# Today's calculation of integral #12

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• May 3rd 2008, 05:25 PM
Krizalid
Today'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 AM
Dystopia
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 AM
Krizalid
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 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
Statement of the DCT: Let http://alt1.mathlinks.ro/Forum/latex...1d97597c41.gif tend pointwise to http://alt1.mathlinks.ro/Forum/latex...af9d98f0f5.gif, with each http://alt1.mathlinks.ro/Forum/latex...1d97597c41.gif absolutely integrable. If we also have http://alt1.mathlinks.ro/Forum/latex...a59512d902.gif with http://alt1.mathlinks.ro/Forum/latex...6e79e2241b.gif (absolutely) integrable, http://alt2.mathlinks.ro/Forum/latex...a3d3013194.gif.
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 http://alt1.mathlinks.ro/Forum/latex...b3a187a868.gif, 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 http://alt1.mathlinks.ro/Forum/latex...6e79e2241b.gif and making the region small.

• May 4th 2008, 09:27 AM
ThePerfectHacker
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.