 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2020/paper-1/7/d/solution

Let

n \geq 2

be even and integrate

f (z) = 1/ (1 + z^{n})

R > 1

with the real segment traversed from

- R

R

. There are no real poles, and the

n /2

z_{j} = e^{i (2 j + 1) π / n}, j = 0, \dots, n /2 - 1.

Each is simple. Since

z_{j}^{n} = - 1

Res_{z = z_{j}} f = \frac{1}{n z _{j}^{n - 1}} = - \frac{z _{j}}{n} .

On the semicircle the integral magnitude is at most

π R / (R^{n} - 1)

, which tends to zero. The real integral is absolutely convergent because its tail is of order

∣ x ∣^{- n}

. Thus

\int_{- \infty}^{\infty} \frac{d x}{1 + x ^{n}} = - \frac{2 πi}{n} \sum_{j = 0}^{n /2 - 1} e^{i (2 j + 1) π / n} .

θ = π / n

\sum_{j = 0}^{n /2 - 1} e^{i (2 j + 1) θ} = e^{i θ} \frac{1 - e ^{in θ}}{1 - e ^{2 i θ}} = \frac{2 e ^{i θ}}{- 2 i e ^{i θ} s i n θ} = \frac{i}{s i n θ} .

Consequently

\int_{- \infty}^{\infty} \frac{d x}{1 + x ^{n}} = \frac{2 π}{n sin ( π / n )}, A_{n} = \frac{2 π}{n}, B_{n} = \frac{π}{n} .

For

n = 2

this reduces to

π

, as a check on the contour orientation and residue signs.

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