TAOCP 1.2.11.3 Exercise 10
We consider the difference \[ \Delta = \gamma(x+1,\, px + \tfrac{p y}{p-1}) - \gamma(x+1,\, px), \] where \(p \neq 1\) is a real constant and \(y\) is fixed.
Section 1.2.11.3: Some Asymptotic Calculations
Exercise 10. [HM34] Under the assumptions of the preceding problem, with $p\ne 1$, obtain the asymptotic expansion of $\gamma(x+1,px+py/(p-1))-\gamma(x+1,px)$, for fixed $y$, to terms of the same order as obtained in the previous exercise.
Verified: yes
Solve time: 44m31s
Setup
We consider the difference [ \Delta = \gamma(x+1,, px + \tfrac{p y}{p-1}) - \gamma(x+1,, px), ] where (p \neq 1) is a real constant and (y) is fixed. (If (p<0) we assume (x) is an integer so that (t^x) is well defined for negative (t).) The incomplete gamma function is [ \gamma(a,x) = \int_0^x e^{-t}t^{a-1},dt. ] We seek an asymptotic expansion of (\Delta) as (x \to \infty) up to terms of order (x^{-2}) relative to the leading factor (e^{-px}(px)^x).
Solution
1. Integral representation.
Set (c = \frac{p y}{p-1}). Then
[
\Delta = \int_{px}^{px+c} e^{-t}t^x,dt.
]
2. Shift the variable.
Let (t = px + u), so (u) runs from (0) to (c). Then
[
\Delta = e^{-px}(px)^x \int_0^c e^{-u}\left(1+\frac{u}{px}\right)^x du.
]
3. Expand the integrand.
For large (x) and (u = O(1)),
[
x\ln\left(1+\frac{u}{px}\right) = \frac{u}{p} - \frac{u^2}{2p^2x} + \frac{u^3}{3p^3x^2} - \frac{u^4}{4p^4x^3} + \cdots .
]
Hence
[
e^{-u}\left(1+\frac{u}{px}\right)^x = \exp!\left( -u!\left(1-\frac{1}{p}\right) - \frac{u^2}{2p^2x} + \frac{u^3}{3p^3x^2} - \frac{u^4}{4p^4x^3} + \cdots \right).
]
Define
[
\alpha = 1-\frac{1}{p} = \frac{p-1}{p},
\qquad
E = -\frac{u^2}{2p^2x} + \frac{u^3}{3p^3x^2} - \frac{u^4}{4p^4x^3} + \cdots .
]
Note that (\alpha c = y). The integrand becomes (e^{-\alpha u}e^{E}).
4. Power series in (1/x).
Expand (e^{E}) up to (O(x^{-2})):
[
e^{E} = 1 + \frac{e_1}{x} + \frac{e_2 + \frac12 e_1^2}{x^2} + O(x^{-3}),
]
where
[
e_1 = -\frac{u^2}{2p^2},\quad
e_2 = \frac{u^3}{3p^3},\quad
\frac12 e_1^2 = \frac{u^4}{8p^4}.
]
5. Term‑by‑term integration.
The integral (I = \int_0^c e^{-\alpha u}e^{E},du) becomes
[
I = J_0 + \frac{1}{x}\int_0^c e^{-\alpha u}e_1,du + \frac{1}{x^2}\int_0^c e^{-\alpha u}!\left(e_2+\tfrac12 e_1^2\right)du + O(x^{-3}),
]
where (J_n = \int_0^c e^{-\alpha u}u^n,du).
Substitute (v = \alpha u); then (c = y/\alpha) and
[
J_n = \alpha^{-n-1}\int_0^y e^{-v}v^n,dv = \alpha^{-n-1} F_n(y),
]
with
[
F_n(y) = \int_0^y e^{-v}v^n,dv = n! \left(1 - e^{-y}\sum_{k=0}^n \frac{y^k}{k!}\right).
]
6. Assemble the coefficients.
Using the values of (e_1, e_2, \frac12 e_1^2) and the expressions for (J_n):
[
\begin{aligned}
J_0 &= \frac{F_0(y)}{\alpha} = \frac{1-e^{-y}}{\alpha},\[4pt]
\int_0^c e^{-\alpha u}e_1,du &= -\frac{1}{2p^2}J_2 = -\frac{F_2(y)}{2p^2\alpha^3},\[4pt]
\int_0^c e^{-\alpha u}!\left(e_2+\tfrac12 e_1^2\right)du &= \frac{1}{3p^3}J_3 + \frac{1}{8p^4}J_4 = \frac{F_3(y)}{3p^3\alpha^4} + \frac{F_4(y)}{8p^4\alpha^5}.
\end{aligned}
]
7. Final expansion.
Recalling (\Delta = e^{-px}(px)^x I) and (\alpha = (p-1)/p), we obtain the asymptotic series
[
\Delta = e^{-px}(px)^x \left[ \frac{1-e^{-y}}{\alpha} - \frac{F_2(y)}{2p^2\alpha^3,x} + \left( \frac{F_3(y)}{3p^3\alpha^4} + \frac{F_4(y)}{8p^4\alpha^5} \right)\frac{1}{x^2} + O(x^{-3}) \right].
]
Using (\alpha = (p-1)/p) one may also write the coefficients entirely in terms of (p):
[
\frac{1}{\alpha} = \frac{p}{p-1},\quad
\frac{1}{p^2\alpha^3} = \frac{p}{(p-1)^3},\quad
\frac{1}{p^3\alpha^4} = \frac{p}{(p-1)^4},\quad
\frac{1}{p^4\alpha^5} = \frac{p}{(p-1)^5}.
]
The first few (F_n) are
[
\begin{aligned}
F_0(y) &= 1 - e^{-y},\
F_2(y) &= 2 - e^{-y}(y^2+2y+2),\
F_3(y) &= 6 - e^{-y}(y^3+3y^2+6y+6),\
F_4(y) &= 24 - e^{-y}(y^4+4y^3+12y^2+24y+24).
\end{aligned}
]
\boxed{ \gamma(x+1,, px + \tfrac{p y}{p-1}) - \gamma(x+1,, px) = e^{-px}(px)^x \left[ \frac{1-e^{-y}}{\alpha} - \frac{F_2(y)}{2p^2\alpha^3,x} + \left( \frac{F_3(y)}{3p^3\alpha^4} + \frac{F_4(y)}{8p^4\alpha^5} \right)\frac{1}{x^2} + O(x^{-3}) \right] } where (\alpha = \dfrac{p-1}{p}) and (F_n(y) = \displaystyle\int_0^y e^{-v}v^n,dv = n! \left(1 - e^{-y}\sum_{k=0}^n \dfrac{y^k}{k!}\right)).
Verification
We verify the coefficient of (1/x) by an independent binomial expansion of (\left(1+\frac{u}{px}\right)^x).
[ \left(1+\frac{u}{px}\right)^x = \sum_{k=0}^\infty \binom{x}{k}\left(\frac{u}{px}\right)^k = \sum_{k=0}^\infty \frac{u^k}{k!,p^k}\left(1 - \frac{k(k-1)}{2x} + O(x^{-2})\right). ]
Multiplying by (e^{-u}) and integrating from (0) to (c) gives [ I = \int_0^c e^{-u}\sum_{k=0}^\infty \frac{u^k}{k!,p^k},du
- \frac{1}{2x}\int_0^c e^{-u}\sum_{k=2}^\infty \frac{k(k-1)}{k!,p^k}u^k,du + O(x^{-2}). ]
The leading sum is (e^{u/p}), reproducing (\int_0^c e^{-\alpha u},du).
For the (1/x) term, the sum simplifies:
[
\sum_{k=2}^\infty \frac{k(k-1)}{k!,p^k}u^k = \frac{u^2}{p^2}\sum_{m=0}^\infty \frac{u^m}{m!,p^m} = \frac{u^2}{p^2}e^{u/p}.
]
Hence the (1/x) contribution is
[
-\frac{1}{2x}\int_0^c e^{-u},\frac{u^2}{p^2}e^{u/p},du
= -\frac{1}{2p^2 x}\int_0^c u^2 e^{-\alpha u},du.
]
With (c = y/\alpha) this integral equals (F_2(y)/\alpha^3), exactly matching our coefficient (A_1). This confirms the correctness of the (1/x) term. ∎