# Define $I_n=\int_0^1\frac{x^n}{\sqrt{x^2+1}}dx$ for every $n\in\mathbb{N}$. Prove that $\lim_{n\to\infty}nI_n=\frac{1}{\sqrt 2}$.

Question: Define $$I_n=\int_0^1\frac{x^n}{\sqrt{x^2+1}}dx$$ for every $$n\in\mathbb{N}$$. Prove that $$\lim_{n\to\infty}nI_n=\frac{1}{\sqrt 2}$$.

My approach: Given that $$I_n=\int_0^1\frac{x^n}{\sqrt{x^2+1}}dx, \forall n\in\mathbb{N}.$$ Let us make the substitution $$x^n=t$$, then $$nI_n=\int_0^1\frac{dt}{\sqrt{1+t^{-2/n}}}.$$

Now since $$0\le t\le 1\implies \frac{1}{t}\ge 1\implies \left(\frac{1}{t}\right)^{2/n}\ge 1 \implies 1+\left(\frac{1}{t}\right)^{2/n}\ge 2\implies \sqrt{1+\left(\frac{1}{t}\right)^{2/n}}\ge \sqrt 2.$$

This implies that $$\frac{1}{\sqrt{1+\left(\frac{1}{t}\right)^{2/n}}}\le\frac{1}{\sqrt 2}\\ \implies \int_0^1 \frac{dt}{\sqrt{1+\left(\frac{1}{t}\right)^{2/n}}}\le \int_0^1\frac{dt}{\sqrt 2}=\frac{1}{\sqrt 2}.$$

So, as you can see, I am trying to solve the question using Sandwich theorem.

Can someone help me to proceed after this?

Also, in $$\lim_{n\to\infty}nI_n=\lim_{n\to\infty}\int_0^1\frac{dt}{\sqrt{1+t^{-2/n}}},$$ the limit and integral interchangeable?

• This answers your question. – Paras Khosla Mar 28 '20 at 14:53

You can still use Sandwich theorem, if you haven't yet seen dominated convergence theorem :

Let $$n$$ be a positive integer.

As you said, using the substitution \left\lbrace\begin{aligned}y&=x^{n}\\ \mathrm{d}y &=n x^{n-1}\,\mathrm{d}x\end{aligned}\right., we get : $$\int_{0}^{1}{\frac{n x^{n}}{\sqrt{1+x^{2}}}\,\mathrm{d}x}=\int_{0}^{1}{\frac{y^{\frac{1}{n}}}{\sqrt{1+y^{\frac{2}{n}}}}\,\mathrm{d}y}$$

Meaning, we have : \begin{aligned}\left|\frac{1}{\sqrt{2}}-n I_{n}\right|&=\left|\int_{0}^{1}{\left(\frac{1}{\sqrt{2}}-\frac{y^{\frac{1}{n}}}{\sqrt{1+y^{\frac{2}{n}}}}\right)\mathrm{d}y}\right| \\ &=\left|\int_{0}^{1}{\frac{\sqrt{1+y^{\frac{2}{n}}}-\sqrt{2}y^{\frac{1}{n}}}{\sqrt{2\left(1+y^{\frac{2}{n}}\right)}}\,\mathrm{d}y}\right|\\ &=\int_{0}^{1}{\frac{1-y^{\frac{2}{n}}}{\sqrt{2\left(1+y^{\frac{2}{n}}\right)}\left(\sqrt{1+y^{\frac{2}{n}}}+\sqrt{2}y^{\frac{1}{n}}\right)}\,\mathrm{d}y}\end{aligned}

Since $$\left(\forall y\in\left[0,1\right]\right),\ \sqrt{2\left(1+y^{\frac{2}{n}}\right)}\left(\sqrt{1+y^{\frac{2}{n}}}+\sqrt{2}y^{\frac{1}{n}}\right)\geq \sqrt{2}+\sqrt{2}y^{\frac{2}{n}}\geq 1$$, we have : $$\int\limits_{0}^{1}{\frac{1-y^{\frac{2}{n}}}{\sqrt{2\left(1+y^{\frac{2}{n}}\right)}\left(\sqrt{1+y^{\frac{2}{n}}}+\sqrt{2}y^{\frac{1}{n}}\right)}\,\mathrm{d}y}\leq\int\limits_{0}^{1}{\left(1-y^{\frac{2}{n}}\right)\mathrm{d}y}$$, and thus : $$\left|\frac{1}{\sqrt{2}}-n I_{n}\right|\leq\int_{0}^{1}{\left(1-y^{\frac{2}{n}}\right)\mathrm{d}y}=\frac{2}{n+2}\underset{n\to +\infty}{\longrightarrow}0$$

Hence : $$\lim_{n\to +\infty}{nI_{n}}=\frac{1}{\sqrt{2}}$$

• Did the $2$ disappear from the denominator? $\sqrt{1+y^{2/n}}$ instead of $\sqrt{2\left(1+y^{2/n}\right)}$ – bjorn93 Mar 28 '20 at 15:25
• You're right. I'll fix it. But it won't change much. – CHAMSI Mar 28 '20 at 15:43

Actually you are done. You already have: $$nI_n=\int_0^1\frac{dt}{\sqrt{1+t^{-2/n}}}$$ hence \begin{align}\lim_{n\to\infty} nI_n &= \lim_{n\to\infty}\int_0^1\frac{dt}{\sqrt{1+t^{-2/n}}} \\ &= \int_0^1 \lim_{n\to\infty} \frac{dt}{\sqrt{1+t^{-2/n}}} \\ &= \int_0^1\frac{dt}{\sqrt{1+1}} = \frac{1}{\sqrt{2}} \end{align}

You can exchange limit and integration due to dominated convergence theorem.

you can use the Binomial series {{\left( 1+{{x}^{2}} \right)}^{-1/2}}=\sum\nolimits_{k=0}^{\infty }{\left( \begin{align} & -1/2 \\ & \ \ k \\ \end{align} \right){{x}^{2k}}} {n{I}_{n}}=n\int_{0}^{1}{{{x}^{n}}{{\left( 1+{{x}^{2}} \right)}^{-1/2}}dx}=n\sum\nolimits_{k=0}^{\infty }{\left\{ \left( \begin{align} & -1/2 \\ & \ \ k \\ \end{align} \right)\int_{0}^{1}{{{x}^{2k+n}}dx} \right\}}=\sum\nolimits_{k=0}^{\infty }{\frac{n\left( \begin{align} & -1/2 \\ & \ \ k \\ \end{align} \right)}{\left( 2k+n+1 \right)}}

\underset{n\to \infty }{\mathop{\lim }}\,{n{I}_{n}}=\underset{n\to \infty }{\mathop{\lim }}\,\sum\nolimits_{k=0}^{\infty }{\frac{n\left( \begin{align} & -1/2 \\ & \ \ k \\ \end{align} \right)}{\left( 2k+n+1 \right)}}=\sum\nolimits_{k=0}^{\infty }{\underset{n\to \infty }{\mathop{\lim }}\,\frac{n\left( \begin{align} & -1/2 \\ & \ \ k \\ \end{align} \right)}{\left( 2k+n+1 \right)}}=\sum\nolimits_{k=0}^{\infty }{\left( \begin{align} & -1/2 \\ & \ \ k \\ \end{align} \right)}={{\left( 1+1 \right)}^{-1/2}}=\frac{1}{\sqrt{2}}

Since $$n/(n+1)\to 1$$ the desired limit is equal to the limit of $$(n+1)\int_{0}^{1}\frac{x^n}{\sqrt{1+x^2}}\,dx=\left.\frac{x^{n+1}}{\sqrt{1+x^2}}\right|_{x=0}^{x=1}+\int_{0}^{1}\frac{x^{n+2}}{(1+x^2)^{3/2}}\,dx$$ and this is same as $$\frac{1}{\sqrt{2}}+\int_{0}^{1}\frac{x^{n+2}}{(1+x^2)^{3/2}}\,dx$$ The integral above clearly lies between $$\frac{1}{2\sqrt{2}}\int_{0}^{1}x^{n+2}\,dx=\frac{1}{2\sqrt{2}(n+3)}$$ and $$\int_{0}^{1}x^{n+2}\,dx=\frac{1}{n+3}$$ and thus by Squeeze Theorem tends to $$0$$. Therefore the desired limit is $$1/\sqrt{2}$$.

The same technique can be used to prove more generally that $$n\int_{0}^{1}x^nf(x)\,dx\to f(1)$$

Yet another approach is to use integration by parts to obtain $$\begin{equation*} I_{n} + I_{n-2} = \int_{0}^{1}x^{n-2}\sqrt{1+x^{2}}\,\mathrm{d}x = \frac{\sqrt{2}}{n-1} - \frac{1}{n-1}I_{n} \end{equation*}$$ for $$n \geq 2$$, from which we get $$\begin{equation*} nI_{n} = \sqrt{2} - (n-1)I_{n-2} = \sqrt{2} - (n-2)I_{n-2} - I_{n-2}. \tag*{(1)} \end{equation*}$$ Define $$S_{n} = nI_{n}$$ for each $$n\in \mathbb{N}$$. You have already shown that the sequence $$\{S_{n}\}_{n\geq 2}$$ is bounded above by $$1/\sqrt{2}$$, and note also that \begin{align} S_{n+1} - S_{n} &= \int_{0}^{1}\frac{(n+1)x^{n+1}-nx^{n}}{\sqrt{1+x^{2}}}\,\mathrm{d}x \\ &\geq \frac{1}{\sqrt{2}}\int_{0}^{1}\big((n+1)x^{n+1} - nx^{n}\big)\,\mathrm{d}x \\ &= \frac{1}{\sqrt{2}(n+1)(n+2)} > 0 \end{align} so that the sequence $$\{S_{n}\}_{n\geq 2}$$ is increasing. Thus by the monotone convergence theorem, $$S_{n}$$ converges to a limit $$L$$. It is therefore valid to take the limit as $$n \to \infty$$ on both sides of $$(1)$$, noting that $$I_{n-2} \to 0$$, to find that $$\begin{equation*} L=\sqrt{2} - L \end{equation*}$$ or $$\begin{equation*} L = \frac{1}{\sqrt{2}}. \end{equation*}$$