 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ia/paper-3/10b/a/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

Let

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

Then

I^{2} = \int_{0}^{\infty} \int_{0}^{\infty} e^{- a (x^{2} + y^{2})} d x d y .

Using polar coordinates on the first quadrant,

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

Since

I > 0

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

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