I'm trying to solve the following problems in the sense of distributions Q8c of example sheet 1 of the Distribution Theory course of part III of the mathematical tripos.
$(e^{2 \pi ix}-1)u'=0$.
In order to solve this problem, I tried working backwards. Take a $\psi \in D(\mathbb{R})$. I want to construct a $\phi \in D(\mathbb{R})$ such that $$ \langle u,\psi \rangle = \langle u,-\frac{\phi}{e^{2 \pi ix}-1} \rangle = \langle (e^{2 \pi ix}-1)u',\phi \rangle =0. $$
My issue is that such a $\phi$ satisfies $\phi(x)=\frac{{ \int_{0}^{x} \psi(y)dy}}{1-e^{2 \pi ix}}$ for $x \neq 0$. This would cause $\phi$ to be undefined on $\mathbb{Z}$.
I'm slightly stuck here. Any help would be greatly appreciated. Thank you.