 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-oxford/2015/prelims/mathematics-iii/7/b/ii/solution

For

I_{1}

t = x^{2} / θ

gives

I_{1} = \int_{0}^{\infty} e^{- t} d t = 1.

For

k \geq 1

, use

d (e^{- x^{2} / θ}) = - (2 x / θ) e^{- x^{2} / θ} d x

I_{k + 2} = - \int_{0}^{\infty} x^{k + 1} d (e^{- x^{2} / θ}) = - [x^{k + 1} e^{- x^{2} / θ}]_{0}^{\infty} + (k + 1) \int_{0}^{\infty} x^{k} e^{- x^{2} / θ} d x = \frac{( k + 1 ) θ}{2} I_{k} .

The boundary term vanishes:

x^{k + 1} \to 0

at zero, and the exponential decay dominates the power at infinity. These estimates also establish convergence of the improper integrals used above.

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