To find the limit $$\lim_{n \rightarrow \infty} \sqrt{n} \int_0^1 \left(\frac{\sin t}{t}\right)^n dt,$$ I attempted to use Laplace's method. We can express the given integral as $$\int_0^1 \exp \left(n \ln \left(\frac{\sin x}{x}\right)\right) dx$$ However, I encountered an issue with this approach. In Laplace's method, we require the function $f(x)=\ln \left(\frac{\sin x}{x}\right)$ to be twice differentiable. Additionally, the global maximum of $f(x)$ within the integration range must be unique and not located at the boundary points. Unfortunately, in this case, the maximum occurs at the boundary point $0$, which prevents me from applying Laplace's method as intended.
I have been thinking about this problem for quite some time, but I haven't been able to come up with any promising ideas. I would really appreciate any guidance or suggestions. Thank you in advance.
Best Answer
You can expand the interval of integration to $[-1,1]$ since $t\mapsto \frac{\sin t}{t}$ is an even positive function on that interval. Then you can apply Laplace's method.
A more direct approach is to use a change of variables and dominated convergence:
Set $I_n=\sqrt{n}\int^1_0\Big(\frac{\sin t}{t}\Big)^n\,dt$. On $[0,1]$ we have that
$$0<1-\frac{t^2}{6}\leq \frac{\sin t}{t}\leq 1-\frac{t^2}{6}+\frac{t^4}{120}$$
The change of variables $u=\sqrt{n} t$ yields $$\int^{\sqrt{n}}_0\Big(1-\frac{u^2}{6n}\Big)^n\,du\leq I_n\leq \int^{\sqrt{n}}_0\Big(1-\frac{u^2}{6n}+\frac{u^4}{120n^2}\Big)^n\,du\leq\int^{\sqrt{n}}_0e^{-\Big(\frac{u^2}{6}-\frac{u^4}{120n}\Big)}\,du$$
Notice that $\big(1-\frac{u^2}{6n}\Big)^n\leq e^{-\frac{u^2}{6}}$.
Notice that on $[0,\sqrt{n}]$, $\frac{u^4}{120 n}\leq \frac{u^2}{120}$ and so, $$\frac{u^2}{6}-\frac{u^4}{120n}\geq\frac{19u^2}{120}$$
This means that $$e^{-\Big(\frac{u^2}{6}-\frac{u^4}{120n}\Big)} \leq e^{-\frac{19u^2}{120}}, \qquad 0\leq u\leq \sqrt{n}$$
Since $f_a(u)=e^{-au^2}\in L_1(\mathbb{R})$ for all $a>0$, an application of dominated convergence gives
$$\lim_nI_n=\int^\infty_0e^{-u^2/6}\,du$$