 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2018/paper-1/12/a/ii/solution

For fixed

x > 0

, integrate first over

y

I = \int_{0}^{\infty} [\int_{x /2}^{x} x e^{- x y} d y] d x = \int_{0}^{\infty} (e^{- x^{2} /2} - e^{- x^{2}}) d x .

The integrand is nonnegative, so this order of the double integral is legitimate even before its finiteness is known. Applying the scaled Gaussian integral

\int_{0}^{\infty} e^{- a x^{2}} d x = π / (2 a)

yields

I = \frac{π}{2} (2 - 1) .

 New to topics? Read the docs here!