Find $\lim _{n \rightarrow+\infty} n \sqrt{n} \int_0^{+\infty} \frac{\sin x^2}{\left(1+x^2\right)^n} d x$

Question

I want to calculate the limit of following integral: $$\lim _{n \rightarrow+\infty} n \sqrt{n} \int_0^{+\infty} \frac{\sin x^2}{\left(1+x^2\right)^n} d x$$ I think a feasible method is through substitution and then using the Dominated Convergence Theorem, but I am not sure how to start. Could someone give me a hint?

Answer 1

The desired expression can be expressed as \begin{align} A_n:=n\sqrt{n}\int^\infty_0\frac{\sin x^2}{(1+x^2)^n}\,dx &\stackrel{x=y/\sqrt{n}}{=}n\int^\infty_0\sin(y^2/n)\Big(1+\frac{y^2}{n}\Big)^{-n}\,dy\\ &=\int^\infty_0\operatorname{sinc}(y^2/n) y^2\Big(1+\frac{y^2}{n}\Big)^{-n}\,dy \end{align}

Notice that

$$0\leq \Big|n \sin(y^2/n)\Big(1+\frac{y^2}{n}\Big)^{-n}\Big|\leq y^2\Big(1+\frac{y^2}{n}\Big)^{-n}\leq \frac{y^2}{\big(1+y^2/2\big)^2}$$ for all $n\geq 2$.

An application of monotone convergence (or dominated convergence) yields $$\lim_nA_n=\int^\infty_0y^2e^{-y^2}\,dy=\frac{\sqrt{\pi}}{4}$$

One an also obtain an asymptotic expansion for addiotnal error terms:

\begin{align} A_n:=n\sqrt{n}\int^\infty_0\frac{\sin x^2}{(1+x^2)^n}\,dx &\stackrel{x=y/\sqrt{n}}{=}n\int^\infty_0\sin(y^2/n)\Big(1+\frac{y^2}{n}\Big)^{-n}\,dy\\ &\sim n\int^\infty_0\sin(y^2/n)e^{-y^2}\,dy\\ &= n\sum^\infty_{m=0}\frac{(-1)^m}{(2m+1)! n^{2m+1}}\int^\infty_0 y^{2(2m+1)}e^{-y^2}\,dy \end{align}

Thus

$$A_n\sim \int^\infty_0y^2e^{-y^2}dy=\frac{\sqrt{\pi}}{4} + o(1/n)$$

Answer 2

Using $\,\,\displaystyle \frac1{(1+x^2)^n}=\frac1{\Gamma(n)}\int_0^\infty t^{n-1}e^{-t(1+x^2)}dt\,$, and changing the order of integration $$I(n)=\Im\,\int_0^\infty \frac{e^{ix^2}}{\left(1+x^2\right)^n} dx=\Re\,\frac1{\Gamma(n)}\int_0^\infty t^{n-1}e^{-t}dt\int_0^\infty e^{ix^2-tx^2}dx$$ $$=\Im\,\frac1{\Gamma(n)}\frac{\sqrt\pi}2\int_0^\infty t^{n-1}e^{-t}\frac{dt}{\sqrt{t-i}}\overset{(n-1)x=t}{=}$$ $$=\Im\,\frac{\sqrt\pi}2\frac{n-1}{(n-1)!}\int_0^\infty e^{-(n-1)x+(n-1)\ln(n-1)-(n-1)\ln x}\frac{dx}{\sqrt{(n-1)x-i}}$$ Denoting for a while $n-1=m$ $$I(m)=\Im\,\frac{\sqrt\pi}2\frac{m^{m+1}}{m!}\int_0^\infty e^{-m(x-\ln x)}\frac{dx}{\sqrt{mx-i}}$$ Using the Laplace' method we see that $\,f(x)=x-\ln x\,$ reaches its minimum at $x=1$. Decomposing $f(x)$ (near $x=1$) and the square root in the denominator (to get the leading imaginary term) $$I(m)\sim\Im\,\frac{\sqrt{2\pi m}}{2\sqrt{2 m}}\frac{m^m e^{-m}}{m!}\int_{-\infty}^\infty e^{-\frac{mt^2}2}\frac{dt}{\sqrt{m(t+1)-i}}$$ $$\sim\frac{\sqrt{2\pi m}}{4\sqrt2\, m}\frac{m^m e^{-m}}{m!}\int_{-\infty}^\infty e^{-\frac{mt^2}2}dt\sim\frac{\sqrt\pi}4\frac1{m\sqrt m}$$ The limit follows.

Answer 3

Substitute $x=\frac{t}{\sqrt{n}}$

Then according to the control convergence theorem (it is not difficult to verify the applicable conditions),$$\operatorname*{lim}_{n\to\infty}\int_{0}^{+\infty}\frac{n\sin\frac{t^{2}}{n}}{(1+\frac{t^{2}}{n})^{n}}\mathrm{d}t=\int_{0}^{+\infty}\lim_{n\to\infty}\frac{n\sin\frac{t^{2}}{n}}{(1+\frac{t^{2}}{n})^{n}}\mathrm{d}t=\int_{0}^{+\infty}t^{2}e^{-t^{2}}\mathrm{d}t=\frac{\sqrt{\pi}}{4}.$$