32D (Asymptotic Methods)

Integration by parts gives the recurrence

γ (x, y) = y^{x - 1} e^{- y} + (x - 1) γ (x - 1, y) .

Iterating

N

times yields

γ (x, y) = y^{x - 1} e^{- y} \sum_{n = 0}^{N - 1} (x - 1) (x - 2) \dots (x - n) y^{- n} + R_{N},

where the empty product for

n = 0

is one and

R_{N} = (x - 1) (x - 2) \dots (x - N) γ (x - N, y) .

For fixed

x

, the remainder has the order of the first omitted term as

y \to \infty

. Therefore

a_{0} (x) = 1, a_{n} (x) = j = 1 \prod n (x - j) (n \geq 1) .

b

For fixed

y

γ (x, y) = Γ (x) - \int_{0}^{y} t^{x - 1} e^{- t} d t .

The omitted integral is at most

y^{x} / x

, which is negligible compared with

Γ (x)

x \to \infty

. Stirling's formula therefore gives

γ (x, y) \sim Γ (x) \sim 2 π x^{x - 1/2} e^{- x} .

Put

y = λ x

and

t = y + s

. Factoring out the endpoint value gives

γ (x, y) = y^{x - 1} e^{- y} \int_{0}^{\infty} exp ((x - 1) lo g (1 + \frac{s}{λ x}) - s) d s .

For fixed

s

(x - 1) lo g (1 + \frac{s}{λ x}) - s = - \frac{λ - 1}{λ} s - \frac{1}{x} (\frac{s}{λ} + \frac{s ^{2}}{2 λ ^{2}}) + O (x^{- 2}) .

Watson's endpoint lemma justifies termwise integration, so with

α = (λ - 1) / λ

\frac{γ ( x , y )}{y ^{x - 1} e ^{- y}} \sim \int_{0}^{\infty} e^{- α s} [1 - \frac{1}{x} (\frac{s}{λ} + \frac{s ^{2}}{2 λ ^{2}})] d s = \frac{1}{α} - \frac{1}{x} (\frac{1}{λ α ^{2}} + \frac{1}{λ ^{2} α ^{3}}) .

Consequently

f (λ) = \frac{λ}{λ - 1}, g (λ) = - \frac{λ ^{2}}{( λ - 1 ) ^{3}} .

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