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Thread: Prove some identities!

  1. December 27th 2010, 07:52 PM #76
    Drexel28
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    Quote Originally Posted by chiph588@ View Post
    Problem (\star\star\star\star): For |x|<1 , compute

    \displaystyle \sum_{s=2}^{\infty}\frac{x^s\zeta(s)}{s}
    Solution:

    EDIT: Already beaten to the answer by simplependulum, but a different method.

    Spoiler:

    We begin by noticing that for x\notin -\mathbb{N} we have that


    \displaystyle \Gamma(-x)=\frac{e^{\gamma x}}{-x}\prod_{n=1}^{\infty}\left(1-\frac{x}{n}\right)^{-1}e^{\frac{-x}{n}}


    (I prove this here). Taking the logarithm of both sides gives


    \displaystyle \log\left(\Gamma(-x)\right)=\gamma x-\log(-x)+\sum_{n=1}^{\infty}\left(-\log\left(1-\frac{x}{n}\right)-\frac{x}{n}\right)


    Or, said differently


    \displaystyle \log\left(\Gamma(1-x)\right)-\gamma x=\sum_{n=1}^{\infty}\left(-\log\left(1-\frac{x}{n}\right)-\frac{x}{n}\right)\quad\mathbf{(1)}


    But, note that since |x|<1 we have that


    \displaystyle \begin{aligned}\sum_{n=1}^{\infty}\left(-\log\left(1-\frac{x}{n}\right)-\frac{x}{n}\right) &=\sum_{n=1}^{\infty}\left(\sum_{m=0}^{\infty}\fra c{x^{m+1}}{n^{m+1} (m+1)}-\frac{x}{n}\right)\\ &= \sum_{n=1}^{\infty}\sum_{m=1}^{\infty}\frac{x^{m+1 }}{n^{m+1}(m+1)}\end{aligned}


    Note though that for each fixed n the inner sum converges absolutely since |x|<1. Thus, we may rearrange the series (see Apostol pg. 202) to get


    \displaystyle \begin{aligned}\sum_{n=1}^{\infty}\sum_{m=1}^{\inf ty}\frac{x^{m+1}}{n^{m+1}(m+1)} &= \sum_{m=1}^{\infty}\frac{x^{m+1}}{m+1}\sum_{n=1}^{ \infty}\frac{1}{n^{m+1}}\\ &= \sum_{m=1}^{\infty}\frac{x^{m+1}\zeta(m+1)}{m+1}\\ &= \sum_{m=2}^{\infty}\frac{x^m \zeta(m)}{m}\end{aligned}


    Plugging this into \mathbf{(1)} gives then that


    \displaystyle \sum_{m=2}^{\infty}\frac{x^m\zeta(m)}{m}=\log\left (\Gamma(1-x)\right)-\gamma x


    for all |x|<1.

    Last edited by Drexel28; December 28th 2010 at 07:24 PM.
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  2. December 27th 2010, 08:28 PM #77
    chiph588@
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    Quote Originally Posted by Drexel28 View Post
    Solution:

    EDIT: Already beaten to the answer by simplependulum, but a different method.

    Spoiler:

    We begin by noticing that for x\notin -\mathbb{N} we have that


    \displaystyle \ln\left(\Gamma(-x)\right)=\frac{e^{\gamma x}}{-x}\prod_{n=1}^{\infty}\left(1-\frac{x}{n}\right)^{-1}e^{\frac{-x}{n}}


    (I prove this here). Taking the logarithm of both sides gives


    \displaystyle \log\left(\Gamma(-x)\right)=\gamma x-\log(-x)+\sum_{n=1}^{\infty}\left(-\ln\left(1-\frac{x}{n}\right)-\frac{x}{n}\right)


    Or, said differently


    \displaystyle \log\left(\Gamma(1-x)\right)-\gamma x=\sum_{n=1}^{\infty}\left(-\ln\left(1-\frac{x}{n}\right)-\frac{x}{n}\right)\quad\mathbf{(1)}


    But, note that since |x|<1 we have that


    \displaystyle \begin{aligned}\sum_{n=1}^{\infty}\left(-\ln\left(1-\frac{x}{n}\right)-\frac{x}{n}\right) &=\sum_{n=1}^{\infty}\left(\sum_{m=0}^{\infty}\fra c{x^{m+1}}{n^{m+1} (m+1)}-\frac{x}{n}\right)\\ &= \sum_{n=1}^{\infty}\sum_{m=1}^{\infty}\frac{x^{m+1 }}{n^{m+1}(m+1)}\end{aligned}


    Note though that for each fixed n the inner sum converges absolutely since |x|<1. Thus, we may rearrange the series (see Apostol pg. 202) to get


    \displaystyle \begin{aligned}\sum_{n=1}^{\infty}\sum_{m=1}^{\inf ty}\frac{x^{m+1}}{n^{m+1}(m+1)} &= \sum_{m=1}^{\infty}\frac{x^{m+1}}{m+1}\sum_{n=1}^{ \infty}\frac{1}{n^{m+1}}\\ &= \sum_{m=1}^{\infty}\frac{x^{m+1}\zeta(m+1)}{m+1}\\ &= \sum_{m=2}^{\infty}\frac{x^m \zeta(m)}{m}\end{aligned}


    Plugging this into \mathbf{(1)} gives then that


    \displaystyle \sum_{m=2}^{\infty}\frac{x^m\zeta(m)}{m}=\log\left (\Gamma(1-x)\right)-\gamma x


    for all |x|<1.

    That's how I did it!

    Quote Originally Posted by simplependulum View Post
    Using the similar method , Prove this formula ( Gauss's Multiplication Formula ) :

    \displaystyle \Gamma(nz) = n^{nz - \frac{1}{2} } (2\pi)^{\frac{1-n}{2} \prod_{k=0}^{n-1} \Gamma{\left( z + \frac{k}{n} \right)} ~ n \in \mathbb{N}

    ps I don't know how many stars it should be given ...
    I've seen a short but messy proof of this fact. I shall spare everyone the headache of reading my solution!
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  3. December 28th 2010, 04:09 PM #78
    Bruno J.
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    Here's an easy one to keep the thread going. Evaluate


    \int_0^\infty \frac{\log x}{1+x^2} dx
    Last edited by Bruno J.; December 28th 2010 at 04:22 PM.
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  4. December 28th 2010, 04:52 PM #79
    Drexel28
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    Quote Originally Posted by Bruno J. View Post
    Here's an easy one to keep the thread going. Evaluate


    \int_0^\infty \frac{\log x}{1+x^2} dx
    Easy way (more general):
    Spoiler:


    Let


    \displaystyle I(a)=\int_0^{\infty}\frac{\log(x)}{x^2+a^2}\text{ }a>0


    Let x=\frac{a}{t} to get


    \displaystyle \begin{aligned}I(a) &= \frac{\log(a)}{a}\int_0^{\infty}\frac{dt}{1+t^2}-\frac{1}{a}\int_0^{\infty}\frac{\log(t}{1+t^2}\tex t{ }dt\end{aligned}



    But, let \displaystyle y=\frac{1}{u} in this second integral to get that


    \displaystyle \int_0^{\infty}\frac{\log(t)}{1+t^2}=-\int_0^{\infty}\frac{\log(u)}{1+u^2}\text{ }du


    So that


    \displaystyle \int_0^{\infty}\frac{\log(t)}{1+t^2}\text{ }dt=0


    Thus,


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






    To spice things up:

    Spoiler:


    Consider


    \displaystyle f(z)=\frac{\log(z)}{z^2+a^2}


    and for 0<\varepsilon<R define the contour \Omega_{\varepsilon,R} to be


    \displaystyle \Omega_{\epsilon,R}=\Gamma_R\cup\gamma_{\varepsilo n}\cup[-R,-\varepsilon]\cup[\varepsilon,R]


    Where \Gamma_R is the half-annulus \Gamma_R=\varepsilon<|z|<R and \gamma_{\varepsilon} is the semi-circle of radius \varepsilon centered at 0. It's evident that


    \displaystyle \left|\int_{\gamma_{\varepsilon}}f(z)\text{ }dz\right|\to0\quad \varepsilon\to 0


    and


    \displaystyle \left|\int_{\Gamma_R}f(z)\text{ }dz\right|\leqslant \frac{|\log(R)|+\pi}{R^2-1}\pi R\to 0\quad R\to\infinity


    But, by the residue theorem


    \displaystyle \int_{\Omega_{R,\varepsilon}}f(z)\text{ }dz=2\pi i \text{Res}\left(f;ia\right)=\frac{\pi}{a}\left(\lo g(a)+\frac{\pi i}{2}\right)



    And thus it follows that


    \displaystyle \int_{-\infty}^{\infty}\frac{\log(z)}{z^2+a^2}\text{ }dz=\frac{\pi}{a}\left(\log(a)+\frac{\pi i}{2}\right)


    But,


    \displaystyle \begin{aligned}\int_{-\infty}^{0}f(z)\text{ }dz &= \int_{-\infty}^{0}\frac{\log(-x)+\pi i}{x^2+a^2}\text{ }dx\\ &= \int_0^{\infty}\frac{\log(x)}{x^2+a^2}\text{ }dx+\frac{\pi^2 i}{2a}\end{aligned}


    From where it evidently follows that


    \displaystyle \int_0^{\infty}\frac{\log(x)}{x^2+a^2}\text{ }dx=\frac{\pi\log(a)}{2a}

    Last edited by Drexel28; December 28th 2010 at 05:06 PM.
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  5. December 28th 2010, 05:04 PM #80
    Bruno J.
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    Good! I prefer your first method, since it works fine and it's much easier.

    Now how about \int_0^1 \frac{\log(1+x)}{1+x^2}?

    By the way, you might want to be more precise in the way you've defined your contour. As it is, \Gamma_R is a domain and not a contour. You also didn't give an orientation.
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  6. December 28th 2010, 05:56 PM #81
    Random Variable
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    let x = \frac{1-t}{1+t}

    EDIT: or let I(a) = \int_{0}^{1} \frac{\ln(ax+1)}{1+x^{2}} \ dx and differentiate
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  7. December 28th 2010, 06:13 PM #82
    chiph588@
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    Quote Originally Posted by Random Variable View Post
    let x = \frac{1-t}{1+t}
    Where did you come up with this substitution?
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  8. December 28th 2010, 06:21 PM #83
    Drexel28
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    Quote Originally Posted by Bruno J. View Post
    Good! I prefer your first method, since it works fine and it's much easier.

    Now how about \int_0^1 \frac{\log(1+x)}{1+x^2}?
    EDIT: Beat by Random Variable
    The classic way to do this is the substitution x=\frac{1-t}{1+t} to get, if I equals our integral, that \displaystyle I=\int_0^1\frac{\log(2)}{t^2+1}-I. So that \displaystyle I=\frac{\log(2)\pi}{8}. Now, solve


    \displaystyle \int_0^{\infty}\frac{\log(1+x)}{1+x^2}\text{ }dx
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  9. December 28th 2010, 06:26 PM #84
    Random Variable
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    Quote Originally Posted by Drexel28 View Post
    EDIT: Beat by Random Variable
    The classic way to do this is the substitution x=\frac{1-t}{1+t} to get, if I equals our integral, that \displaystyle I=\int_0^1\frac{\log(2)}{t^2+1}-I. So that \displaystyle I=\frac{\log(2)\pi}{8}. Now, solve


    \displaystyle \int_0^{\infty}\frac{\log(1+x)}{1+x^2}\text{ }dx
    Sorry. I should only post complete solutions.

    I really can't explain why that substitution works. Can you?
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  10. December 28th 2010, 06:27 PM #85
    chiph588@
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    Quote Originally Posted by Random Variable View Post
    I really can't explain why that substitution works. Can you?
    Then what made you think of it?
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  11. December 28th 2010, 06:34 PM #86
    Drexel28
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    Quote Originally Posted by Random Variable View Post
    Sorry. I should only post complete solutions.

    I really can't explain why that substitution works. Can you?
    I can, what's not to understand. Letting \displaystyle x=\frac{1-t}{1+t} we get (with all the intermediary steps) that


    \displaystyle \begin{aligned}\int_0^1\frac{\log(1+x)}{1+x^2}\tex t{ }dx &= \int_1^0\frac{\log\left(1+\frac{1-t}{1+t}\right)}{1+\frac{(1-t)^2}{(1+t)^2}}\frac{-2}{(1+t)^2}\text{ }dt\\ &= \int_0^1\frac{2\log\left(\frac{2}{x+1}\right)}{(1+ t)^2+(1-t)^2}\\ &= \int_0^1\frac{\log(2)-\log(1+t)}{1+t^2}\text{ }dt\end{aligned}
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  12. December 28th 2010, 06:37 PM #87
    chiph588@
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    Quote Originally Posted by Drexel28 View Post
    \displaystyle \int_0^{\infty}\frac{\log(1+x)}{1+x^2}\text{ }dx
    Spoiler:

    \begin{aligned}\displaystyle \int_0^{\infty}\frac{\log(1+x)}{1+x^2} dx &= \int_0^{1}\frac{\log(1+x)}{1+x^2} dx+\int_1^{\infty}\frac{\log(1+x)}{1+x^2} dx\\ &= \frac18\pi\log2 + \int_0^1 \frac{\log(1+1/x)}{1+x^2} dx \quad \text{ (here we let } x \mapsto 1/x \text{)}\\ &= \frac18\pi\log2 + \int_0^{1}\frac{\log(1+x)}{1+x^2} dx - \int_0^{1}\frac{\log(x)}{1+x^2} dx\\ &= \frac14\pi\log2 + G \end{aligned}

    Where G is Catalan's constant. See here for why.
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  13. December 28th 2010, 06:37 PM #88
    Random Variable
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    No. What I mean is what about the integral would suggest that substitution.
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  14. December 28th 2010, 06:38 PM #89
    Random Variable
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    It's a classic Putnam integral. That's the only reason I knew.
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  15. December 28th 2010, 06:41 PM #90
    Drexel28
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    Quote Originally Posted by chiph588@ View Post
    Spoiler:

    \begin{aligned}\displaystyle \int_0^{\infty}\frac{\log(1+x)}{1+x^2} dx &= \int_0^{1}\frac{\log(1+x)}{1+x^2} dx+\int_1^{\infty}\frac{\log(1+x)}{1+x^2} dx\\ &= \frac18\pi\log2 + \int_0^1 \frac{\log(1+1/x)}{1+x^2} dx \text{(here we let } x \mapsto 1/x \text{)}\\ &= \frac18\pi\log2 + \int_0^{1}\frac{\log(1+x)}{1+x^2} dx - \int_0^{1}\frac{\log(x)}{1+x^2} dx\\ &= \frac14\pi\log2 + G \end{aligned}

    Where G is Catalan's constant. See here for why.

    Right, that's what I was thinking.
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