Page 5 of 7 FirstFirst 1234567 LastLast
Results 61 to 75 of 104

Math Help - MHF Integral Bee!

  1. #61
    Senior Member pankaj's Avatar
    Joined
    Jul 2008
    From
    New Delhi(India)
    Posts
    318
    Quote Originally Posted by NonCommAlg View Post
    3) \int_0^{\infty}\cot^{-1}(ax) \cot^{-1}(bx) \ dx, \ \ a > 0, \ b > 0
    Have fune!
    Please solve
    Follow Math Help Forum on Facebook and Google+

  2. #62
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7
    Quote Originally Posted by pankaj View Post
    Please solve
    okay! start from the identity \cot^{-1}(cx) = \int_c^{\infty} \frac{x}{x^2t^2 + 1} \ dt, \ x > 0, \ c > 0. therefore:

    \int_0^{\infty} \cot^{-1}(ax) \cot^{-1}(bx) \ dx = \int_0^{\infty} \int_a^{\infty} \int_b^{\infty} \frac{x^2}{(x^2y^2 + 1)(x^2z^2 + 1)} \ dy \ dz \ dx

    =\int_a^{\infty} \int_b^{\infty} \int_0^{\infty} \frac{x^2}{(x^2y^2 + 1)(x^2z^2 + 1)} \ dx \ dy \ dz=\frac{\pi}{2} \int_a^{\infty} \int_b^{\infty} \frac{1}{yz(y+z)} \ dy \ dz

    =\frac{\pi}{2} \int_a^{\infty} \frac{\ln(z+b) - \ln b}{z^2} \ dz = \frac{\pi}{2}\left[\frac{1}{a} \ln \left(\frac{a+b}{b} \right) + \frac{1}{b} \ln \left(\frac{a+b}{a} \right) \right].

    in the last integral i used integration by parts: \ln(z+b) - \ln b = u, \ \ \frac{1}{z^2} dz = dv.
    Follow Math Help Forum on Facebook and Google+

  3. #63
    Banned
    Joined
    Aug 2009
    Posts
    143
    Quote Originally Posted by NonCommAlg View Post
    okay! start from the identity \cot^{-1}(cx) = \int_c^{\infty} \frac{x}{x^2t^2 + 1} \ dt, \ x > 0, \ c > 0. therefore:

    \int_0^{\infty} \cot^{-1}(ax) \cot^{-1}(bx) \ dx = \int_0^{\infty} \int_a^{\infty} \int_b^{\infty} \frac{x^2}{(x^2y^2 + 1)(x^2z^2 + 1)} \ dy \ dz \ dx

    =\int_a^{\infty} \int_b^{\infty} \int_0^{\infty} \frac{x^2}{(x^2y^2 + 1)(x^2z^2 + 1)} \ dx \ dy \ dz=frac{\pi}{2}\;\int_a^{\infty}<br /> <br />
\frac{\pi}{2} \int_a^{\infty} \int_b^{\infty} \frac{1}{yz(y+z)} \ dy \ dz

    =\frac{\pi}{2} \int_a^{\infty} \frac{\ln(z+b) - \ln b}{z^2} \ dz = \frac{\pi}{2}\left[\frac{1}{a} \ln \left(\frac{a+b}{b} \right) + \frac{1}{b} \ln \left(\frac{a+b}{a} \right) \right].

    in the last integral i used integration by parts: \ln(z+b) - \ln b = u, \ \ \frac{1}{z^2} dz = dv.
    Here is another method to solve the integral.

    <br />
F(a,b)\;=\;\int_0^\infty \cot^{-1}(ax) \cot^{-1}(bx)\; dx<br />

    <br />
\frac{\partial^2 F}{\partial a \; \partial b}\;=\;\int_0^\infty \frac{x^2}{(a^2x^2+1)(b^2x^2+1)} \; dx\;=\frac{\pi}{2} \frac{1}{ab(a+b)}<br />

    <br />
\frac{\partial F}{\partial a}\;=\; \frac{\pi}{2a} \int_\infty^b \frac{1}{b(a+b)}\;db \;=\; \frac{\pi}{2a^2}\ln\frac{a}{a+b}<br />

    <br />
F(a,b)\;=\; \; \frac{\pi}{2} \int_\infty^a \frac{1}{a^2}\ln\frac{a}{a+b}\, da\;=\;\frac{\pi}{2}\left[\frac{1}{a} \ln \left(\frac{a+b}{b} \right) + \frac{1}{b} \ln \left(\frac{a+b}{a} \right) \right]<br />

    We integrate from \infty because when the integral is 0 when either  a\text{ or } b \text{ is } \infty
    Last edited by luobo; August 22nd 2009 at 02:34 PM.
    Follow Math Help Forum on Facebook and Google+

  4. #64
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7
    Quote Originally Posted by luobo View Post
    Here is a simpler and easier method.

    <br />
F(a,b)\;=\;\int_0^\infty \cot^{-1}(ax) \cot^{-1}(bx)\; dx<br />

    <br />
\frac{\partial^2 F}{\partial a \; \partial b}\;=\;\int_0^\infty \frac{ab}{(a^2x^2+1)(b^2x^2+1)} \; dx=\frac{\pi}{2} \frac{ab}{a+b}<br />

    Then integrate to get F(a,b) .
    i think your differentiation is not correct! anyway, these two methods are actually the same. the only difference is that i personally don't like differentiating with respect to a parameter in the

    integrand. i'm not sure why though! haha
    Follow Math Help Forum on Facebook and Google+

  5. #65
    Banned
    Joined
    Aug 2009
    Posts
    143
    Quote Originally Posted by NonCommAlg View Post
    i think your differentiation is not correct! anyway, these two methods are actually the same. the only difference is that i personally don't like differentiating with respect to a parameter in the

    integrand. i'm not sure why though! haha
    IT IS CORRECT. It turns out to be so simple that it may have surprised you! Check again.
    Follow Math Help Forum on Facebook and Google+

  6. #66
    Super Member Random Variable's Avatar
    Joined
    May 2009
    Posts
    959
    Thanks
    3
    I knew it could be solved by integrating under the integral sign. I just didn't think of integrating under the integral sign twice (which makes sense because there are two parameters).
    Follow Math Help Forum on Facebook and Google+

  7. #67
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7
    Quote Originally Posted by luobo View Post
    IT IS CORRECT. It turns out to be so simple that it may have surprised you! Check again.
    it's not correct! you check again! haha ... the correct one is this: <br />
\frac{\partial^2 F}{\partial a \; \partial b}\;=\;\int_0^\infty \frac{x^2}{(a^2x^2+1)(b^2x^2+1)} \; dx.
    Follow Math Help Forum on Facebook and Google+

  8. #68
    Banned
    Joined
    Aug 2009
    Posts
    143
    Quote Originally Posted by NonCommAlg View Post
    it's not correct! you check again! haha ... the correct one is this: <br />
\frac{\partial^2 F}{\partial a \; \partial b}\;=\;\int_0^\infty \frac{x^2}{(a^2x^2+1)(b^2x^2+1)} \; dx.
    You are talking about the integration procedure, rather than the method itself. Again, the result turns out to be very simple or even simpler in terms of mathematical explanation. The method is correct and simpler anyway (at least in my view, Hahahaha...). Different from you, I like this method better. You may have a different view, that's fine.

    I finally got some time to put other a detailed solution and here is the entire procedure:
    <br />
F(a,b)\;=\;\int_0^\infty \cot^{-1}(ax) \cot^{-1}(bx)\; dx<br />

    <br />
\frac{\partial^2 F}{\partial a \; \partial b}\;=\;\int_0^\infty \frac{x^2}{(a^2x^2+1)(b^2x^2+1)} \; dx\;=\frac{\pi}{2}\; \frac{1}{ab(a+b)}<br />

    <br />
\frac{\partial F}{\partial a}\;=\; \frac{\pi}{2a} \int_\infty^b \frac{1}{b(a+b)}\;db \;=\; \frac{\pi}{2a^2}\ln\frac{a}{a+b}<br />

    <br />
F(a,b)\;=\; \; \frac{\pi}{2} \int_\infty^a \frac{1}{a^2}\ln\frac{a}{a+b}\, da\;=\;\frac{\pi}{2}\left[\frac{1}{a} \ln \left(\frac{a+b}{b} \right) + \frac{1}{b} \ln \left(\frac{a+b}{a} \right) \right]<br />

    We integrate from \infty because when the integral is 0 when either  a\text{ or } b \text{ is } \infty
    Last edited by luobo; August 22nd 2009 at 03:46 PM.
    Follow Math Help Forum on Facebook and Google+

  9. #69
    Banned
    Joined
    Aug 2009
    Posts
    143
    Quote Originally Posted by Random Variable View Post
    I knew it could be solved by integrating under the integral sign. I just didn't think of integrating under the integral sign twice (which makes sense because there are two parameters).
    It happens sometimes. Some times you can even artificially introduce a variable to solve an integral. I bet you know that
    Follow Math Help Forum on Facebook and Google+

  10. #70
    Super Member Random Variable's Avatar
    Joined
    May 2009
    Posts
    959
    Thanks
    3
    It's not my turn, but no new integral has been posted. So I'm going to throw one out there.

     \int^{\infty}_{0} \frac{e^{-ax}-e^{-bx}}{x} \ dx \ \ (a, b >0)
    Follow Math Help Forum on Facebook and Google+

  11. #71
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    13
    That + is actually a -. The solution is quite easy by noting that the integrand equals \int_a^be^{-xt}\,dt.

    (I see you just edited the problem.)
    Follow Math Help Forum on Facebook and Google+

  12. #72
    Super Member
    Joined
    Jan 2009
    Posts
    715
    Quote Originally Posted by Random Variable View Post
    It's not my turn, but no new integral has been posted. So I'm going to throw one out there.

     \int^{\infty}_{0} \frac{e^{-ax}-e^{-bx}}{x} \ dx \ \ (a, b >0)
    Is this method OK?

    we know  \int_0^{\infty} e^{-st} t^{n}~dt = \frac{\Gamma(n+1)}{s^{n+1}}

    so the integral equals to  \frac{ \Gamma(n) }{ a^n} - \frac{\Gamma(n)}{ b^n}  n \to 0

     = \lim_{n\to 0} \frac{ b^n - a^n}{n}
    Follow Math Help Forum on Facebook and Google+

  13. #73
    Super Member
    Joined
    Jan 2009
    Posts
    715
    NonCommAlg :

    I am very sure you have a lot of nice integrals , but not sure if you wanna share
    Follow Math Help Forum on Facebook and Google+

  14. #74
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    13
    Try this one:

    Evaluate \lim_{n\to\infty}\int_0^{\frac\pi3}\frac{\sin^nx}{  \sin^nx+\cos^nx}\,dx.
    Follow Math Help Forum on Facebook and Google+

  15. #75
    Super Member Random Variable's Avatar
    Joined
    May 2009
    Posts
    959
    Thanks
    3
    Quote Originally Posted by simplependulum View Post
    Is this method OK?

    we know  \int_0^{\infty} e^{-st} t^{n}~dt = \frac{\Gamma(n+1)}{s^{n+1}}

    so the integral equals to  \frac{ \Gamma(n) }{ a^n} - \frac{\Gamma(n)}{ b^n}  n \to 0

     = \lim_{n\to 0} \frac{ b^n - a^n}{n}
    I did it the way Krizalid suggested.

    But I guess your way is correct too since  \lim_{n \to 0} \frac{b^{n}-a^{n}} {n} = \ln(b) - \ln(a) = \ln(b/a)
    Follow Math Help Forum on Facebook and Google+

Page 5 of 7 FirstFirst 1234567 LastLast

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: August 31st 2010, 07:38 AM
  2. Replies: 1
    Last Post: June 2nd 2010, 02:25 AM
  3. Replies: 0
    Last Post: May 9th 2010, 01:52 PM
  4. Replies: 0
    Last Post: September 10th 2008, 07:53 PM
  5. Replies: 6
    Last Post: May 18th 2008, 06:37 AM

Search Tags


/mathhelpforum @mathhelpforum