Results 1 to 4 of 4
Like Tree1Thanks
  • 1 Post By GJA

Math Help - Limit of an integral

  1. #1
    Member
    Joined
    Aug 2009
    Posts
    78

    Limit of an integral

    Show that \lim_{n \to \infty} \int_0^1 \frac{x^2e^{x^n}}{1+x^3} dx=\frac{1}{3} \ln 2
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,899
    Thanks
    327
    Awards
    1

    Re: Limit of an integral

    Quote Originally Posted by problem View Post
    Show that \lim_{n \to \infty} \int_0^1 \frac{x^2e^{x^n}}{1+x^3} dx=\frac{1}{3} \ln 2
    I can't solve it anyway, but to clarify: Is e^{x^n}
    = e^{(x^n)}

    or
    = \left ( e^x \right ) ^n

    -Dan
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Aug 2009
    Posts
    78

    Re: Limit of an integral

    It is e^{(x^n)}.
    Last edited by problem; January 4th 2013 at 07:52 AM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    GJA
    GJA is offline
    Member
    Joined
    Jul 2012
    From
    USA
    Posts
    109
    Thanks
    29

    Re: Limit of an integral

    Hi problem,

    We can get at the exercise using the Lebesgue Dominated Convergence Theorem. Let f_{n}=\frac{x^{2}e^{x^{n}}}{1+x^{3}}, and note that f_{n}\geq 0 for all n so that |f_{n}|=f_{n} for all n.

    1) Since we're working on [0,1], x^{n}\leq 1 always. Hence, |f_{n}|=f_{n}\leq \frac{x^{2}e}{1+x^{3}}, which is integrable on [0,1]; i.e. \frac{x^{2}e}{1+x^{3}} is our "dominating function."

    2) Since we're working on  [0,1] , x^{n}\rightarrow 0 as n\rightarrow\infty for all x\neq 1. Furthermore, 1^{n}=1 for all n. Hence, the pointwise limit of the sequence \{f_{n}\} is f(x)=\frac{x^{2}}{1+x^{3}} for x\neq 1 and f(1)=\frac{e}{2}. Now the set \{1\} is of measure 0, so f(1)=\frac{e}{2} is actually irrelevant as far as the integral is concerned.

    Now it is a matter of checking to see if we've satisfied the conditions of Lebesgue's DCT, applying it, then evaluating the integral of the limit function properly to get what we're after.

    Does this get things going in the right direction? Let me know if anything is unclear. Good luck!
    Thanks from topsquark
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. limit with integral
    Posted in the Calculus Forum
    Replies: 1
    Last Post: September 3rd 2011, 08:47 AM
  2. Integral limit
    Posted in the Calculus Forum
    Replies: 5
    Last Post: July 13th 2011, 01:47 AM
  3. The Limit of Integrals Is the Integral of the Limit
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: August 2nd 2010, 03:51 PM
  4. definite integral/ limit of integral
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 22nd 2010, 04:00 AM
  5. Integral limit
    Posted in the Calculus Forum
    Replies: 5
    Last Post: January 17th 2009, 06:47 PM

Search Tags


/mathhelpforum @mathhelpforum