Page 1 of 2 12 LastLast
Results 1 to 15 of 16

Math Help - Challenging improper integral

  1. #1
    Junior Member
    Joined
    Jun 2008
    Posts
    54

    Challenging improper integral

    Prove that

     \int_0^{\infty}x(exp(3x)-1)^{-1/3} dx <br />
=(ln(3)+ \pi/3^{1.5})*\pi/3^{1.5}<br />
    Last edited by mathwizard; June 14th 2008 at 11:48 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Quote Originally Posted by mathwizard View Post
    Prove that

    ∫dx*x*(exp(3x) 1)^(-1/3) from x=0 to infinity =(ln(3)+ pi/3^1.5)*pi/3^1.5
    Considering the integral I would assume the rate of convergence would be very high, and numerically

    \int_0^{5000}\frac{x}{(e^{3x}-1)^{\frac{1}{3}}}dx\ne\bigg(\ln(3)+\frac{\pi}{3^{\  frac{3}{2}}}\bigg)\frac{\pi}{3^{\frac{3}{2}}}

    are you sure this is correct? I am sure you are but I am just wondering
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    I just glanced at it. Gotta go for a little while. But, maybe, just maybe, we can transform it into a beta function form and go from there.

    B(p,q)=\int_{0}^{\infty}\frac{y^{p-1}}{(1+y)^{p+q}}dy

    With the proper subs, this may be doable. Give it a whirl.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7
    Quote Originally Posted by mathwizard View Post
    Prove that

     I=\int_0^{\infty}\frac{x}{\sqrt[3]{e^{3x}-1}} dx <br />
=\frac{\pi}{3\sqrt{3}}\left(\ln 3 + \frac{\pi}{3\sqrt{3}}\right)<br />
    first two simple facts that we will use them later on in the solution:

    (*) if y > 0, then \int_0^{\infty}\frac{dt}{t^3+y}=\frac{1}{\sqrt[3]{y^2}} \int_0^{\infty}\frac{dt}{1+t^3}, and \int_0^{\infty} \frac{dt}{1+t^3} = \frac{2\pi}{3\sqrt{3}}.

    now, in your integral, put e^{3x}-1=\frac{1}{t^3}. then x=\frac{1}{3}\ln\left(1+\frac{1}{t^3}\right), and dx=\frac{-dt}{t(1+t^3)}. thus:

    I=\frac{1}{3}\int_0^{\infty}\frac{\ln(1+\frac{1}{t  ^3})}{1+t^3} \ dt=\frac{1}{3}\int_0^{\infty} \int_0^1\frac{1}{(t^3+y)(t^3+1)} \ dy \ dt

    =\frac{1}{3} \int_0^1 \int_0^{\infty} \frac{1}{(t^3+y)(t^3+1)} \ dt \ dy=\frac{1}{3}\int_0^1 \frac{1}{1-y}\int_0^{\infty} \left(\frac{1}{t^3+y}-\frac{1}{t^3+1}\right) \ dt \ dy. thus by (*):

    I=\frac{2\pi}{9\sqrt{3}}\int_0^1 \frac{1-\sqrt[3]{y^2}}{\sqrt[3]{y^2}(1-y)} \ dy. now let y=z^3. then we will have:

    I=\frac{2\pi}{3\sqrt{3}} \int_0^1 \frac{1+z}{1+z+z^2} \ dz=\frac{2\pi}{3\sqrt{3}}\left[\frac{1}{\sqrt{3}}\tan^{-1}\left(\frac{2z+1}{\sqrt{3}}\right)+\frac{1}{2}\l  n(1+z+z^2) \right]_ 0^1

    =\frac{\pi}{3\sqrt{3}}\left(\ln 3 + \frac{\pi}{3\sqrt{3}} \right). \ \ \ \ \square
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    NCA, you're just a show off

    Beautiful solution. Do you think we could manipulate it somehow and use the Beta as I mentioned?. I gave it a short go, but failed to arrive at anything tangible. You probably can though.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Quote Originally Posted by NonCommAlg View Post
    first two simple facts that we will use them later on in the solution:

    (*) if y > 0, then \int_0^{\infty}\frac{dt}{t^3+y}=\frac{1}{\sqrt[3]{y^2}} \int_0^{\infty}\frac{dt}{1+t^3}, and \int_0^{\infty} \frac{dt}{1+t^3} = \frac{2\pi}{3\sqrt{3}}.

    now, in your integral, put e^{3x}-1=\frac{1}{t^3}. then x=\frac{1}{3}\ln\left(1+\frac{1}{t^3}\right), and dx=\frac{-dt}{t(1+t^3)}. thus:

    I=\frac{1}{3}\int_0^{\infty}\frac{\ln(1+\frac{1}{t  ^3})}{1+t^3} \ dt=\frac{1}{3}\int_0^{\infty} \int_0^1\frac{1}{(t^3+y)(t^3+1)} \ dy \ dt

    =\frac{1}{3} \int_0^1 \int_0^{\infty} \frac{1}{(t^3+y)(t^3+1)} \ dt \ dy=\frac{1}{3}\int_0^1 \frac{1}{1-y}\int_0^{\infty} \left(\frac{1}{t^3+y}-\frac{1}{t^3+1}\right) \ dt \ dy. thus by (*):

    I=\frac{2\pi}{9\sqrt{3}}\int_0^1 \frac{1-\sqrt[3]{y^2}}{\sqrt[3]{y^2}(1-y)} \ dy. now let y=z^3. then we will have:

    I=\frac{2\pi}{3\sqrt{3}} \int_0^1 \frac{1+z}{1+z+z^2} \ dz=\frac{2\pi}{3\sqrt{3}}\left[\frac{1}{\sqrt{3}}\tan^{-1}\left(\frac{2z+1}{\sqrt{3}}\right)+\frac{1}{2}\l  n(1+z+z^2) \right]_ 0^1

    =\frac{\pi}{3\sqrt{3}}\left(\ln 3 + \frac{\pi}{3\sqrt{3}} \right). \ \ \ \ \square
    How hard is it to do difficult definite integrals prior to learning double and triple integration, because it seems very hard
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    10,212
    Thanks
    419
    Awards
    1
    Quote Originally Posted by Mathstud28 View Post
    Considering the integral I would assume the rate of convergence would be very high, and numerically

    \int_0^{5000}\frac{x}{(e^{3x}-1)^{\frac{1}{3}}}dx\ne\bigg(\ln(3)+\frac{\pi}{3^{\  frac{3}{2}}}\bigg)\frac{\pi}{3^{\frac{3}{2}}}

    are you sure this is correct? I am sure you are but I am just wondering
    The difference between the two is only 4.5 \times 10^{-10}~\%! Why would you say this is so bad an estimate??

    -Dan
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Quote Originally Posted by topsquark View Post
    The difference between the two is only 4.5 \times 10^{-10}~\%! Why would you say this is so bad an estimate??

    -Dan
    My 89 gives .18277 and 1.02976?
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    14
    Quote Originally Posted by Mathstud28 View Post

    How hard is it to do difficult definite integrals prior to learning double and triple integration, because it seems very hard
    It's not that hard, it's about practice, lot of practice and learn everyday parameters to construct double integrals.

    I like this way to tackle problems like this one, besides being a nice way to do it, you simplify a lot nasty calculations.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Quote Originally Posted by Krizalid View Post
    It's not that hard, it's about practice, lot of practice and learn everyday parameters to construct double integrals.

    I like this way to tackle problems like this one, besides being a nice way to do it, you simplify a lot nasty calculations.
    By everday paramaters do you mean

    \arctan(x)=\int_0^x\frac{dy}{1+y^2}

    or

    \ln(x)=\int_1^x\frac{dy}{y}

    ?
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    14
    Quote Originally Posted by Mathstud28 View Post

    By everday paramaters do you mean

    \arctan(x)=\int_0^x\frac{dy}{1+y^2}

    or

    \ln(x)=\int_1^x\frac{dy}{y}

    ?
    Yeah, but those are basic ones, you have to find/learn lots of them. (Of course, it depends of your integration skills how to use them to tackle a problem.)
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    10,212
    Thanks
    419
    Awards
    1
    Quote Originally Posted by Mathstud28 View Post
    My 89 gives .18277 and 1.02976?
    \int_0^{5000}\frac{x}{(e^{3x}-1)^{\frac{1}{3}}}dx = 1.0297616606482

    \bigg(\ln(3)+\frac{\pi}{3^{\frac{3}{2}}}\bigg)\fra  c{\pi}{3^{\frac{3}{2}}}  = 1.0297616606528

    For a difference of 4.6 \times 10^{-12} (or so.)

    -Dan
    Follow Math Help Forum on Facebook and Google+

  13. #13
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Quote Originally Posted by topsquark View Post
    \int_0^{5000}\frac{x}{(e^{3x}-1)^{\frac{1}{3}}}dx = 1.0297616606482

    \bigg(\ln(3)+\frac{\pi}{3^{\frac{3}{2}}}\bigg)\fra  c{\pi}{3^{\frac{3}{2}}} = 1.0297616606528

    For a difference of 4.6 \times 10^{-12} (or so.)

    -Dan
    ::grumbles:: stupid 89
    Follow Math Help Forum on Facebook and Google+

  14. #14
    MHF Contributor Reckoner's Avatar
    Joined
    May 2008
    From
    Baltimore, MD (USA)
    Posts
    1,024
    Thanks
    75
    Awards
    1

    Smile

    Quote Originally Posted by Mathstud28 View Post
    ::grumbles:: stupid 89
    Are you sure you put it in right? My 89 gave me the correct (approximate) answer. Perhaps you forgot the parentheses around the \frac13 or the 3x?

    Edit: Ah, it looks like you forgot to take the cube root altogether. Removing that I get the same answer.
    Last edited by Reckoner; June 14th 2008 at 07:37 PM.
    Follow Math Help Forum on Facebook and Google+

  15. #15
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by Reckoner View Post
    Are you sure you put it in right? My 89 gave me the correct (approximate) answer. Perhaps you forgot the parentheses around the \frac13 or the 3x?
    It's like I always say .... A calculator is very, very stupid and must be treated as such ..... It always does exactly what you tell it to do; but this may not always be what you want it to do.
    Follow Math Help Forum on Facebook and Google+

Page 1 of 2 12 LastLast

Similar Math Help Forum Discussions

  1. A particularly challenging integral
    Posted in the Calculus Forum
    Replies: 2
    Last Post: June 21st 2010, 11:06 PM
  2. another challenging integral
    Posted in the Calculus Forum
    Replies: 10
    Last Post: April 20th 2010, 12:10 AM
  3. a challenging integral (perhaps)
    Posted in the Math Challenge Problems Forum
    Replies: 6
    Last Post: March 12th 2010, 10:03 AM
  4. Challenging Integral
    Posted in the Calculus Forum
    Replies: 5
    Last Post: May 9th 2009, 06:04 AM
  5. Challenging Integral?.
    Posted in the Calculus Forum
    Replies: 3
    Last Post: June 7th 2008, 10:56 AM

Search Tags


/mathhelpforum @mathhelpforum