 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2002/paper-1/5/a/solution

Let

I = \int_{0}^{\infty} e^{- x^{2}} d x

. Positivity allows the squared integral to be written as a first-quadrant double integral. In polar coordinates the Jacobian determinant is

r

, so

I^{2} = \int_{0}^{π /2} \int_{0}^{\infty} e^{- r^{2}} r d r d θ = \frac{π}{4}, I = \frac{π}{2} .

x e^{- x^{2}}

vanishing at both endpoints,

\int_{0}^{\infty} x^{2} e^{- x^{2}} d x = \frac{1}{2} \int_{0}^{\infty} e^{- x^{2}} d x = \frac{π}{4} .

The substitution

u = x^{2}

also gives the two antiderivatives

\int x e^{- x^{2}} d x = - \frac{1}{2} e^{- x^{2}} + C, \int x^{3} e^{- x^{2}} d x = - \frac{1}{2} (x^{2} + 1) e^{- x^{2}} + C .

These values also follow from the Gaussian moment recurrence and can be checked by differentiation.

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