For $C>0$, we have $\chi_{[0,C]}\in\mathcal{S}'(\mathbb{R})$. I want to determine the Fourier transform $\mathcal{F}\chi_{[0,C]}$; thus for all $\phi\in\mathcal{S}(\mathbb{R})$, we have
$$\begin{aligned}(\mathcal{F}\chi_{[0,C]})(\phi)&=\chi_{[0,C]}(\mathcal{F}\phi) \\ &=\int_0^C(\mathcal{F}\phi)(\xi)\,d\xi \\ &=\int_0^C\left(\int_{-\infty}^{\infty}\phi(x)e^{-ix\xi}\,dx\right)\,d\xi \\ &=\int_{-\infty}^{\infty}\left(\int_0^C e^{-ix\xi}\,d\xi\right)\phi(x)\,dx \\ &=\int_{-\infty}^\infty\left(\frac{i}{x}e^{-ix\xi}\bigg|_{\xi=0}^C\right)\phi(x)\,dx \\ &=\int_{-\infty}^{\infty}\left(\frac{i}{x}e^{-iCx}-\frac{i}{x}\right)\phi(x)\,dx \\ &=i\int_{-\infty}^\infty\frac{\phi(x)}{x}e^{-iCx}\,dx-i\int_{-\infty}^{\infty}\frac{\phi(x)}{x}\,dx \end{aligned}$$ If $\phi\in\mathcal{D}(\mathbb{R})$ then this would follow, I suppose, through integration by parts. But in this case I don't know how to continue.
Edit:
If I use that $\chi_{[0,C]}\in L^1(\mathbb{R})$, then I can compute $$\begin{aligned} \mathcal{F}\chi_{[0,C]}(\xi)&=\int_{0}^{C}e^{-i\xi x}\,dx \\ &=\frac{i}{\xi}e^{-i\xi x}\bigg|_{x=0}^{C} \\ &=\frac{i}{\xi}(e^{-iC\xi}-1). \end{aligned}$$
Your function is in $L^1(\mathbb R)$ you don't need to use distribution here (because in the case of a distribution $u \in \mathcal S^\prime(\mathbb R)$ that is in $L^1(\mathbb R)$, the distribution $\hat u$ is given by the classical Fourier transform).
So in your case a simple computation gives its Fourier transform (something involving cardinal sine function).