# asymptotic expansions of integral

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

• Hint: Watson's lemma after a change of variables. Jun 14 '20 at 22:53
• macs.hw.ac.uk/~simonm/ae.pdf eq. 4.3. Jun 14 '20 at 23:12
• I will see your indications
– Jane
Jun 14 '20 at 23:25
• @Robert israel thank's for your hint , can you explain me how to use Watson's Lemma ?
– Jane
Jun 15 '20 at 16:26

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 $$n\to +\infty$$.