If $x$ is a Schwartz function and $\theta$ is a tempered distribution given by $$\theta(x)=\int_{0}^{\pi}e^{it}x(t)dt.$$ What is the Fourier transform of $\theta$?
I know $\mathcal{F}\theta(x)=\theta\mathcal{F}(x)$ and $$\theta\mathcal{F}(x)=\frac{1}{(2\pi)^{\frac{d}{2}}}\int_{0}^{\pi}e^{it}dt\int_{\mathbb{R}^d}e^{-i\langle t,s\rangle}x(s)ds.$$
But I am unsure of what to do from here. Can anyone help?