Find the Fourier transform of the function:
$$ I(x) = \int^{1/2}_{0} e^{-(x-t)^2} dt$$ using the theorem about convolution products.
The theorem states that $\mathcal{F} \{f *g\} = \mathcal{F} \{f\} \mathcal{F}\{ g \}$.
I am given the following hint:
Write $I(x)=f*g(x)$, where $f(x)=e^{-x^2}$ and $g(x)=\chi_{[0, 1/2]}$.
Then using the convolution rule you get:
$$\mathcal{F}\{I\}(t)=\frac{1}{\sqrt{2}} e^{-t^2/4} \times \frac{i (1-e^{i t/2})}{\sqrt{2 \pi} t}.$$
But I am stuck here. The same question, but it does not provide a detailed solution: Fourier transform using convolution product
A detailed answer would help me a lot, as I do not have neither notes nor examples explaining this.
Thanks.