Facebook Page
Twitter
RSS
Closed Thread
Page 7 of 7 FirstFirst ... 567
Results 61 to 62 of 62
  1. زيد اليافعي
    MHB Site Helper
    MHB Math Helper
    ZaidAlyafey's Avatar
    Status
    Offline
    Join Date
    Jan 2013
    Location
    KSA
    Posts
    1,666
    Thanks
    3,671 times
    Thanked
    3,857 times
    Thank/Post
    2.315
    Awards
    MHB Math Notes Award (2014)  

MHB Best Ideas (2014)  

MHB Best Ideas (Jul-Dec 2013)  

MHB Analysis Award (Jul-Dec 2013)  

MHB Calculus Award (Jul-Dec 2013)
    #61 Thread Author
    Relation to Hyperfactorial function


    Prove for $n$ is a positive integer

    $$G(n+1) = \frac{(N!)^n}{H(n)}$$

    Where $H(n)$ is the hyperfactorial function

    $$H(n) = \prod^n_{k=1}k^k$$

    proof

    We can prove it by induction for $n=0$ we have, $G(1)=1$,

    suppose that

    $$G(n) = \frac{\Gamma(n)^{n-1}}{H(n-1)}$$

    we want to find

    $$G(n+1) = \Gamma(n)G(n) = \Gamma(n) \frac{\Gamma(n)^{n-1}}{H(n-1)}$$

    Notice that

    $$H(n-1) = \prod^{n-1}k^k = \frac{\prod^{n}k^k}{n^n} = \frac{H(n)}{n^n }$$

    We deduce that

    $$G(n+1) = \Gamma(n)G(n) = \frac{\Gamma(n)^{n}\times n^n}{H(n)} = \frac{(N!)^n}{H(n)}$$

    A related constant


    We define the Glaisher-Kinkelin constant as

    $$A = \lim_{n \to \infty}\frac{H(n)}{n^{n^2/2+n/2+1/12}e^{-n^2/4}}$$

    Prove that

    $$\lim_{n \to \infty}\frac{G(n+1)}{(2\pi)^{n/2}n^{n^2/2-1/12} e^{-3n^2/4}} = \frac{e^{1/12}}{A}$$

    proof

    From the previous result we have

    $$\lim_{n \to \infty}\frac{(N!)^n}{H(n)(2\pi)^{n/2}n^{n^2/2-1/12} e^{-3n^2/4}}$$

    Now use the Stirling approximation

    $$(N!)^n \sim (2\pi)^{n/2}n^{n^2+n/2}e^{-n^2+1/12}$$

    Hence we deduce that

    $$\lim_{n \to \infty}\frac{(2\pi)^{n/2}n^{n^2+n/2}e^{-n^2+1/12}}{H(n)}\times\frac{1}{(2\pi)^{n/2}n^{n^2/2-1/12} e^{-3n^2/4}}$$

    By simplifications we have

    $$e^{1/12}\lim_{n \to \infty}\frac{n^{n^2/2+n/2+1/12}e^{-n^2/4}}{H(n)} = \frac{e^{1/12}}{A}$$

    Exercise

    $$\zeta'(2) = \frac{\pi^2}{6}\left(\log(2\pi)+\gamma-12\log A \right)$$

    We already proved that

    $$\log \left[\frac{G(n+1)}{(2\pi)^{n/2}n^{n^2/2-1/12} e^{-3n^2/4}} \right] \sim \frac{1}{12}-\log A$$

    Let the following

    $$f(n) = \log \left[\frac{G(n+1)}{(2\pi)^{n/2}n^{n^2/2-1/12} e^{-3n^2/4}} \right]$$

    Use the series representation of the Barnes functions

    $$f(n) = \log \left[ \frac{(2\pi)^{n/2}\exp\left(-\frac{n+n^2(1+\gamma)}{2}\right)\prod_{k=1}^{\infty}\left\{\left(1+\frac{n}{k} \right)^k\exp\left(\frac{n^2}{2k}-n\right) \right\}}{(2\pi)^{n/2}n^{n^2/2-1/12} e^{-3n^2/4}} \right]$$

    Which reduces to

    $$f(n) = -\frac{n+n^2(1+\gamma)}{2}+\sum^\infty_{k=1}\left\{k \log \left(1+\frac{n}{k} \right)+\frac{n^2}{2k}-n \right\}\\ -\left( \frac{n^2}{2}-\frac{1}{12}\right)\log(n)+\frac{3n^2}{4}$$

    Differentiate with respect to $n$

    $$f'(n) = -\frac{1}{2}-n-\gamma n+n\psi(n)+\gamma n+1-n\log(n)-\frac{n}{2}+\frac{1}{12n}+\frac{3n}{2}$$

    Note that we already showed that

    $$\frac{d}{dn}\sum^\infty_{k=1}\left\{k \log \left(1+\frac{n}{k} \right)+\frac{n^2}{2k}-n\right\}= n\psi(n)+\gamma n+1$$

    By simplifications we have

    $$f'(n) = n\psi(n)-n\log(n)+\frac{1}{12n}+\frac{1}{2}$$

    Now use that

    $$\psi(n) = \log(n)-\frac{1}{2n}-2\int^\infty_0 \frac{z dz}{(n^2+z^2)(e^{2\pi z}-1)}dz$$

    Hence we deduce that

    $$f'(n) =-2\int^\infty_0 \frac{nz dz}{(n^2+z^2)(e^{2\pi z}-1)}dz+\frac{1}{12n}$$

    Integrate with respect to $n$

    $$f(n) =-\int^\infty_0 \frac{z\log(n^2+z^2)}{(e^{2\pi z}-1)}dz+\frac{1}{12}\log(n)+C$$

    Take the limit $n \to 0$

    $$C = \lim_{n \to 0}f(n)-\frac{1}{12}\log(n)+\int^\infty_0 \frac{z\log(z^2)}{(e^{2\pi z}-1)}dz $$

    Hence we have the limit

    $$\lim_{n \to 0}\frac{G(n+1)}{(2\pi)^{n/2}n^{n^2/2-1/12} e^{-3n^2/4}}-\frac{1}{12}\log(n) = \lim_{n \to 0}\log \frac{G(n+1)}{(2\pi)^{n/2}n^{n^2/2} e^{-3n^2/4}} = 0$$

    Hence we see that

    $$C = 2\int^\infty_0 \frac{z\log(z)}{(e^{2\pi z}-1)}dz $$

    Finally we have

    $$f(n) =-\int^\infty_0 \frac{z\log(n^2+z^2)}{(e^{2\pi z}-1)}dz+\frac{1}{12}\log(n)+ 2\int^\infty_0 \frac{z\log(z)}{(e^{2\pi z}-1)}dz$$

    $$f(n) =-\int^\infty_0 \frac{z\log\left(1+\frac{z^2}{n^2} \right)}{(e^{2\pi z}-1)}dz-\log(n^2)\int^\infty_0 \frac{z}{(e^{2\pi z}-1)}dz+\frac{1}{12}\log(n)\\+ 2\int^\infty_0 \frac{z\log(z)}{(e^{2\pi z}-1)}dz$$

    Also we have

    $$\int^\infty_0 \frac{z}{(e^{2\pi z}-1)}dz = \frac{1}{24}$$

    That simplifies to

    $$f(n)=-\int^\infty_0 \frac{z\log\left(1+\frac{z^2}{n^2} \right)}{(e^{2\pi z}-1)}dz+2\int^\infty_0 \frac{z\log(z)}{(e^{2\pi z}-1)}dz$$

    Take the limit $n \to \infty $

    $$2\int^\infty_0 \frac{z\log(z)}{(e^{2\pi z}-1)}dz = \frac{1}{12}-\log A$$

    Now use that

    $$
    \begin{align}
    2\int^\infty_0 \frac{z\log(z)}{(e^{2\pi z}-1)}dz
    &= 2\int^\infty_0 \frac{z\log(z)}{e^{2\pi z}}\times \frac{1}{1-e^{-2\pi z}}dz\\
    &= 2\sum_{n=0}^\infty \int^\infty_0 e^{-2\pi z(n+1)}z\log(z)\,dz\\
    &= \sum_{n=1}^\infty \frac{\psi(2) − \log(2\pi)+\log(n)}{2\pi^2 n^2}\\
    &=\frac{(\psi(2)-\log(2\pi))\zeta(2)+\zeta'(2)}{2\pi^2}\\
    \end{align}
    $$

    Hence we conclude that

    $$\zeta'(2) =(\log(2\pi)-\psi(2))\zeta(2) + 2\pi^2\left(\frac{1}{12}-\log A \right) = \zeta(2) (\log(2\pi)+\gamma-12 \log A)$$

  2. زيد اليافعي
    MHB Site Helper
    MHB Math Helper
    ZaidAlyafey's Avatar
    Status
    Offline
    Join Date
    Jan 2013
    Location
    KSA
    Posts
    1,666
    Thanks
    3,671 times
    Thanked
    3,857 times
    Thank/Post
    2.315
    Awards
    MHB Math Notes Award (2014)  

MHB Best Ideas (2014)  

MHB Best Ideas (Jul-Dec 2013)  

MHB Analysis Award (Jul-Dec 2013)  

MHB Calculus Award (Jul-Dec 2013)
    #62 Thread Author
    Relation to hurwitz zeta function


    Prove that

    $$\log G(z+1) - z \log \Gamma(z) = \zeta'(-1) - \zeta'(-1,z) $$

    Start by the following

    $$\zeta(s,z) = \frac{z^{-s}}{2}+\frac{z^{1-s}}{s-1}+2\int^\infty_0 \frac{\sin(s\arctan(x/z))}{(z^2+x^2)^{s/2}(e^{2\pi x}-1)}\,dx$$

    Hence we have

    $$\zeta'(-1,z) = -\frac{z\log(z)}{2}+\frac{z^2\log(z)}{2}-\frac{z^2}{4}+\int^\infty_0 \frac{x\log(x^2+z^2)+2z\arctan(x/z)}{(e^{2\pi x}-1)}\,dx$$

    Now use that

    $$\psi(z) = \log(z)-\frac{1}{2z}-2\int^\infty_0 \frac{x }{(z^2+x^2)(e^{2\pi x}-1)}dx$$

    Which implies that

    $$\int^\infty_0 \frac{2zx }{(z^2+x^2)(e^{2\pi x}-1)}dx=z\log(z)-\frac{1}{2}-z\psi(z)$$

    By taking the integral

    $$\int^\infty_0 \frac{x\log(x^2+z^2) -x\log(x^2)}{(e^{2\pi x}-1)}dx=\int^z_0x\log(x)\,dx-\int^z_0x\psi(x)\,dx-\frac{z}{2}$$

    Which simplifies to

    $$\int^\infty_0 \frac{x\log(x^2+z^2) }{(e^{2\pi x}-1)}dx=\zeta'(-1)-\frac{z^2}{4}+\frac{1}{2} z^2 \log(z)-z\log\Gamma(z)+\int^z_0\log\Gamma(x)\,dx-\frac{z}{2}$$

    Also we have

    $$2\int^\infty_0 \frac{x}{(x^2+z^2)(e^{2\pi x}-1)}dx=\log(z)-\frac{1}{2z}-\psi(z)$$

    By integration we have

    $$2\int^\infty_0 \frac{\arctan(x/z)}{(e^{2\pi x}-1)}dx=z+\frac{\log(z)}{2}-z \log(z)+\log\Gamma(z)+C$$

    Let $z \to 1$ to evaluate the constant

    $$2\int^\infty_0 \frac{\arctan(x/z)}{(e^{2\pi x}-1)}dx=z+\frac{\log(z)}{2}-z \log(z)+\log\Gamma(z)-\frac{1}{2}\log(2\pi)$$

    Multiply by $z$

    $$2\int^\infty_0 \frac{z\arctan(x/z)}{(e^{2\pi x}-1)}dx=z^2+\frac{z\log(z)}{2}-z^2 \log(z)+z\log\Gamma(z)-\frac{z}{2}\log(2\pi)$$

    Substitute both integrals our formula

    $$\zeta'(-1,z) =-\frac{z\log(z)}{2}+\frac{z^2\log(z)}{2}-\frac{z^2}{4}+z^2+\frac{z\log(z)}{2}-z^2 \log(z)+z\log\Gamma(z)\\-\frac{z}{2}\log(2\pi)+\zeta'(-1)-\frac{z^2}{4}+\frac{1}{2} z^2 \log(z)-z\log\Gamma(z)+\int^z_0\log\Gamma(x)\,dx-\frac{z}{2}$$

    Which reduces to

    $$\int_{0}^{z} \log \Gamma(x) \, \mathrm dx = \frac{z}{2} \log(2 \pi) + \frac{z(1-z)}{2} - \zeta^{'}(-1) + \zeta^{'}(-1,z)$$

    We also showed that

    $$\int_{0}^{z} \log \Gamma(x) \, \mathrm dx = \frac{z}{2} \log (2 \pi) + \frac{z(1-z)}{2} + z \log \Gamma(z) - \log G(z+1)$$

    By equating the equations we get our result.

    Prove that


    $$\zeta'(-1) = \frac{1}{12}-\log A$$

    Start by

    $$\zeta(s) = \lim_{m \to \infty} \left( \sum_{k=1}^{m} k^{-s} - \frac{m^{1-s}}{1-s} - \frac{m^{-s}}{2} + \frac{sm^{-s-1}}{12} \right) \ , \ \text{Re}(s) >-3.$$

    Differentiate with respect to $s$

    $$\zeta'(s) = \lim_{m \to \infty} \left(- \sum_{k=1}^{m} k^{-s}\log(k) - \frac{m^{1-s}}{(1-s)^2} +\frac{m^{1-s}}{1-s}\log(m)+\frac{m^{-s}}{2}\log(m)\\ + \frac{m^{-s-1}}{12}-\frac{m^{-s-1}}{12}\log(m)\right) $$

    Now let $s \to -1$

    $$\zeta'(-1) = \lim_{m \to \infty} \left(- \sum_{k=1}^{m} k\log(k) - \frac{m^{2}}{4} +\frac{m^{2}}{2}\log(m)+\frac{m}{2}\log(m)\\ + \frac{1}{12}-\frac{1}{12}\log(m)\right) $$

    Take the exponential of both sides

    $$e^{\zeta'(-1)}= e^{1/12}\lim_{m \to \infty} \frac{m^{m^2/2+m/2-1/12} e^{-m^2/4}}{ e^{\sum_{k=1}^{m} k\log(k)}} = e^{1/12}\lim_{m \to \infty} \frac{m^{m^2/2+m/2-1/12} e^{-m^2/4}}{ H(m+1)} = \frac{ e^{1/12}}{A} $$

    We conclude that

    $$\zeta'(-1) = \frac{1}{12}-\log A$$

    Prove that

    $$G\left(\frac{1}{2}\right) = 2^{1/24} \pi^{-1/4}e^{1/8}A^{-3/2}$$

    We know that

    $$\log G(z)+ \log \Gamma(z)- z \log \Gamma(z) = \zeta'(-1) - \zeta'(-1,z) $$

    Note that

    $$\zeta \left(s, \frac{1}{2} \right) = (2^s-1)\zeta(s)$$

    Which implies that

    $$\zeta' \left(-1, \frac{1}{2} \right) = \frac{\log(2)}{2}\zeta(-1)-\frac{1}{2}\zeta'(-1)$$

    Hence we have

    $$\log G\left(\frac{1}{2} \right)+ \frac{1}{2}\log \Gamma\left(\frac{1}{2} \right)= \frac{3}{2}\zeta'(-1)-\frac{\log(2)}{2}\zeta(-1) $$

    Using that we have

    $$G\left(\frac{1}{2}\right) = 2^{1/24}\pi^{-1/4}e^{\frac{3}{2}\zeta'(-1)}$$

    Note that

    $$\zeta(-1) = -\frac{1}{12}$$

    This can be proved by the functional equation of the zeta function.

Similar Threads

  1. Commentary for "Advanced Integration techniques"
    By ZaidAlyafey in forum Commentary threads
    Replies: 0
    Last Post: April 7th, 2013, 04:05
  2. A not very advanced integration technique...
    By chisigma in forum Math Notes
    Replies: 3
    Last Post: February 10th, 2013, 08:12
  3. Commentary for "A not very advanced integration technique..."
    By ZaidAlyafey in forum Commentary threads
    Replies: 5
    Last Post: February 8th, 2013, 14:40
  4. Replies: 5
    Last Post: March 24th, 2012, 13:34

Tags for this Thread

Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •  
Math Help Boards