Results 1 to 5 of 5

Thread: Hard Integration Q

  1. #1
    Junior Member
    Joined
    Sep 2009
    Posts
    56

    Hard Integration Q

    Consider the definite integral $\displaystyle D=\int_{-\pi}^{\pi}\cos\lambda x\cos nx\,dx$ where n is a positive integer and lamda is a positive real number.

    Given also that $\displaystyle D=\left\{\begin{matrix}
    \pi & $for$\,\lambda =n\\
    0 & $for$\,\lambda \, $a positive integer$\,,\lambda \neq n\\
    \frac{(-1)^n2\lambda \sin(\lambda \pi )}{\lambda ^2-n^2} & $for$\,\lambda \,$not an integer$
    \end{matrix}\right.$

    Show that when $\displaystyle 0< \lambda < n,\,|D|< \pi$ and when $\displaystyle \lambda > n+\frac{1}{2},\,|D|< 3$

    Help of any kind would be greatly appreciated.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member
    Joined
    Sep 2009
    Posts
    56
    Would the first part of the question have something to do with limits? We can see that $\displaystyle 0<|\sin(\lambda \pi )|<1$ and so the maximum value for the expression is obviously the limit as lamda approaches n since $\displaystyle \frac{2\lambda}{\lambda ^2-n^2}$ approaches infinite. However, im not sure how to evaluate the limit of the entire expression $\displaystyle \frac{(-1)^n2\lambda \sin(\lambda \pi )}{\lambda ^2-n^2}$.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    May 2008
    From
    Melbourne Australia
    Posts
    217
    Thanks
    29
    I'll attempt the second half.
    Let $\displaystyle \lambda > n+0.5$

    then

    $\displaystyle \lambda^2-n^2 > (n+.5)^2-n^2=n+.25$

    therefore,

    $\displaystyle |\frac{2\lambda \sin(\lambda \pi )}{\lambda ^2-n^2}|<
    |\frac{2(n+.5) \sin(\lambda \pi )}{n+.25}|\leq
    |\frac{2(n+.5) }{n+.25}|$

    Now
    $\displaystyle
    \frac{(n+.5) }{n+.25}$ is a strictly decreasing function and n is an integer greater than or equal to 1, so it is maximum for n=1 or:

    $\displaystyle |\frac{2\lambda \sin(\lambda \pi )}{\lambda ^2-n^2}|<
    \frac{2(1+.5) }{1+.25}<2.4<3$

    You could answer part 1 by assuming $\displaystyle \lambda < n-0.5$ and following the same procedure, then show that D is increasing on the range $\displaystyle n-.5<\lambda\leq n$ and therefore is less than D at n = lambda.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Sep 2009
    Posts
    56
    i think you made a mistake on the third line because: $\displaystyle \lambda ^2-n^2>n+0.25\Rightarrow\frac{2}{\lambda ^2-n^2}<\frac{2}{n+0.25}$ and $\displaystyle \lambda >n+0.5$
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    May 2008
    From
    Melbourne Australia
    Posts
    217
    Thanks
    29

    Not solved - inequality problem

    Oops sorry. If you find a solution can you please post it, this is really bugging me!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Hard Integration Q
    Posted in the Calculus Forum
    Replies: 8
    Last Post: Apr 1st 2010, 09:26 PM
  2. Hard Integration help
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Aug 3rd 2009, 04:34 AM
  3. Integration!!!! Hard!!!
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Feb 5th 2009, 05:18 PM
  4. HARD Integration
    Posted in the Calculus Forum
    Replies: 8
    Last Post: Nov 30th 2008, 08:49 PM
  5. hard integration
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Mar 27th 2008, 04:56 PM

Search Tags


/mathhelpforum @mathhelpforum