 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2022/paper-1/17/c/ii/solution

For

n \geq 2

, write

x^{n} e^{- x^{2}} = x^{n - 1} (x e^{- x^{2}})

- e^{- x^{2}} /2

for the second factor:

I (n) = [- \frac{1}{2} x^{n - 1} e^{- x^{2}}]_{0}^{\infty} + \frac{n - 1}{2} \int_{0}^{\infty} x^{n - 2} e^{- x^{2}} d x .

The boundary term is zero: at zero,

n - 1 > 0

I (n) = \frac{n - 1}{2} I (n - 2), n \geq 2.

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