Results 1 to 5 of 5

Math Help - new challenging integrals

  1. #1
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1

    new challenging integrals

    Here's two integrals I ran into I thought y'all might like a go. Perhaps they are cliche?.

    Show that:

    This one has its own name. The Ahmed Integral.

    \int_{0}^{1}\frac{tan^{-1}}{x(x^{2}+1)}dx=\frac{\text{Catalan}}{2}+\frac{\  pi}{8}ln(2)

    In case, \text{Catalan}=\sum_{k=0}^{\infty}\frac{(-1)^{k}}{(2k+1)^{2}}\approx .915....

    The other:

    \int_{0}^{\infty}\frac{sin(x)}{x^{p}}dx=\frac{\sqr  t{\pi}{\Gamma}(1-\frac{p}{2})}{2^{p}{\Gamma}(\frac{1}{2}+\frac{p}{2  })}
    or some equivalent form.

    I thought these were cool.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    14
    Quote Originally Posted by galactus View Post

    Show that:

    This one has its own name. The Ahmed Integral.

    \int_{0}^{1}\frac{tan^{-1}}{x(x^{2}+1)}dx=\frac{\text{Catalan}}{2}+\frac{\  pi}{8}ln(2)

    In case, \text{Catalan}=\sum_{k=0}^{\infty}\frac{(-1)^{k}}{(2k+1)^{2}}\approx .915....
    Your integral equals \frac{1}{2}\left( \int_{0}^{1}{\frac{\ln \left( 1+{{y}^{2}} \right)}{{{y}^{2}}+1}\,dy}-\ln 2\int_{0}^{1}{\frac{dy}{{{y}^{2}}+1}} \right)-\int_{0}^{1}{\frac{\ln y}{{{y}^{2}}+1}\,dy}

    after some double integration manipulations. Those integrals are not hard to find, from there one gets the result.
    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
    Hey Kriz:

    You are better at this than me. That's for sure. I have to ask. How in the world did you get that ln thing from that integral with arctan?.

    I can do each of those integrals you have, but I never saw it in terms of ln.

    I started out using the series for arctan, but the x^2+1 in there caused me a fit.

    Using tan^{-1}(x)=\sum_{k=0}^{\infty}\frac{(-1)^{k}x^{2k+1}}{2k+1}.

    \int_{0}^{1}\frac{1}{x^{2}+1}\sum_{k=0}^{\infty}\f  rac{(-1)^{k}x^{2k}}{2k+1}

    \sum_{k=0}^{\infty}\frac{(-1)^{k}}{2k+1}\int_{0}^{1}\frac{x^{2k}}{x^{2}+1}..............[3]

    I can see \int x^{2k} dx=\frac{x^{2k+1}}{2k+1}, which

    when multiplied by the existing 2k+1 will result in the Catalan, but I am

    getting hung up because of the x^2+1.

    Now, \sum_{k=0}^{\infty}(-1)^{k}x^{2k}=\frac{1}{1+x^{2}}.

    Then we can write \int_{0}^{1}\sum_{k=0}^{\infty}(-1)^{k}x^{2k}\cdot x^{2k}

    \int_{0}^{1}x^{2k}\sum_{k=0}^{\infty}(-1)^{k}x^{2k}

    When we integrate, we have \frac{x^{2k+1}}{2k+1}\sum_{k=0}^{\infty}(-1)^{k}x^{2k}

    When multiplied with [3], we get the Catalan.

    But what about the rest?. I know it;s there. I can smell it.

    I am making a stupid mistake. I just know it. Every time I tried this thing I ended up in a dead end.

    You probably see it right off, though.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    14
    \int_{0}^{1}{\frac{\arctan x}{x\left( {{x}^{2}}+1 \right)}}\,dx=\int_{0}^{1}{\int_{0}^{x}{\frac{dy\,  dx}{x\left( {{x}^{2}}+1 \right)\left( {{y}^{2}}+1 \right)}}}=\int_{0}^{1}{\int_{y}^{1}{\frac{dx\,dy}  {x\left( {{x}^{2}}+1 \right)\left( {{y}^{2}}+1 \right)}}}.

    That leads the integrals I left before.
    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
    Oh, OK. Good egg. I see now what you done. Clever indeed.

    I also know that \int_{0}^{\frac{\pi}{4}}ln(1+tan(x))dx=\frac{\pi}{  8}ln(2)

    Which is part of our solution. May have to look into how to transform it. I bet there is a connection.

    Also,this integral becomes \int_{0}^{\frac{\pi}{4}}xcot(x)dx, but the Pi/4 makes it more difficult than the usual Pi/2 for this integral.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Integral #4 (and hopefully a bit more challenging)
    Posted in the Math Challenge Problems Forum
    Replies: 11
    Last Post: May 14th 2010, 10:58 AM
  2. Challenging Geometry...
    Posted in the Geometry Forum
    Replies: 1
    Last Post: May 9th 2010, 01:20 PM
  3. Challenging Problem
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: August 15th 2009, 10:53 AM
  4. Replies: 2
    Last Post: May 19th 2008, 07:09 AM
  5. Challenging Dif EQ
    Posted in the Calculus Forum
    Replies: 3
    Last Post: December 11th 2007, 05:01 PM

Search Tags


/mathhelpforum @mathhelpforum