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Math Help - Prove some identities! (Round two)

  1. #76
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by Drexel28 View Post
    So then I suppose it will be impressive when you prove it, right!
    Spoiler:


    Define  \displaystyle f(z) = \frac d{dz}\left(\log\Gamma(z)\right) = \frac{\Gamma'(z)}{\Gamma(z)}

    From the Gamma function duplication formula we see  \displaystyle f(z)+f(z+\tfrac12)+2\log2 = 2f(2z) . Thus  \displaystyle f(\tfrac32) = 2f(2)-f(1)-2\log2 .

    Now we must calculate  \Gamma'(1) and  \Gamma'(2) . Using a definition of  \Gamma , we see  \displaystyle f(s) = -\frac1s-\gamma-\sum_{n=1}^\infty\left(\frac1{n+s}-\frac1n\right)

    From this it's easy to see  \displaystyle f(1) = -\gamma and  \displaystyle f(2) = 1-\gamma \quad\text{\footnotesize(telescoping sum)} .

    So we see  f(\tfrac32) = 2-\gamma-2\log2

    -----

    We know  \displaystyle \Gamma(z+1) = z\Gamma(z) \implies f(x+1) + f(x)+\frac1x . Thus  f(\tfrac12) = f(\tfrac32)-2 = -\gamma-2\log2 .

    Finally  \displaystyle \Gamma'\left(\frac12\right) = \Gamma\left(\frac12\right)(-\gamma-2\log2) = -\sqrt{\pi}(\gamma+2\log2) .

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  2. #77
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by chiph588@ View Post
    Spoiler:


    Define  \displaystyle f(z) = \frac d{dz}\left(\log\Gamma(z)\right) = \frac{\Gamma'(z)}{\Gamma(z)}

    From the Gamma function duplication formula we see  \displaystyle f(z)+f(z+\tfrac12)+2\log2 = 2f(2z) . Thus  \displaystyle f(\tfrac32) = 2f(2)-f(1)-2\log2 .

    Now we must calculate  \Gamma'(1) and  \Gamma'(2) . Using a definition of  \Gamma , we see  \displaystyle f(s) = -\frac1s-\gamma-\sum_{n=1}^\infty\left(\frac1{n+s}-\frac1n\right)

    From this it's easy to see  \displaystyle f(1) = -\gamma and  \displaystyle f(2) = 1-\gamma \quad\text{\footnotesize(telescoping sum)} .

    So we see  f(\tfrac32) = 2-\gamma-2\log2

    -----

    We know  \displaystyle \Gamma(z+1) = z\Gamma(z) \implies f(x+1) + f(x)+\frac1x . Thus  f(\tfrac12) = f(\tfrac32)-2 = -\gamma-2\log2 .

    Finally  \displaystyle \Gamma'\left(\frac12\right) = \Gamma\left(\frac12\right)(-\gamma-2\log2) = -\sqrt{\pi}(\gamma+2\log2) .

    See, it wasn't so bad!
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  3. #78
    Super Member Random Variable's Avatar
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    I don't understand how you used the duplication formula. Could you explain? I'm stupid.

    Quote Originally Posted by chiph588@ View Post
    Spoiler:


    Define  \displaystyle f(z) = \frac d{dz}\left(\log\Gamma(z)\right) = \frac{\Gamma'(z)}{\Gamma(z)}

    From the Gamma function duplication formula we see  \displaystyle f(z)+f(z+\tfrac12)+2\log2 = 2f(2z) . Thus  \displaystyle f(\tfrac32) = 2f(2)-f(1)-2\log2 .

    Now we must calculate  \Gamma'(1) and  \Gamma'(2) . Using a definition of  \Gamma , we see  \displaystyle f(s) = -\frac1s-\gamma-\sum_{n=1}^\infty\left(\frac1{n+s}-\frac1n\right)

    From this it's easy to see  \displaystyle f(1) = -\gamma and  \displaystyle f(2) = 1-\gamma \quad\text{\footnotesize(telescoping sum)} .

    So we see  f(\tfrac32) = 2-\gamma-2\log2

    -----

    We know  \displaystyle \Gamma(z+1) = z\Gamma(z) \implies f(x+1) + f(x)+\frac1x . Thus  f(\tfrac12) = f(\tfrac32)-2 = -\gamma-2\log2 .

    Finally  \displaystyle \Gamma'\left(\frac12\right) = \Gamma\left(\frac12\right)(-\gamma-2\log2) = -\sqrt{\pi}(\gamma+2\log2) .

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  4. #79
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by Random Variable View Post
    I don't understand how you used the duplication formula. Could you explain. I'm stupid.
    You're not stupid, math is hard!

     \displaystyle 2\sqrt{\pi}\Gamma(2z) = 2^{2z}\Gamma(z)\Gamma(z+\tfrac12) \implies \log(2\sqrt{\pi})+\log\Gamma(2z) = \log 2^{2z} + \log\Gamma(z) + \log\Gamma(z+\tfrac12)

     \displaystyle \frac d{dz}\left(\log(2\sqrt{\pi})+\log\Gamma(2z)\right) = \frac d{dz}\left(2z\log 2 + \log\Gamma(z) + \log\Gamma(z+\tfrac12)\right)

     \displaystyle \implies 2f(2z) = 2\log2+f(z)+f(z+\tfrac12)
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  5. #80
    Super Member Random Variable's Avatar
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    More "fun" with the polygamma function.


    EDIT: Show that  \displaystyle \int^{\infty}_{0} \sin (x) x^{a-1} \ dx = \Gamma(a) \sin \big(\frac{\pi a}{2} \big) \ 0<a<1 .

    Then use the result to find  \displaystyle \int^{\infty}_{0} \sin(x^{2}) \ln^{2} x \ dx .
    Last edited by Random Variable; January 17th 2011 at 05:09 PM.
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  6. #81
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Random Variable View Post
    More "fun" with the polygamma function.


    EDIT: Show that  \displaystyle \int^{\infty}_{0} \sin (x) x^{a-1} \ dx = \Gamma(a) \sin \big(\frac{\pi a}{2} \big) \ 0<a<1 .

    Then use the result to find  \displaystyle \int^{\infty}_{0} \sin(x^{2}) \ln^{2} x \ dx .
    I'll leave the second part to someone else:

    Spoiler:



    Let

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


    Note then that



    \displaystyle I(a)=\frac{1}{\Gamma(1-a)}\int_0^{\infty}\sin(x)\int_0^{\infty} y^{-a}e^{-xy}dy dx


    Revering the order of integration gives


    \displaystyle I(a)=\frac{1}{\Gamma(1-a)}\int_0^{\infty}\int_0^{\infty}\sin(x) y^{-a}e^{-xy} dx dy=\frac{1}{\Gamma(1-a)}\int_0^{\infty}\frac{y^{-a}}{y^2+1}\text{ }dy


    This last integral is relatively easy (we've done it in this or the last thread I think) and thus we get that


    \displaystyle \begin{aligned}I(a) &=\frac{1}{\Gamma(1-a)}\frac{\pi}{2\cos\left(\frac{\pi a}{2}\right)}\\ &=\frac{\Gamma(a)}{\Gamma(a)\Gamma(1-a)}\frac{\pi}{2\cos\left(\frac{\pi a}{2}\right)}\\ &= \frac{\Gamma(a)}{\frac{\pi}{\sin(\pi a)}}\frac{\pi}{2\cos\left(\frac{\pi a}{2}\right)}\\ &= \Gamma(a)\frac{\sin(\pi a)}{2\cos\left(\frac{\pi a}{2}\right)}\\ &= \Gamma(a)\sin\left(\frac{\pi a}{2}\right)\end{aligned}

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  7. #82
    Super Member Random Variable's Avatar
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    I think the integral also converges for  -1<a<0 . But a=0 seems to be a problem even if you take the limit as a goes to zero.

    EDIT: I didn't make that clear. I meant before you manipulated your answer, should it be  \cos \Big( \frac{\pi a}{2} \Big) or  \sin \Big(\frac{\pi a}{2} \Big)  ?



    Quote Originally Posted by Drexel28 View Post
    I'll leave the second part to someone else:

    Spoiler:



    Let

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


    Note then that



    \displaystyle I(a)=\frac{1}{\Gamma(1-a)}\int_0^{\infty}\sin(x)\int_0^{\infty} y^{-a}e^{-xy}dy dx


    Revering the order of integration gives


    \displaystyle I(a)=\frac{1}{\Gamma(1-a)}\int_0^{\infty}\int_0^{\infty}\sin(x) y^{-a}e^{-xy} dx dy=\frac{1}{\Gamma(1-a)}\int_0^{\infty}\frac{y^{-a}}{y^2+1}\text{ }dy


    This last integral is relatively easy (we've done it in this or the last thread I think) and thus we get that


    \displaystyle \begin{aligned}I(a) &=\frac{1}{\Gamma(1-a)}\frac{\pi}{2\cos\left(\frac{\pi a}{2}\right)}\\ &=\frac{\Gamma(a)}{\Gamma(a)\Gamma(1-a)}\frac{\pi}{2\cos\left(\frac{\pi a}{2}\right)}\\ &= \frac{\Gamma(a)}{\frac{\pi}{\sin(\pi a)}}\frac{\pi}{2\cos\left(\frac{\pi a}{2}\right)}\\ &= \Gamma(a)\frac{\sin(\pi a)}{2\cos\left(\frac{\pi a}{2}\right)}\\ &= \Gamma(a)\sin\left(\frac{\pi a}{2}\right)\end{aligned}

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  8. #83
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Random Variable View Post
    I think the integral also converges for  -1<a<0 . But a=0 seems to be a problem even if you take the limit as a goes to zero.
    It converges for \text{Re}(a)\in(-1,1). When, a=0 this is the famous integral \displaystyle \int_0^{\infty}\frac{\sin(x)}{x}\text{ }dx=\frac{\pi}{2}.


    And should it be  \cos \Big( \frac{\pi a}{2} \Big) or  \sin \Big(\frac{\pi a}{2} \Big)  ?
    Are you talking about for the integral extended to negative values, or for my integral? For mine it is supposed to be \cos\left(\frac{\pi a}{2}\right)? Would you like a proof?
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  9. #84
    Super Member Random Variable's Avatar
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    I'm well aware of that integral. But does  \lim_{a \to 0} \Gamma(a) \sin \Big( \frac{\pi a}{2} \Big) = \frac{\pi}{2} ?

    And I thought  \displaystyle  \int^{\infty}_{0} \frac{y^{-a}}{1+y^{2}} \ dx = \frac{\frac{\pi}{2}}{\sin \frac{\pi (1-a)}{2}} = \frac{\frac{\pi}{2}}{\cos \frac{\pi a}{2}}. Nevermind.
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  10. #85
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Random Variable View Post
    I'm well aware of that integral. But does  \lim_{a \to 0} \Gamma(a) \sin \Big( \frac{\pi a}{2} \Big) = \frac{\pi}{2} ?
    Go back to extending \Gamma analytically to \mathbb{C} by the reflection formula! In particular, in my methodology above we see that \displaystyle \Gamma(a)\sin\left(\frac{\pi a}{2}\right)=\frac{\pi}{2}\frac{\Gamma(1-a)}{\cos\left(\frac{\pi a}{2}\right)}. What happens if you plug a=0 in there?
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  11. #86
    MHF Contributor chiph588@'s Avatar
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    For  \displaystyle k\in\mathbb{Z} , evaluate  \displaystyle \sum_{n=1}^\infty (-1)^nn^{-1+2\pi ik/\log2}

    Not too difficult, but the result is somewhat surprising if you've never seen it.
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  12. #87
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by chiph588@ View Post
    For  \displaystyle k\in\mathbb{Z} , evaluate  \displaystyle \sum_{n=1}^\infty (-1)^nn^{-1+2\pi ik/\log2}

    Not too difficult, but the result is somewhat surprising if you've never seen it.
    This was surprising!

    Spoiler:



    Merely note that for k\ne 0 we have that


    \displaystyle \sum_{n=1}^{\infty}(-1)^n n^{-1+\frac{2\pi i k}{\log(2)}}=\left(1-2^{\frac{2\pi i k}{\log(2)}\right)\zeta\left(1-\frac{2\pi i k}{\log(2)}\right)=\left(1-1\right)\zeta\left(1-\frac{2\pi i k}{\log(2)}\right)=0


    and for k=0 this is just the alternating harmonic series which evaluates to \log(2). It follows that



    \displaystyle \sum_{n=1}^{\infty}(-1)^{n}n^{-1+\frac{2\pi i k}{\log(2)}}=\begin{cases}0 & \mbox{if}\quad k\ne0\\ \log(2) & \mbox{if}\quad k=0\end{cases}

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  13. #88
    MHF Contributor Drexel28's Avatar
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    Try this:


    \displaystyle \sum_{n=0}^{\infty}\frac{(a)_n(b)_n}{(c)_nn!}\quad c-b-a>0
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  14. #89
    MHF Contributor Bruno J.'s Avatar
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    Quote Originally Posted by Drexel28 View Post
    Try this:


    \displaystyle \sum_{n=0}^{\infty}\frac{(a)_n(b)_n}{(c)_nn!}\quad c-b-a>0
    That's just Gauss' hypergeometric series...

    This thread has started as "Prove some identities", and has become "Use this transcendental function's properties as proven in Whittaker and Watson". There's nothing wrong with it, but it'd be fun to go back to some relatively simple stuff.

    Here's an easy problem I made up : show that, for any positive integer n, we have

    \displaystyle \sum_{k=2}^\infty\lfloor \log_k n\rfloor = \sum_{k=1}^\infty \lfloor n^{1/k}-1\rfloor
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  15. #90
    MHF Contributor Bruno J.'s Avatar
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    I didn't mean to kill this thread! Somebody must have solved the problem I gave but forgotten to post their solution... right? ^^
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