Find $\displaystyle \lim_{n\to\infty}\frac{n}{2^{n}}\int_{0}^{1}{x^{2n }\left( x^{2}+1 \right)^{n}\,dx}.$

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- May 3rd 2008, 05:25 PMKrizalidToday's calculation of integral #12
Find $\displaystyle \lim_{n\to\infty}\frac{n}{2^{n}}\int_{0}^{1}{x^{2n }\left( x^{2}+1 \right)^{n}\,dx}.$

- May 4th 2008, 04:56 AMDystopia
Let $\displaystyle I_{m,n} = \int^{1}_{0} x^{m} (x^{2} + 1)^{n} \; \mathrm{d}x$

Then by using integration by parts,

$\displaystyle I_{m,n} = \frac{2^{m}}{m+1} - \frac{2n}{m+1} I_{m+2,n-1}$

Setting m = 2n, and using this repeatedly gives an expression for the integral:

$\displaystyle I_{2n,n} = 2^{n}\left[\frac{1}{2n+1} - \frac{n}{(2n+1)(2n+3)} + \cdots + (-1)^{n} \frac{n!}{(2n+1)(2n+3)\cdots(4n+1)} \right]$

Where the general term is $\displaystyle (-1)^{r} \frac{n(n-1) \cdots (n+1-r)}{(2n+1)(2n+3) \cdots (2n + 2r + 1)}$

So $\displaystyle \lim_{n \to \infty} \frac{n}{2^{n}} \int^{1}_{0} x^{2n}(x^{2}+1)^{n}\;\mathrm{d}x = \frac{1}{2}-\frac{1}{4}+\frac{1}{8} - \cdots = \frac{1}{3}$ - May 4th 2008, 07:51 AMKrizalid
That's right.

I saw this problem in another forum yesterday. Here's another solution:

Let $\displaystyle \lambda =\int_{0}^{1}{x^{2n}\left( x^{2}+1 \right)^{n}\,dx}.$ Now let $\displaystyle x\to x^2$ and the integral becomes $\displaystyle \lambda =\frac{1}{2}\int_{0}^{1}{(1+x)^{n}x^{n-1/2}\,dx},$ let $\displaystyle x\to1-x$ and the integral becomes $\displaystyle \lambda =2^{n-1}\int_{0}^{1}{\bigg\{ 1-\frac{x}{2} \bigg\}^{n}(1-x)^{n-1/2}\,dx},$ let $\displaystyle x\to nx$ and the integral becomes $\displaystyle \lambda =\frac{2^{n-1}}{n}\int_{0}^{n}{\bigg\{ 1-\frac{x}{2n} \bigg\}^{n}\bigg\{ 1-\frac{x}{n} \bigg\}^{n-1/2}\,dx}.$

Now the limit becomes $\displaystyle \lim_{n\to\infty}\frac12\int_{0}^{n}{\bigg\{ 1-\frac{x}{2n} \bigg\}^{n}\bigg\{ 1-\frac{x}{n} \bigg\}^{n-1/2}\,dx}.$ Finally, by Dominated Convergence Theorem the limit equals $\displaystyle \frac{1}{2}\int_{0}^{\infty }{e^{-x/2}\cdot e^{-x}\,dx}=\frac{1}{2}\int_{0}^{\infty }{e^{-3x/2}\,dx}=\frac{1}{2}\cdot\frac{2}{3}=\frac{1}{3}.$ - May 4th 2008, 08:17 AMIsomorphism
Hey

**Krizalid**,

:eek:Dominated convergence theorem!!

What is this Dominated convergence theorem? (The wiki article is greek to me, with "measure spaces" (Crying))

Can you,please, state the conditions and the theorem and illustrate an interesting example case where this theorem is powerful?

Thank you,

Isomorphism

- May 4th 2008, 08:42 AMKrizalid
See in MathLinks forum, I learnt it from there.

- May 4th 2008, 09:09 AMIsomorphism
For my future reference:

Quote:

Originally Posted by**jmerry**

- May 4th 2008, 09:27 AMThePerfectHacker
It basically means if $\displaystyle f_n\to f$ you cannot pass the limit through because of possible pointwise convergence.

If $\displaystyle f_n\to f$ uniformly it is okay, but Krizalid also passes the limit to the $\displaystyle n$ term getting $\displaystyle \infty$ in the upper limit.