 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2009/paper-1/16/d/solution

The integrand in the preceding expression is nonnegative. Tonelli's theorem permits squaring as a quadrant integral; then polar coordinates give

Γ (1/2)^{2} = 4 \int_{0}^{\infty} \int_{0}^{\infty} e^{- (u^{2} + v^{2})} d u d v = 4 \int_{0}^{π /2} \int_{0}^{\infty} e^{- r^{2}} r d r d θ = π .

The defining integral is positive, fixing the sign of the square root. Therefore

Γ (1/2)^{2} = π, Γ (1/2) = π .

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