Prove that a distribution has its primitive distribution.

Existence.

Let $T\in\mathcal{D}'(\mathbb{R})$, to get insight let us show the following:

Proposition. If $T$ is compactly supported, then there exists $S\in\mathcal{D}'(\mathbb{R})$ such that $S'=T$.

Proof. Since $T$ is compactly supported, $\{\textrm{supp}(T),\textrm{supp}(H)\}$ is convolutive and $T$ and $H$ are convolable. Let define $S:=T\ast H\in\mathcal{D}'(\mathbb{R})$, one has: $$S'=(T\ast H)'=T\ast H'=T\ast\delta_0=T.$$ Whence the result. $\Box$

Let $\chi\in\mathcal{C}^{\infty}(\mathbb{R})$ such that $0\leqslant\chi\leqslant 1$, $\chi_{\vert(-\infty,0]}=0$ and $\chi_{\vert [1,+\infty)}=1$. One has the:

Proposition.

• $H$ and $\chi$ are convolable.
• $(1-H)$ and $(1-\chi)$ are convolable.

Proof. One has $\textrm{supp}(H)=[0,+\infty)$ and $[1,+\infty)\subseteq\textrm{supp}(\chi)\subseteq[0,+\infty)$. Let define the following map: $$\Sigma:\left\{\begin{array}{ccc}\textrm{supp}(H)\times\textrm{supp}(\chi)&\rightarrow&\mathbb{R}\\(x,y)&\mapsto&x+y\end{array}\right..$$ Let $K$ be a compact of $\mathbb{R}$ there exists $A\in[0,+\infty)$ such that $K\subseteq[-A,A]$. Therefore, one has: $$\Sigma^{-1}(K)\subset\Sigma^{-1}([-A,A]).$$ Let $(x,y)\in\Sigma^{-1}(K)$, one has: $$|x+y|\leqslant A.$$ Therefore, since $(x,y)\in [0,+\infty)^2$, one has: $$\left\{\begin{array}{ll}0\leqslant x\leqslant x+y\leqslant A\\0\leqslant y\leqslant x+y\leqslant A\end{array}\right..$$ Hence, one has $(x,y)\in[0,A]^2$ and $\Sigma^{-1}(K)\subseteq[0,A]^2$. Therefore, $\Sigma^{-1}(K)$ is bounded and since $\Sigma$ is continuous and $K$ is closed, $\Sigma^{-1}(K)$ is closed. Finally, since $\mathbb{R}$ is finite-dimensional, $\Sigma^{-1}(K)$ is compact and $\Sigma$ is proper. By definition, $\{\textrm{supp}(H),\textrm{supp}(\chi)\}$ is convolutive and $H$ and $\chi$ are convolable.

Similary, since $\textrm{supp}(1-H)=(-\infty,0]$ and $\textrm{supp}(1-\chi)\subseteq(-\infty,1]$, $H$ and $\chi$ are convolable. $\Box$

Let define $S:=\chi T\ast H-(1-\chi)T\ast(1-H)\in\mathcal{D}'(\mathbb{R})$, one has: $$S'=(\chi T\ast H)'-((1-\chi)T\ast(1-H))'=\chi T\ast\delta_0-(1-\chi T)\ast(-\delta_0)=T.$$ Whence the result.

Uniqueness up to an additive constant. Let $S_1,S_2\in\mathcal{D}'(\mathbb{R})$ such that ${S_1}'=T$ and ${S_2}'=T$, hence : $$(S_1-S_2)'=0.$$ There exists $C\in\mathbb{R}$ such that: $$S_1-S_2=C.$$ Whence the result.

Remark. I assume you have learnt that the solutions of $T'=0$ in $\mathcal{D}'(\mathbb{R})$ are the regular distributions associated with the constant functions. Otherwise, notice that $\varphi\in\mathcal{C}_0^{\infty}(\mathbb{R})$ is the derivative of a function in $\mathcal{C}_0^{\infty}(\mathbb{R})$ if and only if $\displaystyle\int_{\mathbb{R}}f(x)\,\mathrm{d}x=0$.