# Asymptotic expansion using gamma functions and Stirling's formula

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!

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.

• Thank you! I guess you're right. Just applying Laplace'es method is more efficient as things are already in polynomials form. Commented Apr 19 at 10:59
• You are welcome. Did you find my answer to your other question useful?
– Gary
Commented Apr 19 at 12:33
• Absolutely yes! With your help I've been able to do all of them. Thank you so much! Commented Apr 19 at 13:59
• For the question using Stirling's formula and gamma functions, I haven't studied Bernoulli numbers, but since I only need to find the first 2 terms, so I used your idea and got the first two terms explicitly :) Commented Apr 19 at 14:01

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.