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

u = x^{2}

d u = 2 x d x

. The convergent improper integral is

\int_{0}^{\infty} x e^{- x^{2}} d x = \frac{1}{2} \int_{0}^{\infty} e^{- u} d u = \frac{1}{2} .

ii

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.

Start with the supplied Gaussian integral

I (0) = π /2

and part (i)'s

I (1) = 1/2

. Repeated use of the Gaussian moment recurrence gives

I (2) = \frac{π}{4}, I (3) = \frac{1}{2}, I (4) = \frac{3 π}{8}, I (5) = 1, I (6) = \frac{15 π}{16} .

The even indices inherit the

π

factor from

I (0)

, while the odd indices inherit rational values from

I (1)

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