Discussion
Fourier cosine transform of \(f(t) = e^{-t^{2}}\) is :
Answer: D
The Fourier cosine transform is \(\dfrac{1}{\sqrt{2}}e^{-s^{2}/4}\).
With the convention \(F_{c}(s) = \sqrt{2/\pi}\int_{0}^{\infty} f(t)\cos st\,dt\), and the standard integral
\[ \begin{aligned} \int_{0}^{\infty} e^{-t^{2}}\cos st\,dt &= \frac{\sqrt{\pi}}{2}e^{-s^{2}/4} \\ F_{c}(s) &= \sqrt{\frac{2}{\pi}}\cdot\frac{\sqrt{\pi}}{2}e^{-s^{2}/4} = \frac{1}{\sqrt{2}}e^{-s^{2}/4} \end{aligned} \]
The coefficient \(1/\sqrt{2}\) in every option shows this convention. The exponent must be \(-s^{2}/4\); the other exponents come from mis-scaling the Gaussian (\(-s^{2}/2\) is the transform of \(e^{-t^{2}/2}\)).
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