Asymptotic expansion using gamma functions and Stirling's formula

Question

Question: obtain an expression for the $n$th term of an asymptotic expansion, valid as $ \lambda \to \infty$ for the integral

$$I(\lambda) = \int_0^1 t^{2\alpha}e^{-\lambda(t^2+t^3)} dt $$

where $ \alpha >-1/2 $. Also, obtain the first two terms of an asymptotic expansion, valid as $ \lambda \to\infty $ for the integral

$$I(\lambda) = \int_0^1 t^{2\alpha}e^{-\lambda(t^2-t^3)} dt $$

where $ -1/2<\alpha <0$.

(hint: use the gamma function and Stirling's formula).

What I've done: I've only tried the first part and got stuck. I tried to relate it to gamma functions, by simply letting $(t^2 + t^3) = u$. This looks appropriate because $(t^2 + t^3)$ is monotonic, but to change variables I would need to find the inverse of this function, which is quite nasty.

So, how shall I proceed? Any help is greatly appreciated!

Answer 1

I cannot see how the gamma function or Stirling's formula could help you. By Laplace's method, $$ \int_0^1 {t^{2\alpha } {\rm e}^{ - \lambda (t^2 + t^3 )} {\rm d}t} \sim\lambda^{ - \alpha - 1/2} \sum\limits_{n = 0}^\infty {\Gamma \bigg( {\frac{{n + 1}}{2} + \alpha } \bigg)\frac{{b_n }}{{\lambda^{n/2} }}} $$ as $\lambda\to+\infty$, where $$ b_n = \frac{1}{2}\frac{1}{{2\pi {\rm i}}}\oint_{(0 + )} {(1 + t)^{ - \frac{{n + 1}}{2} - \alpha } \frac{{{\rm d}t}}{{t^{n + 1} }}} = \frac{{1}}{2}\binom{ - \frac{{n + 1}}{2} - \alpha }{n} = \frac{( - 1)^n}{2} \binom{\frac{{3n - 1}}{2} + \alpha }{n}. $$ Further simplification yields $$ \int_0^1 {t^{2\alpha } {\rm e}^{ - \lambda (t^2 + t^3 )} {\rm d}t} \sim \frac{1}{2}\lambda^{ - \alpha - 1/2} \sum\limits_{n = 0}^\infty {\frac{{\Gamma \big( {\frac{{3n + 1}}{2} + \alpha } \big)}}{{n!}}\frac{( - 1)^n}{{\lambda^{n/2} }}} $$ as $\lambda\to+\infty$.

For the second integral, split it at $t=\frac{2}{3}$ and consider the two integrals separately.

Answer 2

I've found a solution by ignoring the hint and simply using Laplace's method (thanks to Gary). I'll post it below:

Set $\phi(t)=-t^2-t^3$ and $f(t)=t^{2\alpha}$. The only absolute maximum is $t = 0$, so contributions of all orders come from this. Then the integral becomes (under asymptotic equivalence):

$\int_0^\epsilon t^{2\alpha} e^{-\lambda t^2}\Sigma_{n=0}^\infty (-\lambda t^3)^n/n!$

(exchange the order of summation and integration by Fubini, and then do integrals by the Gamma function)

= $\Sigma_{n=0}^\infty \Gamma(1/2+\alpha+3n/2) (-1)^n \lambda^{-1/2-n/2-\alpha} / 2n! $

For the second part, set $\phi(t) = t^3 - t^2$, then absolute maxima are t = 0 and 1, with t = 0 being a local maximum (but t = 1 is not). So, contributions of all orders will come from nbhds of these 2 points, but the l.o.t. is determined by the nbhd of t = 0. To find the first 2 terms, find the first 2 orders around t = 0 and the l.o.t around t = 1, then sum up.