 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2018/paper-1/17/c/solution

Differentiate

x / (1 + x^{2})^{n}

and integrate between zero and infinity. Its boundary values vanish for every integer

n \geq 1

0 = \int_{0}^{\infty} [\frac{1}{( 1 + x ^{2} ) ^{n}} - \frac{2 n x ^{2}}{( 1 + x ^{2} ) ^{n + 1}}] d x .

Since

x^{2} / (1 + x^{2})^{n + 1} = (1 + x^{2})^{- n} - (1 + x^{2})^{- n - 1}

, this becomes

0 = E_{n} - 2 n (E_{n} - E_{n + 1})

E_{n + 1} = \frac{2 n - 1}{2 n} E_{n} .

The starting value is

E_{1} = [arctan x]_{0}^{\infty} = π /2

. Therefore

E_{2} = π /4

E_{3} = 3 π /16

, and

\int_{0}^{\infty} \frac{d x}{( 1 + x ^{2} ) ^{4}} = E_{4} = \frac{5}{6} \frac{3}{4} \frac{1}{2} \frac{π}{2} = \frac{5 π}{32} .

x = tan t

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