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Math Help - Digamma Integral Representation

  1. #1
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    Digamma Integral Representation

    Hi,

    I'm trying to figure out how the integral representation of the Digamma function works.

    I understand that the Digamma function is the logarithmic derivative of the Gamma function. Therefore

    \Psi(z)=\frac{\Gamma'(z)}{\Gamma(z)}

    Since \Gamma(z)=\int^\infty_0e^{-t}t^{z-1}dt

    then I would think
    \Gamma'(z)=\int^\infty_0e^{-t}t^{z-1}\ln(t)dt

    Anyway, the integral expression for the Digamma function is given by

    \Psi(z)=\int^\infty_0\frac{e^{-t}}{t}-\frac{e^{-zt}}{1-e^{-t}}dt

    How would I derive this formula?
    Should I instead look at another representation of the Gamma function and work with that?

    Thank you.
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  2. #2
    MHF Contributor chisigma's Avatar
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    May be that the right way is starting from this 'infinite product'...

    \frac{1}{\Gamma(z)} = z \cdot e^{\gamma z} \cdot \prod_{n=1}^{\infty} (1+\frac{z}{n})\cdot e^{-\frac{z}{n}} (1)

    Now compute the logarithm of (1), change sign, then differentiate...

    Kind regards

    \chi \sigma
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  3. #3
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    We have  \ln(\frac{b}{a}) = \int_0^{\infty} \frac{ e^{-ax} - e^{-bx} }{x}~dx

    Let  a = 1 ~,~ b=t

    Therefore ,

     \Gamma'(z) = \int_0^{\infty} e^{-t} t^{z-1} \int_0^{\infty} \frac{ e^{-x} - e^{-tx} }{x}~dx dt

    By reversing the order , we have

     \int_0^{\infty} \frac{1}{x} \int_0^{\infty} e^{-t} t^{z-1} ( e^{-x} - e^{-tx}) ~dtdx

    Consider  \int_0^{\infty} e^{-t} e^{-tx} t^{z-1}~dt

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

    Sub.  (x+1)t = u

    we find that it is equal to

     \Gamma(z) \frac{1}{(x+1)^z}

    Back to the derivative of Gamma function ,

    If we continue we obtain

     \Gamma(z) \int_0^{\infty}\frac{1}{x} \left[ e^{-x} - \frac{1}{(x+1)^z} \right] ~dx

    Sub.  x+1 = e^{u} for the second integral ,

    we have  \int_0^{\infty}\frac{1}{x(x+1)^z}~dx = \int_0^{\infty} \frac{ e^{-zx} }{1 - e^{-x} }~dx

    Therefore ,

     \frac{ \Gamma'(z) }{\Gamma(z) } = \int_0^{\infty} \left(\frac{e^{-x}}{x} - \frac{ e^{-zx} }{1 - e^{-x} }\right)~dx

    Edit: Here is what i guess , not sure if it is true .

    If we differentiate w.r.t.  z again , i find something interesting ...


     \frac{ \Gamma(z) \Gamma''(z) - [\Gamma'(z)]^2 }{ \Gamma^2(z)}= \int_0^{\infty} \frac{ x e^{-zx}}{ 1 - e^{-x} } ~dx

    Sub.  z = 1 , from L.H.S , giving

     \Gamma''(1)- [\Gamma'(1)]^2

    Since  \Gamma'(1) = - \gamma , what it gives us is also equal to  \Gamma''(1)- \gamma^2  , while from R.H.S we have

      \int_0^{\infty} \frac{x}{ e^x - 1}~dx =  \Gamma(2) \zeta(2) =  \frac{\pi^2}{6} so is that ture

     \Gamma''(1) = \gamma^2 + \frac{\pi^2}{6} ?

    This method might be a way to obtain  \Gamma^{(n)}(1) by differentiating the integral  n-1 times without a great detour ...
    Last edited by simplependulum; May 12th 2010 at 11:28 PM.
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  4. #4
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    Quote Originally Posted by simplependulum View Post

    Sub.  x+1 = e^{u} for the second integral ,

    we have  \int_0^{\infty}\frac{1}{x(x+1)^z}~dx = \int_0^{\infty} \frac{ e^{-zx} }{1 - e^{-x} }~dx

    Therefore ,

     \frac{ \Gamma'(z) }{\Gamma(z) } = \int_0^{\infty} \left(\frac{e^{-x}}{x} - \frac{ e^{-zx} }{1 - e^{-x} }\right)~dx
    Thank you!

    I basically understand the whole proof until this part.
    I'm not quite sure how this substitution works out into that integral.

    As for this at the end of your post:

    \Gamma''(1) = \gamma^2 + \frac{\pi^2}{6}

    It all seems correct. It's pretty interesting seeing the \zeta(2) in there.

    Again thank you.
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  5. #5
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    Quote Originally Posted by Anthonny View Post
    Thank you!

    I basically understand the whole proof until this part.
    I'm not quite sure how this substitution works out into that integral.

    As for this at the end of your post:

    \Gamma''(1) = \gamma^2 + \frac{\pi^2}{6}

    It all seems correct. It's pretty interesting seeing the \zeta(2) in there.

    Again thank you.
    Let me see ...

    \int_0^{\infty}\frac{1}{x(x+1)^z}~dx = \int_0^{\infty} \frac{ e^{-zx} }{1 - e^{-x} }~dx

    Sub.  x + 1 = e^u

    dx = e^u du

    the bound changes to  (0,\infty)

    so it is equal to

     \int_0^{\infty} \frac{ 1}{ (e^u - 1) (e^u)^z } (e^u du)

     = \int_0^{\infty} \frac{ e^{-zu}}{ e^u - 1 } (e^u du)

     = \int_0^{\infty} \frac{ e^{-zu}}{ e^u ( 1- e^{-u}) } (e^u du)

     = \int_0^{\infty} \frac{ e^{-zu} }{ 1- e^{-u} }~du

    Then change the "dummy" variable

     =  \int_0^{\infty} \frac{ e^{-zx} }{ 1- e^{-x} }~dx



    To have a more formal proof , i think we need to show that

     \int_0^{\infty} [\frac{e^{-x}}{x} - \frac{1}{x(x+1)^z}]~dx

     =  \int_0^{\infty} [\frac{e^{-x}}{x} - \frac{ e^{-zx}}{ 1 - e^{-x}}]~dx

    because the integral \int_0^{\infty} \frac{e^{-x}}{x} ~dx is divergent .
    Last edited by simplependulum; May 16th 2010 at 12:06 AM.
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  6. #6
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    Quote Originally Posted by simplependulum View Post

     =  \int_0^{\infty} \frac{ e^{-zx} }{ 1- e^{-x} }~dx


    Thanks! I think I just overlooked the substitution process. I completely understand it now though.

    Anyway, now I understand the integral representation for \Psi(z)

    However, I have one more question (and this question is open to anyone). I want to see how

    \Psi(z)=H_{n-1}-\gamma

    Where H_{n} are the Harmonic numbers and \gamma is the Euler-Mascheroni constant.

    I want to derive this relationship using the integral representation of the Digamma function. I tried doing substitution, but I'm stuck. Here is my work so far:

    For

    \Psi(x)=\int_0^{\infty} [\frac{e^{-t}}{t} - \frac{ e^{-xt}}{ 1 - e^{-t}}]~dt

    Let u=e^{-t}, therefore -\ln(u)=t and -\frac{1}{u}du=dt

    Substituting, I get

    \int^0_1 [\frac{u}{-\ln(u)}-\frac{e^{-x\ln(u)}}{1-u}][-\frac{1}{u}]du

    which simplifies to

    -\int^1_0 [\frac{1}{\ln(u)}+\frac{u^{-x}u^{-1}}{1-u}]du

    I know that \gamma=\int^1_0 [\frac{1}{\ln(u)}+\frac{1}{1-u}du

    However, I'm not sure what to do from here. I also know the integral representation for the Harmonic numbers is given by:

    H_n=\int^1_0 \frac{1-x^n}{1-x}dx, but I don't see that occurring in the integral.

    Of course, my substitution might not be the right choice. The main thing to do is to get from the Digamma integral representation into a form in which the Harmonic integral representation and an integral representing \gamma arises.

    Thank you.
    Last edited by Anthonny; May 16th 2010 at 10:29 AM.
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  7. #7
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    You have done a great job !

    Your substitution is perfect !

    so  -\gamma = \Psi(1) = \int_0^{\infty} [\frac{e^{-t}}{t} - \frac{ e^{-t}}{ 1 - e^{-t}}]~ dt


    Also  H_n = \int_0^1 \frac{1 - x^n}{1-x}~dx

    We need to make a further substitution ,

     x = e^{-t}

     dx = -e^{-t}~dt

     H_n = \int_0^{\infty} \frac{1 - e^{-nt}}{1 - e^{-t}} e^{-t}~dt

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

    Therefore ,

     H_n - \gamma = \int_0^{\infty} \frac{e^{-t} - e^{-(n+1)t}}{1 - e^{-t}}~dt + \int_0^{\infty} [\frac{e^{-t}}{t} - \frac{ e^{-t}}{ 1 - e^{-t}}]~ dt

     = \int_0^{\infty} [\frac{ e^{-t}}{t} - \frac{e^{-(n+1)t}}{1 - e^{-t}}]~dx

     = \Psi(n+1) !!

    Or  \Psi(n) = H_{n-1} - \gamma

     n > 1
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  8. #8
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    Quote Originally Posted by simplependulum View Post

     = \Psi(n+1) !!

    Or  \Psi(n) = H_{n-1} - \gamma

     n > 1

    Thanks!

    I see how to apply the substitution even further and understand your explanation.

    Thanks on the help on the Digamma function.
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