 Solution

ID: past-exam-of-the-options-ii-course-of-the-university-of-oxford/2006/d/4/b/solution

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

Integration by parts has zero boundary term: the sine vanishes at zero and the Gaussian tends to zero at infinity. Hence

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

The same Gaussian majorant and dominated convergence give continuity at zero, where the allowed Gaussian integral gives

F (0) = π /2

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

; continuity at zero fixes that constant. Thus

F (t) = \frac{π}{2} e^{- t^{2}} (0 \leq t < b) .

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