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Math Help - Prove some identities! (Round two)

  1. #31
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Random Variable View Post
    It would appear that the answer is just a multiple of that impossible-looking integral.
    I'm a little busy now, but I'll think about it. The methodology is to note that \displaystyle \sin\left(\pi\sqrt{z}\right)=\pi\sqrt{z}\prod_{n=1  }^{\infty}\left(1-\frac{z}{n^2}\right) and thus \displaystyle \log\left(\sin\left(\pi\sqrt{z}\right)\right)-\log\left(\pi\sqrt{z}\right)=\sum_{n=1}^{\infty}\l  og\left(1-\frac{z}{n^2}\right) but expanding the logarithm gives \displaystyle \log\left(\sin\left(\pi\sqrt{z}\right)\right)-\log\left(\pi\sqrt{z}\right)\right)=-\sum_{n=1}^{\infty}\sum_{m=1}^{\infty}\frac{z^m}{n  ^{2m}m} doing the quick verification that the inner sum can be interchanged this implies that \displaystyle \log\left(\sin\left(\pi\sqrt{z}\right)\right)-\log\left(\pi\sqrt{z}\right)\right)=\sum_{m=1}^{\i  nfty}\frac{\zeta(2m)}{m}z^m. Integrating both sides from 0 to 1 etc. etc
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  2. #32
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by Random Variable View Post
    It would appear that the answer is just a multiple of that impossible-looking integral.
     \displaystyle \log(\pi x) - \log(\sin\pi x) = \sum_{n=1}^\infty \frac{\zeta(2n)}{n}x^{2n} \implies \log(\pi \sqrt{x}) - \log(\sin(\pi \sqrt{x})) = \sum_{n=1}^\infty \frac{\zeta(2n)}{n}x^{n}  \displaystyle \implies \sum_{n=1}^\infty \frac{\zeta(2n)}{n(n+1)} = \int_0^1 \log(\pi\sqrt{x}) dx - \int_0^1 \log(\sin(\pi \sqrt{x})) dx

    For the second integral, let  t=\pi\sqrt{x} \implies dt = \frac\pi{2\sqrt{x}}dx \implies dx = \frac2{\pi^2} t dt

    Thus the second integral equals  \displaystyle \frac2{\pi^2}\int_0^\pi t\log(\sin t) dt .
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  3. #33
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by chiph588@ View Post
     \displaystyle \log(\pi x) - \log(\sin\pi x) = \sum_{n=1}^\infty \frac{\zeta(2n)}{n}x^{2n} \implies \log(\pi \sqrt{x}) - \log(\sin(\pi \sqrt{x})) = \sum_{n=1}^\infty \frac{\zeta(2n)}{n}x^{n}  \displaystyle \implies \sum_{n=1}^\infty \frac{\zeta(2n)}{n(n+1)} = \int_0^1 \log(\pi\sqrt{x}) dx - \int_0^1 \log(\sin(\pi \sqrt{x})) dx

    For the second integral, let  t=\pi\sqrt{x} \implies dt = \frac\pi{2\sqrt{x}}dx \implies dx = \frac2{\pi^2} t dt

    Thus the second integral equals  \displaystyle \frac2{\pi^2}\int_0^\pi t\log(\sin t) dt .
    Gah! I made a stupid Calc I error substituting! Haha! Anways, let's finish this

    Spoiler:

    Let \displaystyle S=\sum_{n=1}^{\infty}\frac{\zeta(2n)}{n(n+1)} then we have shown that



    \displaystyle S=\log(\pi)-\frac{1}{2}-I



    Where


    \displaystyle I=\frac{2}{\pi^2}\int_0^{\pi}t\log(\sin(t))\text{ }dt


    Letting \pi-y=t in this integral gives


    \displaystyle I=\frac{2}{\pi^2}\int_0^\pi\left(\pi-y\right)\log\left(\sin(t)\right)\text{ }dt=\frac{2}{\pi}\int_0^{\pi}\log\left(\sin(t)\rig  ht)\text{ }dt-I


    And thus


    \displaystyle \pi I=\int_0^{\pi}\log\left(\sin(t)\right)\text{ }dt


    Let \displaystyle t=\frac{\pi}{2}-y to get that


    \displaystyle \pi I=\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}\log\left(\cos(y)\right)\t  ext{ }dy


    But, since \cos(y) is symmetric on \displaystyle \left[\frac{-\pi}{2},\frac{\pi}{2}\right] we get that


    \displaystyle \frac{\pi}{2}I=\int_0^{\frac{\pi}{2}}\log\left(\co  s(y)\right)\text{ }dy


    Letting \displaystyle y=\frac{\pi}{2}-u gives that


    \displaystyle \frac{\pi}{2}I=\int_0^{\frac{\pi}{2}}\log\left(\si  n(u)\right)\text{ }du



    This integral is well-known (see here fore example) and so we get


    \displaystyle I=\frac{\frac{-\pi\log(2)}{2}}{\frac{\pi}{2}}=-\log(2)


    So, finally we may conclude that


    \displaystyle S=\log(2\pi)-\frac{1}{2}

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  4. #34
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by Drexel28 View Post
    Gah! I made a stupid Calc I error substituting! Haha! Anways, let's finish this

    Spoiler:

    Let \displaystyle S=\sum_{n=1}^{\infty}\frac{\zeta(2n)}{n(n+1)} then we have shown that



    \displaystyle S=\log(\pi)-\frac{1}{2}-I



    Where


    \displaystyle I=\frac{2}{\pi^2}\int_0^{\pi}t\log(\sin(t))\text{ }dt


    Letting \pi-y=t in this integral gives


    \displaystyle I=\frac{2}{\pi^2}\int_0^\pi\left(\pi-y\right)\log\left(\sin(t)\right)\text{ }dt=\frac{2}{\pi}\int_0^{\pi}\log\left(\sin(t)\rig  ht)\text{ }dt-I


    And thus


    \displaystyle \pi I=\int_0^{\pi}\log\left(\sin(t)\right)\text{ }dt


    Let \displaystyle t=\frac{\pi}{2}-y to get that


    \displaystyle \pi I=\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}\log\left(\cos(y)\right)\t  ext{ }dy


    But, since \cos(y) is symmetric on \displaystyle \left[\frac{-\pi}{2},\frac{\pi}{2}\right] we get that


    \displaystyle \frac{\pi}{2}I=\int_0^{\frac{\pi}{2}}\log\left(\co  s(y)\right)\text{ }dy


    Letting \displaystyle y=\frac{\pi}{2}-u gives that


    \displaystyle \frac{\pi}{2}I=\int_0^{\frac{\pi}{2}}\log\left(\si  n(u)\right)\text{ }du



    This integral is well-known (see here fore example) and so we get


    \displaystyle I=\frac{\frac{-\pi\log(2)}{2}}{\frac{\pi}{2}}=-\log(2)


    So, finally we may conclude that


    \displaystyle S=\log(2\pi)-\frac{1}{2}

    Wow! You made that integral look easy!
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  5. #35
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by chiph588@ View Post
    Wow! You made that integral look easy!
    I actually found a much better way.
    Spoiler:



    Looking at my last solution see that we have


    \displaystyle \pi I=\int_0^\pi \log(\sin(t))\text{ }dt


    and


    \displaystyle \frac{\pi}{2}I=\int_0^{\frac{\pi}{2}}\log\left(\si  n(t)\right)\text{ }dt=\int_0^{\frac{\pi}{2}}\log\left(\cos(t)\right)  \text{ }dt


    Thus, from the equality immediately preceding this sentence


    \displaystyle \begin{aligned}\pi I &= \int_0^{\frac{\pi}{2}}\left(\log(\sin(t)\right)+\l  og\left(\cos(t)\right)\right)\text{ }dt\\ &= \int_0^{\frac{\pi}{2}}\log\left(\frac{1}{2}\sin\le  ft(2t\right)\right)\\ &= \frac{-\pi\log(2)}{2}+\int_0^{\frac{\pi}{2}}\log\left(\si  n(2t)\right)\text{ }dt\\ &= \frac{-\pi\log(2)}{2}+\frac{1}{2}\int_0^\pi \log(\sin(w))\text{ }dw\quad\quad 2t\mapsto w\\ &= \frac{-\pi\log(2)}{2}+\frac{\pi}{2}I\end{aligned}


    Solving we find that \displaystyle I=-\log(2) as desired. Thus, we've completely circumvented the need to calculate \displaystyle \int_0^{\frac{\pi}{2}}\log\left(\sin(x)\right)\tex  t{ }dx directly, and have even found an alternate way of computing it.


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  6. #36
    MHF Contributor chiph588@'s Avatar
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    This isn't too hard to calculate, but I think it's a good integral to know. Compute  \displaystyle \int_0^\infty \frac{x^n}{(ax+b)^m} dx .
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  7. #37
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by chiph588@ View Post
    This isn't too hard to calculate, but I think it's a good integral to know. Compute  \displaystyle \int_0^\infty \frac{x^n}{(ax+b)^m} dx .
    Of course assuming m>n+1. I'll assume that b\ne 0 since this is trivial otherwise. Similarly, I'll assume that \frac{a}{b}>0.

    Spoiler:



    Let


    \displaystyle I=\int_0^{\infty}\frac{x^n}{(ax+b)^m}\text{ }dx


    Then,


    \displaystyle b^mI=\int_0^{\infty}\frac{x^n}{(1+cx)^m}\text{ }dx


    where \displaystyle c=\frac{a}{b}. Let 1+cx=z so that



    \displaystyle c^{n+1}b^mI=\int_1^{\infty}\frac{(z-1)^n}{z^m}\text{ }dz


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



     \displaystyle \begin{aligned}c^{n+1}b^m I &=\int_0^1 u^{m-2}\left(\frac{1}{u}-1\right)^n\text{ }du\\ &=\int_0^1 u^{m-2-n}(1-u)^n\text{ }du\\ &=B\left(m-1-n,n+1\right)\end{aligned}


    and thus


    \displaystyle I=\frac{b^{m-n-1}}{a^{n+1}}B\left(m-1-n,m+1\right)

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  8. #38
    MHF Contributor Drexel28's Avatar
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    This one's not too bad.

    Compute:


    \displaystyle \int_0^{\infty}x\sin\left(ax^2\right)\sin\left(2bx  \right)\text{ }dx
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  9. #39
    Super Member Random Variable's Avatar
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    Are you sure that integral converges?
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  10. #40
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Random Variable View Post
    Are you sure that integral converges?
    Assuming I didn't miscalculate something? Anyways, try this integral. It's much more surprising (albeit not too hard)!


    \displaystyle \int_0^1\log\left(\Gamma(x)\right)\cos\left(2\pi n x\right)\text{ }dx,\quad n\in\mathbb{N}
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  11. #41
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    Quote Originally Posted by Drexel28 View Post

    \displaystyle \int_0^1\log\left(\Gamma(x)\right)\cos\left(2\pi n x\right)\text{ }dx,\quad n\in\mathbb{N}


     \displaystyle I= \int_0^1 \ln[\Gamma(x)]\cos(2\pi nx)~dx



    \displaystyle  = \frac{1}{2} [ \int_0^1 \ln[\Gamma(x)]\cos(2\pi nx)~dx + \int_0^1 \ln[\Gamma(x)]\cos(2\pi nx)~dx ]



    Sub.  x = 1-t on the second integral and change the dummy variable , we have



     \displaystyle I= \frac{1}{2} [ \int_0^1 \ln[\Gamma(x)]\cos(2\pi nx)~dx + \int_0^1 \ln[\Gamma(1-x)]\cos(2\pi nx)~dx



     \displaystyle = \frac{1}{2} \int_0^1 \ln[ \Gamma(x) \Gamma(1-x) ] \cos(2\pi nx ) ~dx



     \displaystyle = \frac{1}{2} \int_0^1 [ \ln(\pi) - \ln[\sin(\pi x)] ] \cos(2\pi nx ) ~dx




    \displaystyle  = - \frac{1}{2\pi} \int_0^{\pi} \ln[\sin(x)] \cos(2nx)~dx



    Integration by parts



    \displaystyle  = - \frac{1}{4n\pi} \left[ \ln[\sin(x)] \sin(2nx)\right]_0^\pi + \frac{1}{4n\pi} \int_0^{\pi} \frac{\sin(2nx)\cos(x)}{\sin(x)}~dx



     \displaystyle  = \frac{1}{8n\pi} \int_0^{\pi} \frac{\sin(2n+1)x + \sin(2n-1)x}{\sin(x)}~dx



    \displaystyle   = \frac{1}{8n\pi} (2\pi) = \frac{1}{4n}



    Problem :

    How do you prove this

     \displaystyle \frac{\Gamma(x) \Gamma(y) }{\Gamma(x+y)} = \int_0^1 t^{x-1} (1-t)^{y-1} ~dt

    My solution is to make a substitution , followed by changing the order of a double integral .
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  12. #42
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by simplependulum View Post

    Problem :

    How do you prove this

     \displaystyle \frac{\Gamma(x) \Gamma(y) }{\Gamma(x+y)} = \int_0^1 t^{x-1} (1-t)^{y-1} ~dt

    My solution is to make a substitution , followed by changing the order of a double integral .
    Let \displaystyle B(x,y)=\int_0^1 t^{x-1}(1-t)^{y-1}\text{ }dt. It's easy to show that \displaystyle B(x,y)=\frac{x+y}{y}B(x,y+1) and more so by induction that \displaystyle B(x,y)=\frac{(x+y)_n}{n! n^{x+y-1}}\frac{n! n^{y-1}}{(y)_n}\int_0^n t^{x-1}\left(1-\frac{t}{n}\right)^{n+y-1}\text{ }dt. But, by definition (I guess it really depends upon your definition of the Gamma function, but this is a common one) it follows that \displaystyle B(x,y)=\frac{\Gamma(y)}{\Gamma(x+y)}\lim_{n\to\inf  ty}\int_0^n t^{x-1}\left(1-\frac{t}{n}\right)^{y+n-1}\text{ }dt but one can verify (using the LDC theorem for example) that this integral tends to \displaystyle \int_0^{\infty}t^{x-1}e^{-t}\text{ }dt from where the conclusion follows.
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  13. #43
    Super Member Random Variable's Avatar
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     \int^{1}_{0} t^{x-1} (1-t)^{y-1} \ dt


    let  t = \sin^{2} \theta

     = 2 \int^{\pi /2}_{0} (\sin^{2} \theta)^{r-1} (1-\sin^{2} \theta)^{s-1} \sin \theta \cos \theta \ d \theta

     = 2 \int^{\pi /2}_{0} (\sin \theta)^{2r-1} (\cos \theta)^{2s-1} \ d \theta


     \Gamma(x) = \int^{\infty}_{0} t^{x-1} e^{-t} \ dt

    let  t = u^{2}

     \Gamma (x) = \int^{\infty}_{0}  u^{2x-2} e^{-u^{2}} (2 u \ du)

     = 2 \int^{\infty}_{0}  u^{2x-1} e^{-u^{2}} \ du


    then  \Gamma(x) \Gamma(y) = 4 \int^{\infty}_{0}  u^{2x-1} e^{-u^{2}} \ du \int^{\infty}_{0}  v^{2y-1} e^{-v^{2}} \ dv

     = 4 \int^{\infty}_{0} \int^{\infty}_{0} u^{2x-1} v^{2y-1} e^{-(u^{2}+v^{2})} \ du \ dv

    change to polar coordinates

     \Gamma(x) \Gamma (y) = 4 \int^{\pi / 2}_{0} \int^{\infty}_{0} (r \cos \theta)^{2x-1} \ (r \sin \theta)^{2y-1} e^{-r^{2}} r \ dr \ d \theta

     = 4 \int^{\pi /2}_{0} \int^{\infty}_{0}   (cos \theta)^{2x-1} \ (\sin \theta)^{2y-1} \ r^{2(x+y)-1} e^{-r^{2}} \ dr \ d \theta

     =  2 \Gamma(x+y) \int^{\pi /2}_{0} (cos \theta)^{2x-1} \ (\sin \theta)^{2y-1} \ d \theta

     =  \Gamma (x+y) \int^{1}_{0} t^{x-1} (1-t)^{y-1} \ dt
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  14. #44
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by Drexel28 View Post
    This one's not too bad.

    Compute:


    \displaystyle \int_0^{\infty}x\sin\left(ax^2\right)\sin\left(2bx  \right)\text{ }dx
    Here's a thought:  \displaystyle x\sin(ax^2)\sin(2bx) = \frac{x}2 (\cos(a x^2-2 b x)-\cos(a x^2+2 b x))

     = \frac{x}2\left(\cos\left(\frac{(a x+b)^2}a-\frac{b^2}a\right)-\cos\left(\frac{(a x+b)^2}a-\frac{b^2}a\right)\right)

    Now split the cosines up into its exponentials.

    I could be wrong, and won't have time to work it out... I'll be away from a computer for a week.
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  15. #45
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    Here is my method :

    \displaystyle   I = \int_0^1 t^{x-1} (1-t)^{y-1}~dt


    Sub.  \displaystyle  t = \frac{1}{u+1}

     \displaystyle  I = \int_0^{\infty} \frac{u^{y-1}}{(u+1)^{x+y}}~du


    Consider  \displaystyle  \Gamma(x+y) = \int_0^{\infty} e^{-z} z^{x+y-1}~dz

    Sub.  z = (u+1)s ,  u as parameter , we have

     \displaystyle  \frac{\Gamma(x+y)}{(u+1)^{x+y}} = \int_0^{\infty}  e^{-(u+1)s} s^{x+y-1}~ds


    Let's come back to the original problem , we thus have

     \displaystyle  I = \frac{1}{\Gamma(x+y)}~ \int_0^{\infty} \int_0^{\infty} e^{-(u+1)s} u^{y-1} s^{x+y-1}~dsdu

     \displaystyle  = \frac{1}{\Gamma(x+y)}~ \int_0^{\infty} \left[ \frac{\Gamma(y)}{s^y}~ e^{-s}  s^{x+y-1} \right] ~ds

     \displaystyle  = \frac{\Gamma(y)}{\Gamma(x+y)}~\Gamma(x)

     \displaystyle  = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}
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