$\lim_{n\to\infty}\int_0^{\pi/2} \frac{\sin^n(x)}{1+x^2} \, dx$

Question

$$\lim_{n\to\infty}\int_0^{\pi/2} \frac{\sin^n(x)}{1+x^2} \, dx$$

Is it right answer ? $$ \begin{cases} 0, & x \ne \pi/2\\[8pt] \dfrac{2\pi}{4 + \pi^2}, & x = \pi/2 \end{cases} $$

Answer 1

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

Laplace Method:

\begin{align} \lim_{n\to\infty}\int_{0}^{\pi/2}{\sin^n\pars{x} \over 1 + x^{2}}\,\dd x & = \lim_{n\to\infty}\int_{0}^{\pi/2}{\cos^n\pars{x} \over 1 + \pars{\pi/2 - x}^{2}}\,\dd x = \lim_{n\to\infty}\int_{0}^{\pi/2}{\exp\pars{n\ln\pars{\cos\pars{x}}} \over 1 + \pars{\pi/2 - x}^{2}}\,\dd x \\[5mm] & = \lim_{n\to\infty}\int_{0}^{\infty}{\exp\pars{-nx^{2}/2} \over 1 + \pars{\pi/2 - 0}^{2}}\,\dd x = \lim_{n\to\infty}\pars{% \root{\pi \over 2}\,{4 \over \pi^{2} + 4}\,{1 \over \root{n}}} = \bbx{\ds{0}} \end{align}

Answer 2

Consider the sequence of functions $f_n (x) = {\sin^n (x)}/({1+x^2})$ on $[0,2\pi]$

If $a \neq \pi/2, 3\pi/2$ for some $k\in\mathbb{Z}$, then $|\sin(a)| < 1$, so as $n \to \infty$ we have $f_n (a) \to 0$. Thus, pointwise almost everywhere, as $n\to\infty$ we have $f_n \to 0$.

Let $g(x) = 1/(1+x^2)$. Note that $|f_n (x)| \le |g(x)|$ and $g(x)$ is integrable on $[0,2\pi]$. Hence, by the dominated convergence theorem, it follows $$\lim_{n\to\infty} \int_{0}^{\pi/2} \frac{\sin^n (x)}{1+x^2} \, dx = \lim_{n\to\infty} \int_{0}^{\pi/2} f_n (x) \, dx = \int_{0}^{\pi/2} \lim_{n\to\infty} f_n (x) \, dx = \int_{0}^{\pi/2} 0 \, dx = 0.$$

Answer 3

The answer should not depend on anything called $x$, since $x$ is a bound variable, not a free variable, in the integral. For example: $$ \sum_{i=1}^3 (i^2 + j^2) = \overbrace{(1^2+j^2) + (2^2 + j^2) + (3^2 + j^2)}^{\begin{smallmatrix} \text{No “$i$'' appears here, since $i$ is} \\[3pt] \text{a bound variable in this sum.} \\[3pt] {} \end{smallmatrix}} $$

---

You have $\lim\limits_{n\to\infty} \dfrac{\sin^n x}{1+x^2} = \begin{cases} 0 & \text{if }x\ne\dfrac\pi2, \\[8pt] \dfrac 1 {1+\left(\frac\pi2\right)^2} = \dfrac 4 {4+\pi^2} & \text{if } x = \dfrac\pi2. \end{cases} $

And then: $$ \overbrace{\lim_{n\to\infty} \int_0^{\pi/2} \frac{\sin^n x}{1+x^2} \,dx = \int_0^{\pi/2} \lim_{n\to\infty} \frac{\sin^n x}{1+x^2} \, dx}^{\Large\text{Is this valid?}\phantom{\frac 11}} $$

Interchanging a limit and an integral as above is not always valid, but it is valid when the all of the functions in the sequence are "dominated" by one function whose integral is finite. That is Lebesgue's dominated convergence theorem. We have $$ \frac{-1}{1+x^2} \le \frac{\sin^n x}{1+x^2} \le \frac 1 {1+x^2} $$ and we note that

• The function $\dfrac 1 {1+x^2}$ does not depend on the index $n$, and
• Its integral from $0$ to $\pi/2$ is finite.

Hence the dominated convergence theorem can be applied. You then have an integral of a function that is $0$ except at one point. The integral of that function is $0$.