 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2020/paper-1/12/a/i/solution

Let

I

be the positive Gaussian integral. Its improper integral converges because

e^{- x^{2}} \leq e^{- ∣ x ∣}

for

∣ x ∣ \geq 1

. Multiplying two copies and using the polar coordinates area element gives

I^{2} = \int_{R^{2}} e^{- (x^{2} + y^{2})} d x d y = \int_{0}^{2 π} \int_{0}^{\infty} e^{- r^{2}} r d r d θ = 2 π [- \frac{1}{2} e^{- r^{2}}]_{0}^{\infty} = π .

Nonnegativity justifies the product double integral and the change of integration order. Since

I > 0

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

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