asymptotic expansions of integral

Question

Let $u_n=\displaystyle\int_0^1f(x)^{n-1}dx$ and $f(x)=1+ax-bx^2$ with $a>2b$, $a>b$, $a\geq 0$ . Prove that $$u_n\sim\dfrac{(f(1))^n}{nf'(1)}$$

I don't know how to do this problem. any ideas to solve it?

edit: with hint of Robert Israel , put $y= 1+ax-bx^2$ and let $\alpha = 1+a-b$ then after calculation $$u_n=\int_1^{\alpha}\frac {y^{n-1}}{\sqrt {a^2+4b-4by}} dy$$ put $t=\frac {y^n}{\alpha ^n}$ then $$u_n=\frac {\alpha^n}{n}\int_{\alpha^{-n}}^{1}\frac 1{\sqrt{a^2+4b-4b\alpha t^{\frac 1n}}}\sim \frac {\alpha^n}{n}\int_{0}^{1}\frac 1{\sqrt{a^2+4b-4b\alpha }}\ dt =\frac {\alpha^n}{n (a-2b)}$$ which ends the proof. But i can't see how to use Laplace's method or Watson's lemma

Answer 1

First note that, for $x \in \left[ {0,1} \right]$, $f'(x) = a - 2bx \geq a-2b > 0$ and $f(0)=1$. So $f(x)\geq 1$ on $\left[ {0,1} \right]$. Accordingly, we can write $$ \int_0^1 {f^{n - 1}(x) dx} = \int_0^1 {e^{(n-1)\log f(x)} dx} . $$ Since $f'(x) > 0$, the exponent is increasing on the interval of integration and reaches a maximum at the endpoint $x=1$. Since the saddle point is at $x=\frac{a}{2b}>1$, this is the linear endpoint case of Laplace's method (see, e.g, Eq. (4.3) in http://www.macs.hw.ac.uk/~simonm/ae.pdf). Thus, $$ \int_0^1 {e^{(n - 1)\log f(x)} dx} \sim \frac{1}{{n - 1}}\frac{{f(1)}}{{f'(1)}}e^{(n - 1)\log f(1)} \sim \frac{1}{n}\frac{{f^n (1)}}{{f'(1)}} $$ as $n\to +\infty$.