Today's calculation of integral #12

**Krizalid** · May 3rd 2008, 05:25 PM

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

**Dystopia** · May 4th 2008, 04:56 AM

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 $\lim_{n \to \infty} \frac{n}{2^{n}} \int^{1}_{0} x^{2n}(x^{2}+1)^{n}\;\mathrm{d}x = \frac{1}{2}-\frac{1}{4}+\frac{1}{8} - \cdots = \frac{1}{3}$

**Krizalid** · May 4th 2008, 07:51 AM

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 $\frac{1}{2}\int_{0}^{\infty }{e^{-x/2}\cdot e^{-x}\,dx}=\frac{1}{2}\int_{0}^{\infty }{e^{-3x/2}\,dx}=\frac{1}{2}\cdot\frac{2}{3}=\frac{1}{3}.$

**Isomorphism** · May 4th 2008, 08:17 AM

Originally Posted by Krizalid

[snip]

Finally, by Dominated Convergence Theorem the limit equals

[/snip]

Hey Krizalid,

Dominated convergence theorem!!

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

)

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

Thank you,
Isomorphism

**Krizalid** · May 4th 2008, 08:42 AM

See in MathLinks forum, I learnt it from there.

**Isomorphism** · May 4th 2008, 09:09 AM

For my future reference:

Originally Posted by jmerry

Statement of the DCT: Let

tend pointwise to

, with each

absolutely integrable. If we also have

with

(absolutely) integrable,

.
That could be almost any kind of integral, including infinite sums. It's also closely related to Fubini's theorem. In the usual Lebesgue context, we relax "pointwise" to "pointwise almost everywhere". For improper Riemann integrals, that's "pointwise except on a discrete set".
The idea of the proof is to look at

, and split this into two parts. On one part, we make it small with the pointwise convergence. On the other, we make it small by comparing to

and making the region small.

**ThePerfectHacker** · May 4th 2008, 09:27 AM

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.

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