Why is $\int_{0}^{1}{(1+x^2)^n dx} \sim \frac{2^n}{n} $?

Question

Why is $\int_{0}^{1}{(1+x^2)^n dx} \sim \frac{2^n}{n}$?

By $a_n \sim b_n$ I mean that $\lim_{n \rightarrow \infty} \frac{a_n}{b_n} = 1$.

I don't know how to do this problem. I have tried to apply binomial theorem and I got $$\int_{0}^{1}{(1+x^2)^n dx} = \int_0^1 \sum_{k=0}^n{\binom{n}{k}x^{2k}dx} = \sum_{k=0}^n \int_0^1{ \binom{n}{k}x^{2k}dx} = \sum_{k=0}^n \frac {\binom{n}{k}}{2k+1}$$ But I don't know what I could do with this, nor if it is a correct approach.

$\sum_{k=0}^n \binom{n}{k}=2^n$ however I don't know why $\sum_{k=0}^n \frac {\binom{n}{k}}{2k+1}\approx \frac{2^n}{n}$ — quantus14, May 24 '20 at 23:23
You are talking about asymptotic equivalence, not approximation. I made the necessary edit in notation. — gen-ℤ ready to perish, May 24 '20 at 23:26
@quantus see my solution if it correctly answer the question — James Moriarty, May 25 '20 at 00:17
for what it's worth, this means the area bounded by $y=0, x=1, x=0, $ and $y=(1+x^2)^n$ approaches the area bounded by $y=0, x=1, $ and the tangent line to $y=(1+x^2)^n$ at $x=1$ — J. W. Tanner, May 25 '20 at 00:55

score 8 · Accepted Answer · answered May 25 '20 at 13:41

Let $u=(1+x^2)/2$, so that $du=x\,dx=\sqrt{2u-1}\,dx$. It follows that

$${n\over2^n}\int_0^1(1+x^2)^n\,dx=n\int_{1/2}^1{u^n\over\sqrt{2u-1}}\,du=n\int_{1/2}^1u^n\,du+n\int_{1/2}^1u^n\left({1\over\sqrt{2u-1}}-1\right)\,du$$

Now

$$n\int_{1/2}^1u^n\,du={n\over n+1}\left(1-\left(1\over2\right)^{n+1}\right)\to1(1-0)=1$$

So it remains to show that

$$n\int_{1/2}^1u^n\left({1\over\sqrt{2u-1}}-1\right)\,du\to0$$

Note that

$$0\le{1\over\sqrt{2u-1}}-1={1-\sqrt{2u-1}\over\sqrt{2u-1}}={2(1-u)\over\sqrt{2u-1}(1+\sqrt{2u-1})}\le{1-u\over\sqrt{2u-1}}$$

so it's enough to show that

$$n\int_{1/2}^1{u^n(1-u)\over\sqrt{2u-1}}\,du\to0$$

It's convenient to let $u=1-v$ and, taking $n=N^3$ to be sufficiently large, rewrite the integral as

$$\begin{align} N^3\int_0^{1/2}{v(1-v)^{N^3}\,dv\over\sqrt{1-2v}} &=N^3\int_0^{1/N^2}{v(1-v)^{N^3}\,dv\over\sqrt{1-2v}} +N^3\int_{1/N^2}^{1/2}{v(1-v)^{N^3}\,dv\over\sqrt{1-2v}}\\ &\le{N^3\over\sqrt{1-1/N^2}}\int_0^{1/N^2}v\,dv+{N^3\over2}\left(1-{1\over N^2}\right)^{N^3}\int_0^{1/2}{dv\over\sqrt{1-2v}}\\ &={1\over2N\sqrt{1-2/N^2}}+{N^3\over2}\left(\left(1-{1\over N^2}\right)^{N^2} \right)^N\\ &\le{1\over N}+N^3(e^{-1/2})^N\\ &\to0+0=0 \end{align}$$

(Note, these calculations do not require $n$ to be an integer.)

Ryszard Szwarc · Answer 2 · 2024-05-26T17:32:56.607

7

Solution 1

Let $I_n=\displaystyle\int\limits_0^1(1+x^2)^{n}\,dx.$ For $n\ge 1$ we have $$ I_n\ge\int\limits_0^12x(1+x^2)^{n-1}\,dx=\left.{1\over n}(1+x^2)^n\right\vert_0^1={2^n-1\over n}$$ On the other hand $$I_n\le \int\limits_0^1(1+x)(1+x^2)^{n-1}\,dx=I_{n-1}+{2^n-1\over 2n}\le I_{n-1}+{2^{n-1}\over n}$$ For $n\ge 6$ we have $$ {2^{n-1}\over n}<{2^n\over n-2}-{2^{n-1}\over n-3}$$ Thus the sequence $I_n-{2^n\over n-2}$ is decreasing for $n\ge 5,$ i.e. $$I_n-{2^n\over n-2}< I_5-{2^5\over 3},\quad n>5$$ Summarizing we got $${2^n-1\over n}<I_n<{2^n\over n-2}+I_5-{2^5\over 3},\ n> 5$$

Solution 2 (inspired by the answer of Claude Lebovici)

We have $$I_n=\sum_{k=0}^n{n\choose k}{1\over 2k+1}\ge \sum_{k=0}^n{n\choose k}{1\over 2(k+1)}\\ ={1\over 2(n+1)}\sum_{k=0}^n{n+1\choose k+1}={1\over 2(n+1)}(2^{n+1}-1)$$ On the other hand $$I_n= \sum_{k=0}^n{n\choose k}{1\over 2(k+1)}+ \sum_{k=0}^n{n\choose k}{1\over 2(k+1)(2k+1)}\\ \le {2^n\over n+1}+{1\over 2}+\sum_{k=1}^n{n\choose k}{1\over 2(k+1)(k+2)}$$ $$\sum_{k=1}^n{n\choose k}{1\over 2(k+1)(k+2)} ={1\over 2(n+1)(n+2)}\sum_{k=1}^n{n+2\choose k+2}\\ \le {2^{n+1}\over (n+1)(n+2)}$$ Summarizing we have obtained $${1\over 2(n+1)}(2^{n+1}-1)\le I_n\le {2^n\over n+1}+{1\over 2}+{2^{n+1}\over (n+1)(n+2)}$$

Remark The lower estimate in solution $1$ is stronger than in solution $2.$ However the upper estimate in solution $2$ is stronger than in solution $1.$

edited May 26 '24 at 17:32

answered May 25 '24 at 14:11

Ryszard Szwarc

44,675

You are welcome. FIY, I found the link using Approach$0$. – Anne Bauval May 25 '24 at 14:15
I had an answer ready to go using the gaussian hypergeometric function. But you deleted and so I did withut saving my answer !!!. I shall try to come back – Claude Leibovici May 25 '24 at 14:22
Nice. Typo in first line? – River Li May 25 '24 at 14:30
@ClaudeLeibovici I am terribly sorry. I was unaware that you were in the process of answering. I wonder if you can retrieve your answer when I undelete my question temporarily. Just let me know and we can try the experiment. – Ryszard Szwarc May 25 '24 at 14:35
@RiverLi Thanks. Corrected. – Ryszard Szwarc May 25 '24 at 14:35
1

Don't worry, please ! I shall redo it probably tomorrow here. Cheers :-) – Claude Leibovici May 25 '24 at 14:41
@ClaudeLeibovici I look forward to seeing your answer, although Roland Garros starts tomorrow :) – Ryszard Szwarc May 25 '24 at 14:43
I have been limited by the fact that I did not want to use or reuse any material already given in this post. – Claude Leibovici May 26 '24 at 01:49
There is one thing which is amuzing. Let $x=\sinh(t)$ and then $$I_n=\frac{i}{2} \left(B_2\left(n+1,\frac{1}{2}\right)-\frac{\sqrt{\pi } \Gamma (n+1)}{\Gamma \left(n+\frac{3}{2}\right)}\right)$$ – Claude Leibovici May 26 '24 at 09:33

James Moriarty · Answer 3 · 2024-11-30T23:58:05.637

A partial progress

We make the substitution $x=\tan(y)$. Then $dx=\sec^2(y)dy$.

Now $$\int_{0}^{1}(1+x^2)^ndx=\int_{0}^{\frac{\pi}{4}}\sec^{2n}(y)\sec^2(y)dy=\int_{0}^{\frac{\pi}{4}}\sec^{2(n+1)}(y)dy$$

Maybe it will be helpful.

Here is the proof by proceeding in your way.

$$\sum_{k=0}^{n}\frac{n\binom{n}{k}}{2k+1}>\sum_{k=0}^{n}\frac{n\binom{n}{k}}{2k+2}=\frac{n}{2(n+1)}\sum_{k=0}^{n}\frac{n+1}{k+1}\binom{n}{k}=\frac{n}{2(n+1)}\sum_{k=0}^{n}\binom{n+1}{k+1}=\frac{n}{2(n+1)}(2^{n+1}-1)$$

Hence $$\lim_{n\rightarrow\infty}\sum_{k=0}^{n}\frac{n\binom{n}{k}}{2^n(2k+1)}\geq\lim_{n\rightarrow\infty}\frac{n}{2(n+1)}\left(\frac{2^{n+1}}{2^n}-\frac{1}{2^n}\right)=1$$

Again we see that,

$$\sum_{k=0}^{n}\frac{\binom{n}{k}}{2k+1}<1+\left(\frac{1}{2}\sum_{k=1}^{n}\frac{1}{k}\binom{n}{k}\right)$$

From this you can show that $$\lim_{n\rightarrow\infty}\sum_{k=0}^{n}\frac{n\binom{n}{k}}{2^n(2k+1)}\leq1$$

Hence the result will follow.

score 3 · Answer 4 · answered May 25 '20 at 00:19

3

You can also write $ \int_0^1 (1+x^2)^ndx = \int_0^1 e^{n\log(1+x^2)}dx $ and use a generalization of Laplace's method to handle the boundary case.

answered May 25 '20 at 00:19

mr_snazzly

455

score 3 · Answer 5 · answered May 28 '24 at 02:19

We perform a change of variables in the integral from $x$ to $t$ using the transformation $1 + x^2 = 2{\rm e}^{ - t} $, leading to $$ \int_0^1 {(1 + x^2 )^n {\rm d}x} = 2^n \int_0^{\log 2} {{\rm e}^{ - nt} \frac{{{\rm e}^{ - t} }}{{\sqrt {2{\rm e}^{ - t} - 1} }}{\rm d}t} . $$ We split the integral at $t=\log\sqrt{2}$ into two parts and estimate each part separately. For the upper part, we have $$ 2^n \int_{\log \sqrt 2 }^{\log 2} {{\rm e}^{ - nt} \frac{{{\rm e}^{ - t} }}{{\sqrt {2{\rm e}^{ - t} - 1} }}{\rm d}t} \le 2^{n/2} \int_{\log \sqrt 2 }^{\log 2} {\frac{{{\rm e}^{ - t} }}{{\sqrt {2{\rm e}^{ - t} - 1} }}{\rm d}t} = \sqrt {\sqrt 2 - 1} \,2^{n/2} = \mathcal{O}(2^{n/2} ) $$ as $n\to+\infty$. To estimate the lower part, we first note that for small $t$, $$ \frac{{{\rm e}^{ - t} }}{{\sqrt {2{\rm e}^{ - t} - 1} }} = \sum\limits_{k = 0}^\infty {a_k \frac{{t^k }}{{k!}}} ,\quad \quad a_k = \sum\limits_{j = 0}^k {\left( {\frac{{(2j)!}}{{2^j j!}}} \right)^2 S(k,2j)} , $$ where $S(k,j)$ are the Stirling numbers of the second kind (see $\text{A}014304$). By Watson's lemma $$ 2^n \int_0^{\log \sqrt 2 } {{\rm e}^{ - nt} \frac{{{\rm e}^{ - t} }}{{\sqrt {2{\rm e}^{ - t} - 1} }}{\rm d}t} \sim \frac{{2^n }}{n}\sum\limits_{k = 0}^\infty {\frac{{a_k }}{{n^k }}} $$ as $n\to+\infty$. Therefore, $$ \int_0^1 {(1 + x^2 )^n {\rm d}x} \sim \frac{{2^n }}{n}\sum\limits_{k = 0}^\infty {\frac{{a_k }}{{n^k }}} = \frac{{2^n }}{n}\left( {1 + \frac{1}{{n^2 }} + \frac{1}{{n^3 }} + \frac{3}{{n^4 }} + \frac{{16}}{{n^5 }} + \ldots } \right) $$ as $n\to+\infty$.

Would you be interested by https://math.stackexchange.com/questions/4925408/approximate-solution-of-xx-x-nxn ? Cheers — Claude Leibovici, May 31 '24 at 14:20

score 2 · Answer 6 · answered May 26 '24 at 01:47

2

$$I_n=\int_0^1 (1+x^2)^n \,dx=\sum_{k=0}^n \frac {\binom{n}{k}}{2k+1}=\,_2F_1\left(\frac{1}{2},-n;\frac{3}{2};-1\right)=\, _2F_1\left(-n,\frac{1}{2};\frac{3}{2 };-1\right)$$ grows exponentially. I have not been able to derive its exact asymptotic.

But, since $$\frac 1 {2k+1}=\sum_{m=0}^\infty (-1)^m (2k)^{-(m+1)}$$

consider $$J_n= \frac n {2^{n}}\, I_n-1$$ For $n \geq 3$, they are $$\left\{\frac{1}{35},\frac{17}{315},\frac{37}{693},\frac{45}{1001},\frac{229 }{6435},\frac{3013}{109395},\frac{49 09}{230945},\frac{5347}{323323},\frac {8821}{676039},\frac{176933}{169009 75},\cdots\right\}$$

answered May 26 '24 at 01:47

Claude Leibovici

289,558

Thanks. I do not quite see the advantage of having $2k$ in the denominator in place of $2k+1.$ In my opinion $2(k+1)$ is more handy as $${{n\choose k}\over 2(k+1)}= {1\over 2(n+1)}{n+1\choose k+1}$$ – Ryszard Szwarc May 26 '24 at 06:34
@RyszardSzwarc. By the end, me neither ! Cheers :-) – Claude Leibovici May 26 '24 at 06:36
Your answer inspired me to give an alternative proof based on the formula $\sum {n\choose k}{1\over 2k+1}.$ Shall I extend my answer ? – Ryszard Szwarc May 26 '24 at 06:40
@RyszardSzwarc. For sure, do it – Claude Leibovici May 26 '24 at 06:50

Felix Marin · Answer 7 · 2024-06-11T06:00:31.357

$\underline{\large Laplace’s Method}$: \begin{align} & \color{#44f}{\int_{0}^{1}\left(1 + x^{2}\right)^{n} {\rm d}x} = \int_{0}^{1}\left[1 + \left(1 - x\right)^{2}\right]^{n}{\rm d}x \\[5mm] & \stackrel{{\rm as}\ n\ \to\ \infty}{\sim}\,\,\, 2^{n}\int_{0}^{\infty}{\rm e}^{n\ln\left(1 - x\right)}\,\,\,{\rm d}x \\[5mm] & \stackrel{{\rm as}\ n\ \to\ \infty}{\sim}\,\,\, 2^{n}\int_{0}^{\infty}{\rm e}^{-nx}\,{\rm d}x = \color{#44f}{{\Large 2^{n}} \over {\Large n}} \end{align}

score 1 · Answer 8 · answered May 25 '24 at 15:49

Alternative solution.

First, using the integral substitution $x = \sqrt{y}$, we have \begin{align*} \int_0^1 (1 + x^2)^n \,\mathrm{d} x &= \int_0^1 \frac{(1 + y)^n}{2\sqrt{y}}\, \mathrm{d} y\\ &\ge \int_0^1 \frac{(1 + y)^n}{2}\left(1 - \frac{y-1}{2}\right)\, \mathrm{d} y\\ &= \frac{n + 3}{(n + 1)(n + 2)}\cdot 2^n - \frac{3n + 7}{4(n + 1)(n + 2)}.\tag{1} \end{align*} (Note: We have the Taylor expansion $\frac{1}{\sqrt{y}} = 1 - \frac12(y - 1) + \frac38(y - 1)^2 + \cdots$. We can prove that $\frac{1}{\sqrt{y}} \ge 1 - \frac12(y - 1)$ on $[0, 1]$.)

Second, we have \begin{align*} \int_0^1 (1 + x^2)^n \,\mathrm{d} x &= \int_0^{1/2} (1 + x^2)^n \,\mathrm{d} x + \int_{1/2}^1 (1 + x^2)^n \,\mathrm{d} x\\ &= \int_0^{1/2} (1 + x^2)^n \,\mathrm{d} x + \int_{1/4}^1 \frac{(1 + y)^n}{2\sqrt{y}}\,\mathrm{d} y\\ &\le \frac12(1 + 1/4)^n + \int_{1/4}^1 \frac{(1 + y)^n}{2}(3 - 2y)\,\mathrm{d} y\\ &= \frac{n+6}{(n+1)(n+2)}\cdot 2^n -\frac{25(n+3)}{16(n+1)(n+2)}(5/4)^n + \frac12 (5/4)^n \tag{2} \end{align*} where we use the integral substitution $x = \sqrt{y}$, and $\frac{1}{\sqrt{y}} \le 3 - 2y$ on $[1/4, 1]$.

From (1) and (2), we have $$\int_0^1 (1 + x^2)^n \,\mathrm{d} x \sim \frac{2^n}{n}.$$

score 1 · Answer 9 · answered Jun 01 '24 at 02:56

More than inspired by @Gary's answer $$I_n=\int_0^1 {(1 + x^2 )^n {\rm d}x} = 2^n \int_0^{\log( 2)} \frac{e^{-(n+1) t}}{\sqrt{2 e^{-t}-1}}\,dt$$ $$I_n=\frac{2^n }{2 n+1} \,\,_2F_1\left(1,-n;\frac{1}{2}-n;\frac{ 1}{2}\right)$$ $$\,_2F_1\left(1,-n;\frac{1}{2}-n;\frac{1}{2}\right)=2+\frac{1}{n}+\frac{2}{n^2}+O\left(\frac{1}{n^3}\right)$$ $$I_n \sim \frac {2 ^n}n \,\,\frac{2 n^2+n+2}{n (2 n+1)}=\frac {2 ^n}n \,\,\Bigg(1+\frac{1}{n^2}+O\left(\frac{1}{n^3}\right) \Bigg)$$

Why is $\int_{0}^{1}{(1+x^2)^n dx} \sim \frac{2^n}{n} $?

9 Answers9

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