 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2014/paper-1/4/v/solution

Differentiating

\int e^{- a x^{2}} d x = π / a

with respect to

a > 0

gives

\int x^{2} e^{- a x^{2}} d x = π / (2 a^{3/2})

. Therefore

E = \int_{- \infty}^{\infty} x^{2} e^{- 2 x^{2}} d x = \frac{π}{4 2} .

From part (iii),

∣ f (k) ∣^{2} = (π /4) k^{2} e^{- k^{2} /2}

\frac{1}{2 π} \int_{- \infty}^{\infty} ∣ f (k) ∣^{2} d k = \frac{1}{8} \int_{- \infty}^{\infty} k^{2} e^{- k^{2} /2} d k = \frac{2 π}{8} = \frac{π}{4 2},

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