TAOCP 1.2.11.3 Exercise 7
Let J(x)=\int_0^{y x^{1/4}} e^{-u}\left(1+\frac ux\right)^x\,du, where $y$ is fixed and $x\to\infty$.
Section 1.2.11.3: Some Asymptotic Calculations
Exercise 7. [HM30] In the evaluation of $I_2$, we had to consider
$$ \int_0^y e^{-u}\left(1+\frac{u}{x}\right)^x du. $$
Give an asymptotic representation of
$$ \int_0^{yx^{1/4}} e^{-u}\left(1+\frac{u}{x}\right)^x du $$
to terms of order $O(x^{-2})$, when $y$ is fixed and $x$ is large.
Verified: no
Solve time: 6m22s
Let
$$ J(x)=\int_0^{y x^{1/4}} e^{-u}\left(1+\frac ux\right)^x,du, $$
where $y$ is fixed and $x\to\infty$.
We seek an asymptotic expansion through terms of order $O(x^{-2})$.
1. Expansion of the integrand
Write
$$ e^{-u}\left(1+\frac ux\right)^x = \exp!\left( -u+x\log!\left(1+\frac ux\right) \right). $$
Using
$$ \log(1+z) = z-\frac{z^2}{2}+\frac{z^3}{3}-\frac{z^4}{4} +\frac{z^5}{5}-\frac{z^6}{6} +O(z^7), $$
with $z=u/x$, we obtain
$$ -u+x\log!\left(1+\frac ux\right) = -\frac{u^2}{2x} +\frac{u^3}{3x^2} -\frac{u^4}{4x^3} +\frac{u^5}{5x^4} -\frac{u^6}{6x^5} +O!\left(\frac{u^7}{x^6}\right). $$
Set
$$ E= -\frac{u^2}{2x} +\frac{u^3}{3x^2} -\frac{u^4}{4x^3} +\frac{u^5}{5x^4} -\frac{u^6}{6x^5} +O!\left(\frac{u^7}{x^6}\right). $$
Since $u\le yx^{1/4}$,
$$ E=O(x^{-1/2}) $$
uniformly on the interval of integration.
Let
$$ a_1=-\frac{u^2}{2x},\qquad a_2=\frac{u^3}{3x^2},\qquad a_3=-\frac{u^4}{4x^3},\qquad a_4=\frac{u^5}{5x^4}. $$
The term $a_5=-u^6/(6x^5)$ contributes after integration only
$$ O!\left(x^{(6+1)/4-5}\right) = O(x^{-13/4}), $$
so it is below the required accuracy and may be absorbed into the remainder.
We therefore expand
$$ e^E = 1+E+\frac{E^2}{2} +\frac{E^3}{6} +\frac{E^4}{24} +\frac{E^5}{120} +R_6 . $$
Since $E=O(x^{-1/2})$,
$$ R_6=O(E^6)=O(x^{-3}) $$
uniformly. Hence
$$ \int_0^{yx^{1/4}}R_6,du = O(x^{-3}),O(x^{1/4}) = O(x^{-11/4}), $$
which is $o(x^{-2})$.
Now compute all monomials that can contribute at least $O(x^{-2})$ after integration.
From $E$:
$$ a_1+a_2+a_3+a_4. $$
From $E^2/2$:
$$ \frac12 a_1^2+a_1a_2+a_1a_3+\frac12 a_2^2. $$
These equal
$$ \frac{u^4}{8x^2} -\frac{u^5}{6x^3} +\frac{u^6}{8x^4} +\frac{u^6}{18x^4}. $$
From $E^3/6$:
$$ \frac16 a_1^3+\frac12 a_1^2a_2, $$
giving
$$ -\frac{u^6}{48x^3} +\frac{u^7}{24x^4}. $$
From $E^4/24$:
$$ \frac1{24}a_1^4 = \frac{u^8}{384x^4}. $$
Terms from $E^5$ and all omitted products contribute after integration only $O(x^{-9/4})$ or smaller.
Therefore
$$ \begin{aligned} e^{-u}\left(1+\frac ux\right)^x ={}& 1 -\frac{u^2}{2x} +\frac{u^3}{3x^2} +\frac{u^4}{8x^2} -\frac{u^4}{4x^3} -\frac{u^5}{6x^3} -\frac{u^6}{48x^3} \ &+ \frac{u^5}{5x^4} +\left(\frac18+\frac1{18}\right)\frac{u^6}{x^4} +\frac{u^7}{24x^4} +\frac{u^8}{384x^4} +R(u,x), \end{aligned} $$
with
$$ \int_0^{yx^{1/4}}R(u,x),du = O(x^{-9/4}). $$
The $u^5/x^4$ and $u^6/x^4$ terms integrate to $O(x^{-5/2})$ and $O(x^{-9/4})$ respectively, so they are already below the required precision and may be dropped.
Thus
$$ \begin{aligned} e^{-u}\left(1+\frac ux\right)^x ={}& 1 -\frac{u^2}{2x} +\frac{u^3}{3x^2} +\frac{u^4}{8x^2} -\frac{u^4}{4x^3} -\frac{u^5}{6x^3} -\frac{u^6}{48x^3} \ &+ \frac{u^7}{24x^4} +\frac{u^8}{384x^4} +\widetilde R(u,x), \end{aligned} $$
where
$$ \int_0^{yx^{1/4}}\widetilde R(u,x),du = O(x^{-9/4}). $$
2. Term-by-term integration
Let
$$ L=yx^{1/4}. $$
Since
$$ \int_0^L u^n,du=\frac{L^{n+1}}{n+1}, $$
we obtain:
$$ \int_0^L 1,du = yx^{1/4}, $$
$$ -\frac1{2x}\frac{L^3}{3} = -\frac{y^3}{6}x^{-1/4}, $$
$$ \frac1{3x^2}\frac{L^4}{4} = \frac{y^4}{12}x^{-1}, $$
$$ \frac1{8x^2}\frac{L^5}{5} = \frac{y^5}{40}x^{-3/4}, $$
$$ -\frac1{4x^3}\frac{L^5}{5} = -\frac{y^5}{20}x^{-7/4}, $$
$$ -\frac1{6x^3}\frac{L^6}{6} = -\frac{y^6}{36}x^{-3/2}, $$
$$ -\frac1{48x^3}\frac{L^7}{7} = -\frac{y^7}{336}x^{-5/4}, $$
$$ \frac1{24x^4}\frac{L^8}{8} = \frac{y^8}{192}x^{-2}, $$
$$ \frac1{384x^4}\frac{L^9}{9} = \frac{y^9}{3456}x^{-7/4}. $$
All remaining contributions are $O(x^{-9/4})$.
3. Final asymptotic expansion
Collecting terms in descending order,
$$ \boxed{ \begin{aligned} J(x) ={}& yx^{1/4} -\frac{y^3}{6}x^{-1/4} +\frac{y^5}{40}x^{-3/4} +\frac{y^4}{12}x^{-1} -\frac{y^7}{336}x^{-5/4} \[1ex] &-\frac{y^6}{36}x^{-3/2} +\left(-\frac{y^5}{20} +\frac{y^9}{3456}\right)x^{-7/4} +\frac{y^8}{192}x^{-2} +O(x^{-9/4}). \end{aligned} } $$
Since $O(x^{-9/4})=o(x^{-2})$, this is the required asymptotic representation through terms of order $O(x^{-2})$.