 Solution

ID: past-exam-of-the-perturbation-methods-and-applied-complex-variables-course-of-the-university-of-oxford/2010/a/2/ii/solution

Let

g_{x} (t) = (1 + x t)^{- 1}

. Repeated integration by parts using the factor

e^{- t}

gives

I (x) = \sum_{n = 0}^{N} g_{x}^{(n)} (0) + \int_{0}^{\infty} e^{- t} g_{x}^{(N + 1)} (t) d t .

g_{x}^{(n)} (t) = (- 1)^{n} n! x^{n} (1 + x t)^{- n - 1}

. Therefore

I (x) = \sum_{n = 0}^{N} (- 1)^{n} n! x^{n} + R_{N} (x), ∣ R_{N} (x) ∣ \leq (N + 1)! x^{N + 1} (x > 0) .

For every fixed

N

, this is

o (x^{N})

x ↓ 0

, proving

I (x) \sim n \geq 0 \sum (- 1)^{n} n! x^{n} .

For each fixed

x > 0

the displayed infinite series diverges, since successive term magnitudes have ratio

(n + 1) x \to \infty

. Its usefulness is precisely that of an asymptotic expansion, not a convergent power series.

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