 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2013/paper-2/20/c/solution

Use differentiation under the integral sign, justified by the integrable bound

∣ x sin (t x) e^{- x^{2} /2} ∣ \leq x e^{- x^{2} /2}

. This gives

F^{'} (t) = - \int_{0}^{\infty} x sin (t x) e^{- x^{2} /2} d x = \int_{0}^{\infty} sin (t x) d (e^{- x^{2} /2}) .

Integration by parts makes the boundary term vanish: sine vanishes at zero and the Gaussian tends to zero at infinity. Therefore

F^{'} (t) = - t \int_{0}^{\infty} cos (t x) e^{- x^{2} /2} d x = - tF (t) .

e^{t^{2} /2}

makes

(e^{t^{2} /2} F)^{'} = 0

, so

F (t) = C e^{- t^{2} /2}

. At

t = 0

, the given half-line Gaussian integral fixes

C = π /2

. Thus

F (t) = \frac{π}{2} e^{- t^{2} /2} .

This is the even cosine-transform form of the Fourier transform of a Gaussian.

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