# Fourier transform using convolution product

The question asks to find the Fourier transform of the function:

$$I(x) = \int^{1/2}_{0} e^{-(x-t)^2} dt$$ using the theorem about convolution products. I know that the theorem states that $$\mathcal{F} \{f *g\} = \mathcal{F} \{f\} \mathcal{F}\{ g \}$$.

My textbook does not do a great job explaining Fourier transforms so if anyone could give a detailed explanation of how to solve a problem such as this one it would be greatly appreciated.

• You have something wrong. Because I is not function of x. Maybe it is t? – Pasha Jan 8 '20 at 15:31
• It must be $dt$ instead of $dx$. – S. Maths Jan 8 '20 at 16:15
• Yes, sorry that was a typo. – MathsBBB Jan 8 '20 at 17:56

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:
$$\mathcal{F}\{I\}(t)=\frac{1}{\sqrt{2}} e^{-t^2/4} \times \frac{i (1-e^{i t/2})}{\sqrt{2 \pi} t}.$$
• I think it would be useful to mention about the special case here by square term in the exponent. It decreases in both positive and negative direction and Fourier transform converges. Usually we have step function multiplied by exponential function and therefore resultant Transform integral is unilateral, i.e. from zero to infinity. For normal exponential $e^{at}$, without multiplying it with unit step function, the bilateral Fourier transform would diverge. – Pasha Jan 8 '20 at 16:33