Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 1 / 7A (Further Complex Methods)

Table of contents
- a 7A
  - Solution a
- b 7A
  - Solution b

a

Solution

With

u = x^{4}

, the change of variables formula gives

\int_{0}^{1} \frac{d x}{1 - x ^{4}} = \frac{1}{4} \int_{0}^{1} u^{- 3/4} (1 - u)^{- 1/2} d u = \frac{1}{4} B (\frac{1}{4}, \frac{1}{2}) .

The beta--gamma identity,

Γ (1/2) = π

, and the Gamma reflection formula give

\frac{1}{4} B (\frac{1}{4}, \frac{1}{2}) = \frac{Γ ( 1/4 ) π}{4Γ ( 3/4 )}, Γ (1/4) Γ (3/4) = π 2 .

Therefore

\int_{0}^{1} \frac{d x}{1 - x ^{4}} = \frac{Γ ( 1/4 ) ^{2}}{32 π} .

b

Solution

Use the suggested substitution

x = \frac{t}{2 - t ^{2}} .

[0, 1]

to itself and satisfies

d x = \frac{2 d t}{( 2 - t ^{2} ) ^{3/2}}, 1 - x^{4} = \frac{4 ( 1 - t ^{2} )}{( 2 - t ^{2} ) ^{2}} .

Hence

\int_{0}^{1} \frac{d x}{1 - x ^{4}} = \frac{1}{2} \int_{0}^{1} \frac{d t}{( 1 - t ^{2} ) ( 1 - t ^{2} /2 )} = \frac{1}{2} K (1/ 2),

where

K

is the complete elliptic integral of the first kind. Part (a) and the Gamma function recurrence

Γ (5/4) = Γ (1/4) /4

now give

K (1/ 2) = \frac{Γ ( 1/4 ) ^{2}}{4 π} = \frac{4Γ ( 5/4 ) ^{2}}{π} .

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