# Today's calculation of integral #12

• May 3rd 2008, 05:25 PM
Krizalid
Today's calculation of integral #12
Find $\lim_{n\to\infty}\frac{n}{2^{n}}\int_{0}^{1}{x^{2n }\left( x^{2}+1 \right)^{n}\,dx}.$
• May 4th 2008, 04:56 AM
Dystopia
Let $I_{m,n} = \int^{1}_{0} x^{m} (x^{2} + 1)^{n} \; \mathrm{d}x$

Then by using integration by parts,

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

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

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

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

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

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

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

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

Originally Posted by Krizalid
[snip]

Finally, by Dominated Convergence Theorem the limit equals

[/snip]

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 AM
Krizalid
See in MathLinks forum, I learnt it from there.
• May 4th 2008, 09:09 AM
Isomorphism
For my future reference:

Quote:

Originally Posted by jmerry
It basically means if $f_n\to f$ you cannot pass the limit through because of possible pointwise convergence.
If $f_n\to f$ uniformly it is okay, but Krizalid also passes the limit to the $n$ term getting $\infty$ in the upper limit.