# Thread: Today's calculation of integral #12

1. ## Today's calculation of integral #12

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

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

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

4. 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

5. See in MathLinks forum, I learnt it from there.

6. 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.

7. 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.