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Math Help - Prove some identities! (Round two)

  1. #16
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by simplependulum View Post
     \int_0^1 \int_0^1 \frac{\ln^s(xy)}{xy - 1  }~dx dy

    Sub.  xy = u (  y being constant ) we have

     \int_0^1 \int_0^1 \frac{\ln^s(xy)}{xy - 1  }~dx~dy

     = \int_0^1 \int_0^y \frac{\ln^s(u)}{u-1} \frac{du}{y} ~dy

     = \int_0^1 \int_u^1 \frac{dy}{y}~ \frac{\ln^s(u)}{u-1}~du

     = - \int_0^1 \frac{\ln^{s+1}(u)}{u-1} ~du

    Sub.  u = e^{-t}

     = - \int_0^{\infty} \frac{(-1)^{s+1} t^{s+1} }{ 1 - e^t }~dt

     = (-1)^{s+1} \int_0^{\infty} \frac{t^{s+1}}{e^t - 1 } ~dt

     = (-1)^{s+1} \Gamma(s+2) \zeta(s+2)
    Right, it's not that much harder.
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  2. #17
    Super Member Random Variable's Avatar
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    I've come across the following identity multiple times, but I've never been able to prove it.

    Show  \displaystyle \int_{0}^{\infty} \frac{1}{1+x^{a}} \ dx = \frac{\pi}{a} \csc \Big(\frac{\pi}{a}}\Big) \ \text{for} \ a>1
    Last edited by Random Variable; December 31st 2010 at 11:44 AM.
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  3. #18
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Random Variable View Post
    I've come across the following identity multiple times, but I've never been able to proof it.

    Show  \displaystyle \int_{0}^{\infty} \frac{1}{1+x^{a}} \ dx = \frac{\pi}{a} \csc \Big(\frac{\pi}{a}}\Big) \ \text{for} \ a>0
    It should be a>1.


    Here is one method. There is another I know using contour integration if you're interested.


    Method 1:


    Spoiler:



    Let


    \displaystyle I(a)=\int_0^{\infty}\frac{dx}{1+x^a}



    Let 1+x^a=u and so du=a x^{a-1} or dx=\frac{1}{a}\left(u-1\right)^{\frac{1}{a}-1}. Thus,


    \displaystyle I(a)=\frac{1}{a}\int_1^{\infty}\frac{(u-1)^{\frac{1}{a}-1}}{u}\text{ }du


    Let z=\frac{1}{u} to get that


    \displaystyle \begin{aligned}I(a) &=\frac{1}{a}\int_0^1 z^{\frac{-1}{a}}\left(1-z\right)^{\frac{1}{a}-1}\\ &= \frac{1}{a}B\left(1-\frac{1}{a},\frac{1}{a}\right)\\ &= \frac{1}{a}\frac{\Gamma\left(\frac{1}{a}\right)\Ga  mma\left(1-\frac{1}{a}\right)}{\Gamma(1)}\\ &=\frac{1}{a}\Gamma\left(\frac{1}{a}\right)\Gamma\  left(1-\frac{1}{a}\right)\\ &= \frac{\pi}{a\sin\left(\frac{\pi}{a}\right)}\end{al  igned}

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  4. #19
    Super Member Random Variable's Avatar
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    Well, that was awfully simple.

    And yeah, the integral obviously diverges for  a \le 1

    simplependulum posted some problems on the bottom of the last page that you'll most likely find much more interesting.

    Quote Originally Posted by Drexel28 View Post
    It should be a>1.


    Here is one method. There is another I know using contour integration if you're interested.


    Method 1:


    Spoiler:



    Let


    \displaystyle I(a)=\int_0^{\infty}\frac{dx}{1+x^a}



    Let 1+x^a=u and so du=a x^{a-1} or dx=\frac{1}{a}\left(u-1\right)^{\frac{1}{a}-1}. Thus,


    \displaystyle I(a)=\frac{1}{a}\int_1^{\infty}\frac{(u-1)^{\frac{1}{a}-1}}{u}\text{ }du


    Let z=\frac{1}{u} to get that


    \displaystyle \begin{aligned}I(a) &=\frac{1}{a}\int_0^1 z^{\frac{-1}{a}}\left(1-z\right)^{\frac{1}{a}-1}\\ &= \frac{1}{a}B\left(1-\frac{1}{a},\frac{1}{a}\right)\\ &= \frac{1}{a}\frac{\Gamma\left(\frac{1}{a}\right)\Ga  mma\left(1-\frac{1}{a}\right)}{\Gamma(1)}\\ &=\frac{1}{a}\Gamma\left(\frac{1}{a}\right)\Gamma\  left(1-\frac{1}{a}\right)\\ &= \frac{\pi}{a\sin\left(\frac{\pi}{a}\right)}\end{al  igned}

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  5. #20
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Random Variable View Post
    Well, that was awfully simple.

    And yeah, the integral obviously diverges for  a \le 1

    simplependulum posted some problems on the bottom of the last page that you'll most likely find much more interesting.
    The one with the AGM is well known but a pain to verify, so I won't do that one. The other, I haven't been able to find a nice answer for it yet :S
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  6. #21
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Random Variable View Post
    What's required to switch the order of summation? Absolute convergence of the inner sum?
    Yes, this can be seen (for example) on page 202 of Apostol.
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  7. #22
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    Yes , AGM is one of the most famous identities , I have a proof of it but of course it is not due to me , I was inspired to give a complete proof by this problem from American Mathematical Monthly (AMM 6672 ) :


    Show that \displaystyle  I(a,b) = \int_0^{\pi/2} \frac{d\theta}{\sqrt{a^2\cos^2{\theta} + b^2 \sin^2{\theta} }} =  \int_0^{\pi/2} \frac{d\phi}{ \sqrt{(a\cos^2{\phi} + b\sin^2{\phi})(a\sin^2{\phi} + b\cos^2{\phi}) } }  . Then use this result to show that :


    \displaystyle   I(a,b) = I\left( \sqrt{ab} , \frac{a+b}{2} \right)



    Oh , I have made a very big mistake on the first problem , sorry about that

    The integrand should be squared :

     \displaystyle \int_0^1 \int_0^1 \int_0^1 \frac{dxdydz}{(1 + x^2 + y^2 + z^2)^2 }
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  8. #23
    Super Member Random Variable's Avatar
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    Show that  \displaystyle \int^{\infty}_{0} \frac{J_{1}(x)}{\sqrt{a^{2}+x^{2}}} \ dx = \frac{1- e^{-a}}{a}

    where  \displaystyle J_{1}(x) is the first order Bessel function of the first kind.



    The proof is fairly long, but simple and pretty cool at the same time.
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  9. #24
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    Quote Originally Posted by Random Variable View Post
    Show that  \displaystyle \int^{\infty}_{0} \frac{J_{1}(x)}{\sqrt{a^{2}+x^{2}}} \ dx = \frac{1- e^{-a}}{a}

    where  \displaystyle J_{1}(x) is the first order Bessel function of the first kind.



    The proof is fairly long, but simple and pretty cool at the same time.
    Not having heard about Bessel function , I thought I would not be able to solve this until I found this formula on a website :

     \displaystyle J_1(x) = \frac{1}{\pi} \int_0^{\pi} \cos[t - x\sin(t)]~dt

      \displaystyle = \frac{1}{\pi}  \int_0^{\pi} \left( \cos(t)\cos[x\sin(t)] + \sin(t)\sin[x\sin(t)] \right) ~dt

      \displaystyle = \frac{1}{\pi}  \int_0^{\pi} \sin(t)\sin[x\sin(t)] ~dt

    Therefore , your integral

      \displaystyle I =  \frac{1}{\pi} \int_0^{\infty} \int_0^{\pi} \frac{ \sin(t)\sin[x\sin(t)] }{\sqrt{a^2 + x^2} }~dtdx

    My method is to switch the order of the integral , followed by substitution , then switch back its order :

    Sub.   \displaystyle x\sin(t) = y we have

      \displaystyle I = \frac{1}{\pi} \int_0^{\pi} \int_0^{\infty} \frac{\sin(y)}{\sqrt{a^2 + y^2/\sin^2(t) }}~dydt

     \displaystyle  = \frac{1}{\pi} \int_0^{\pi} \int_0^{\infty} \frac{\sin(t)\sin(y)}{\sqrt{a^2 \sin^2(t) + y^2 }}~dydt

    Then switching back !

     \displaystyle  =  \frac{1}{\pi }\int_0^{\infty} \int_0^{\pi} \frac{\sin(t)\sin(y)}{\sqrt{ a^2 + y^2 - a^2 \cos^2(t) }}~dtdy

      \displaystyle = \frac{2}{a \pi} \int_0^{\infty} \sin(y) \sin^{-1}{\left( \frac{a}{\sqrt{a^2 + y^2 }}\right)}~dy

     \displaystyle  = \frac{2}{a \pi} \int_0^{\infty} \sin(y) \tan^{-1}(a/y) ~ dy

    Integration by parts ,

      \displaystyle I = \frac{2}{a \pi}  \left[- \cos(y)  \tan^{-1}(a/y) \right]_0^{\infty} - \frac{2}{\pi} \int_0^{\infty}  \frac{\cos(y)}{y^2 + a^2 }~dy

      \displaystyle = \frac{1}{a} - \frac{2}{\pi} ~ \frac{\pi}{2a} e^{-a}

      \displaystyle = \frac{1}{a} ( 1 - e^{-a} )
    Last edited by simplependulum; January 2nd 2011 at 01:30 AM.
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  10. #25
    Super Member Random Variable's Avatar
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    Changing the order of integration solely for the purpose of making a substitution? Where you do come up with these ideas?

    The solution I had in mind doesn't use the integral representation of the Bessel function.


    let  \displaystyle I(b) = \int^{\infty}_{0} \frac{b J_{1}(bx)}{\sqrt{a^{2}+x^{2}}} \ dx \ , b \ge 0


    then using  \displaystyle \frac{d}{dx} [x^{p} J_{p}(x)] = x^{p}J_{p-1}(x)


     \displaystyle I'(t) = \int^{\infty }_{0} \frac{bx J_{0}(bx)}{\sqrt{a^{2}+x^{2}}} \ dx =\int^{\infty}_{0} \frac{t J_{0}(t)}{\sqrt{c^{2}+t^{2}}} \ dt  where  \displaystyle t=bx and  \displaystyle c=ab


    By defintion  \displaystyle J_{0}(t) satisfies  \displaystyle tf''(t) + f'(t) + tf(t) = 0


    Now here comes the cool part.


    let   \displaystyle u(c) = \int^{\infty}_{0} \frac{t J_{0}(t)}<br />
{\sqrt{a^{2}+x^{2}}} \ dt

     \displaystyle v(c) = \int^{\infty}_{0} \frac{J^{'}_{0}(t)}<br />
{\sqrt{a^{2}+x^{2}}} \ dt

    and  \displaystyle w(c) = \int^{\infty}_{0} \frac{t J^{''}_{0}(t)}<br />
{\sqrt{a^{2}+x^{2}}} \ dt


    This directly implies that  \displaystyle u + v + w = 0 .

    Then using the facts that  \displaystyle J_{0}(0) = 1, \ J_{0}(\infty) = 0, \ \frac{d}{dx} [x^{-p} J_{p}(x)] = -x^{-p} J_{p+1}(x), \ J_{1}(0) = 0 and   \displaystyle J_{1} (\infty) = 0 , you'll find that \displaystyle  u'(c) = -1 - cv(c) and  \displaystyle v'(c) = \frac{w(c)}{c}

    Solve the system of equations simultaneously to get homogenous second order linear ODE  \displaystyle u''(c) = u which has the general solution  \displaystyle u(c) = Ae^{c} + Be^{-c}

     A must be zero because u(c) is a decreasing function, and  B must be 1 because Bessel functions are normalized.

     \displaystyle u(c) = e^{-c} = e^{-ab}

    but  \displaystyle I'(b) = u(c)

    so  I(b) = \frac{1-e^{-ab}}{a} ( since  I(0) = 0 )

    The integral we want is  I(1) = \frac{1-e^{-a}}{a}
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  11. #26
    Super Member Random Variable's Avatar
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    So that this thread remains active, try the following fairly easy problems:

     \displaystyle \text{Show that} \int_{0}^{\frac{\pi}{2}} \sin^{2n} x \ dx = \frac{\pi (2n)!}{2^{2n+1} (n!)^{2}} \ \text{for} \ n = 0,1,2,...

     \displaystyle \text{and} \int_{0}^{\frac{\pi}{2}} \sin^{2n+1} x \ dx = \frac{2^{2n} (n!)^{2}}{(2n+1)!} \ \text{for} \ n = 0,1,2,... \ .
    Last edited by Random Variable; January 3rd 2011 at 02:59 PM.
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  12. #27
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by Random Variable View Post
    So that this thread remains active, try the following fairly easy problems:

     \displaystyle \text{Show that} \int_{0}^{\frac{\pi}{2}} \sin^{2n} x \ dx = \frac{\pi (2n)!}{2^{2n+1} (n!)^{2}} \ \text{for} \ n = 0,1,2,...

     \displaystyle \text{and} \int_{0}^{\frac{\pi}{2}} \sin^{2n+1} x \ dx = \frac{2^{2n} (n!)^{2}}{(2n+1)!} \ \text{for} \ n = 0,1,2,... \ .
    Two words: Beta Function (by the way, what a powerful tool, huh?)

    Here's a nice one: compute  \displaystyle \sum_{n=1}^\infty \frac{\zeta(2n)}{n(n+1)}
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  13. #28
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by chiph588@ View Post
    Two words: Beta Function (by the way, what a powerful tool, huh?)

    Here's a nice one: compute  \displaystyle \sum_{n=1}^\infty \frac{\zeta(2n)}{n(n+1)}
    Is this really what you meant to put? I feel like this might not have a nice closed form considering that it's equivalent to integrating \displaystyle \int_0^1\frac{\log(\sin(\pi x))}{\sqrt{x}}\text{ }dx.
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  14. #29
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by Drexel28 View Post
    Is this really what you meant to put? I feel like this might not have a nice closed form considering that it's equivalent to integrating \displaystyle \int_0^1\frac{\log(\sin(\pi x))}{\sqrt{x}}\text{ }dx.
    Yes I did mean to put that. The integral you posted is a bit off though. You should get something similar and yes it's a hard one to compute.
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  15. #30
    Super Member Random Variable's Avatar
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    It would appear that the answer is just a multiple of that impossible-looking integral.

    Quote Originally Posted by chiph588@ View Post
    Yes I did mean to put that. The integral you posted is a bit off though. You should get something similar and yes it's a hard one to compute.
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