pi-x)$. Namely $g(x) = x$ if $x\in[\pi, 2\pi]$ and $g(x)=0$ for $x\in R\setminus [\pi, 2\pi]$. So $$ \int_{\pi}^{2\pi}x\phi(x)dx = \int_{-\infty}^{+\infty}g(x)\phi(x)dx. $$

Answer 2

Although I think @NikitaEvseev's answer is the most natural, perhaps the small exercise of rewriting integration over an interval $[a,b]$ as a distribution is worthwhile. That is, consider $$ u(\varphi) \;=\; \int_a^b \varphi(x)\;dx $$ Edit: simpler than what I wrote before: $$ u(\varphi) \;=\; \int_a^b \varphi(x)\,dx \;=\; \int_{-\infty}^\infty (H(x-a) - H(x-b))\cdot \varphi(x)\;dx $$ That is, if translates and reflections of the Heaviside function are allowed, then it's just this (with $\varphi(x)$ replaced by $x\cdot \varphi(x)$).

But then I'm confused as to why the posing of the question in the linked-to source made it seem as though something more complicated should happen.