Results 1 to 7 of 7

Math Help - Today's calculation of integral #12

  1. #1
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    12

    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}.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Newbie
    Joined
    Apr 2008
    Posts
    12
    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}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    12
    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}.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Lord of certain Rings
    Isomorphism's Avatar
    Joined
    Dec 2007
    From
    IISc, Bangalore
    Posts
    1,465
    Thanks
    6
    Quote Originally Posted by Krizalid View Post
    [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


    Follow Math Help Forum on Facebook and Google+

  5. #5
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    12
    See in MathLinks forum, I learnt it from there.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Lord of certain Rings
    Isomorphism's Avatar
    Joined
    Dec 2007
    From
    IISc, Bangalore
    Posts
    1,465
    Thanks
    6
    For my future reference:

    Quote 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.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    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.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Today's calculation of integral #14
    Posted in the Calculus Forum
    Replies: 8
    Last Post: May 13th 2008, 03:42 PM
  2. Today's calculation of integral #2
    Posted in the Calculus Forum
    Replies: 3
    Last Post: April 1st 2008, 12:03 PM
  3. Today's calculation of integral #4
    Posted in the Calculus Forum
    Replies: 2
    Last Post: April 1st 2008, 03:48 AM
  4. Today's calculation of integral #3
    Posted in the Calculus Forum
    Replies: 5
    Last Post: March 31st 2008, 06:13 PM
  5. Today's calculation of integral #1
    Posted in the Calculus Forum
    Replies: 3
    Last Post: March 31st 2008, 05:12 PM

Search Tags


/mathhelpforum @mathhelpforum