limit with a parametrical integral $ \lim_{n \to \infty} n^2 \int_{0}^{\infty} \frac{sin(x)}{(1 + x)^n} dx $

Question

I need help in calculating this strange limit.

$$ \lim_{n \to \infty} n^2 \int_{0}^{\infty} \frac{sin(x)}{(1 + x)^n} dx $$

Answer 1

We can substitute $x=\frac{t}{n}$:

$$\lim_{n \to \infty} n^2 \int_{0}^{\infty} \frac{\sin(x)}{(1 + x)^n} dx=\lim_{n \to \infty} n^2 \int_{0}^{\infty} \frac{\sin(t/n)}{(1 + \frac tn)^n}\frac{1}{n}dt$$

Now in the denominator we get $(1+\frac{t}{n})^n$, and we're taking the limit $n\to\infty$; this becomes $e^t$, so that we get (also take the $\frac 1n$ out of the integral):

$$\lim_{n \to \infty} n \int_{0}^{\infty} \frac{\sin(t/n)}{e^t}dt$$

Now also see that $\lim_{n\to\infty}n\sin(\frac{t}{n})=t$ so that we get

$$\lim_{n \to \infty}\int_{0}^{\infty} \frac{t}{e^t}dt$$

Now the $n$'s are gone so we can write

$$\int_{0}^{\infty} \frac{t}{e^t}dt$$

Can you take it from here?

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \lim_{n \to \infty}\bracks{n^{2}\int_{0}^{\infty} {\sin\pars{x} \over \pars{1 + x}^{n}}\,\dd x} & = \lim_{n \to \infty}\bracks{n^{2}\int_{0}^{\infty} \sin\pars{x}\exp\pars{-n\ln\pars{1 + x}}\,\dd x} \\[5mm] & = \lim_{n \to \infty}\bracks{n^{2}\int_{0}^{\infty} x\,\exp\pars{-nx}\,\dd x} = \lim_{n \to \infty}\int_{0}^{\infty} x\,\exp\pars{-x}\,\dd x \\[5mm] & = \bbx{1} \end{align}

See Laplace's Method.