Results 1 to 8 of 8

Math Help - Simple integral

  1. #1
    Junior Member
    Joined
    Mar 2007
    From
    Missouri
    Posts
    54

    Simple integral

    Simple integral from 1 to infinity of (ln x)/(x^2).
    Follow Math Help Forum on Facebook and Google+

  2. #2
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by Possible actuary View Post
    Simple integral from 1 to infinity of (ln x)/(x^2).
    We proceed By Parts:

    Again, this is an improper integral

    lim{N-->infinity} int{lnx/x^2}dx from 1 to N = -x^-1*lnx + int{x^-1 * x^-1}dx
    .....................= lim{N--> infinity} -lnx/x - x^-1 + C
    .....................= lim{N-->infinity} -lnN/N - 1/N + ln1 + 1
    .....................= 0 + ln1 + 1
    .....................= ln1 + 1
    .....................= 1.................................since ln(1)=0 (thanks ecMathGeek)

    So i went through this kind of fast neglecting the finer details. ask away if you're unclear about something

    You know how to do integration by parts right?
    Last edited by Jhevon; March 23rd 2007 at 07:34 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member ecMathGeek's Avatar
    Joined
    Mar 2007
    Posts
    436
    Quote Originally Posted by Possible actuary View Post
    Simple integral from 1 to infinity of (ln x)/(x^2).
    Use Integration by Parts on this:
    INT (udv) = uv - INT(vdu) ... I'm hoping you're familiar with this.

    At the end, we will also need to use limits in order to integrate from 1 to infinity.

    Let u = lnx <--> du = 1/xdx
    Let dv = 1/x^2dx <--> v = -1/x


    So, INT lnx/x^2 dx becomes:
    -lnx/x - INT (-1/x^2)
    -lnx/x - 1/x
    -(lnx + 1)/x


    Now, plugging in the limits of integration, using a limit for the upper limit of infinity:

    lim[n->infinity] -(lnx + 1)/x from (1 to n)
    lim[n->infinity] -(ln(n) + 1)/n + (ln(1) + 1)/1


    Now we need to use L'Hopitals to simplify the limit

    lim[n->infinity] (-1/x)/1 + 1 = 0/1 + 1 = 1
    Follow Math Help Forum on Facebook and Google+

  4. #4
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by ecMathGeek View Post
    Use Integration by Parts on this:
    INT (udv) = uv - INT(vdu) ... I'm hoping you're familiar with this.

    At the end, we will also need to use limits in order to integrate from 1 to infinity.

    Let u = lnx <--> du = 1/xdx
    Let dv = 1/x^2dx <--> v = -1/x


    So, INT lnx/x^2 dx becomes:
    -lnx/x - INT (-1/x^2)
    -lnx/x - 1/x
    -(lnx + 1)/x


    Now, plugging in the limits of integration, using a limit for the upper limit of infinity:

    lim[n->infinity] -(lnx + 1)/x from (1 to n)
    lim[n->infinity] -(ln(n) + 1)/n + (ln(1) + 1)/1


    Now we need to use L'Hopitals to simplify the limit

    lim[n->infinity] (-1/x)/1 + 1 = 0/1 + 1 = 1
    i think you forgot the ln(1) at the end friend...or did i mess up again?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member ecMathGeek's Avatar
    Joined
    Mar 2007
    Posts
    436
    Quote Originally Posted by Jhevon View Post
    i think you forgot the ln(1) at the end friend...or did i mess up again?
    I was just checking over our work to see if either of us messed up. I'm not sure just yet, but in case you forgot, ln(1) = 0
    Follow Math Help Forum on Facebook and Google+

  6. #6
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by ecMathGeek View Post
    I was just checking over our work to see if either of us messed up. I'm not sure just yet, but in case you forgot, ln(1) = 0
    yup, you're right. i messed up again silly me
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Senior Member ecMathGeek's Avatar
    Joined
    Mar 2007
    Posts
    436
    Quote Originally Posted by Jhevon View Post
    i think you forgot the ln(1) at the end friend...or did i mess up again?
    You didn't state that you used L'Hopital's Rule on lim[n->infinity] of ln(n)/n. But besides that, our work is the same.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by ecMathGeek View Post
    You didn't state that you used L'Hopital's Rule on lim[n->infinity] of ln(n)/n. But besides that, our work is the same.
    Yeah, i kind of did it intuitively. i know that N increases a lot faster than ln(N) so the fraction would go to 0. but you're right, i should have done L'Hopital's for the benifit of the class
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. A simple integral
    Posted in the Calculus Forum
    Replies: 5
    Last Post: August 9th 2010, 08:30 AM
  2. simple integral
    Posted in the Calculus Forum
    Replies: 5
    Last Post: August 4th 2010, 09:09 PM
  3. Simple Integral Question (I think it's simple)
    Posted in the Calculus Forum
    Replies: 7
    Last Post: February 13th 2010, 02:37 PM
  4. Need help on simple Integral...
    Posted in the Calculus Forum
    Replies: 2
    Last Post: November 30th 2009, 06:55 AM
  5. Help with simple (i think...) integral
    Posted in the Calculus Forum
    Replies: 5
    Last Post: October 15th 2008, 06:33 PM

Search Tags


/mathhelpforum @mathhelpforum