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Math Help - Prove some identities! (Round two)

  1. #61
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Random Variable View Post
    Sorry. I have another one.

     \displaystyle \int^{\infty}_{0} \frac{x^{a}}{(1+e^{x})^{2}} \ dx \ , \ a>0
    Thanks!

    Spoiler:


    Let


    \displaystyle I(a)=\int_0^{\infty}\frac{x^a}{\left(1+e^x\right)^  2}\text{ }dx


    Note then that


    \displaystyle I(a)=\int_0^{\infty}\frac{e^{-2x}x^a}{\left(1+e^{-x}\right)^2}\text{ }dx


    Noticing though (this isn't in full rigor, but it can be justified via limits) that for our bounds we have that


    \displaystyle \frac{1}{(1+e^{-x})^2}=\sum_{n=1}^{\infty} ne^{-(n-1)x}(-1)^n


    We may conclude that


    \displaystyle \begin{aligned}I(a) &= \int_0^{\infty}\frac{e^{-2x}x^a}{(1+e^{-x})^2}\text{ }dx\\ &= \int_0^{\infty}e^{-2x}x^a\sum_{n=1}^{\infty}ne^{-(n-1)x}(-1)^n\text{ }dx\\ &=\sum_{n=1}n(-1)^n\int_0^{\infty}x^a e^{-(n+1)x}\text{ }dx\\ &= \Gamma(a+1)\sum_{n=1}^{\infty}\frac{n(-1)^n}{(n+1)^{a+1}}\\ &=\Gamma(a+1)\left\{\sum_{n=1}^{\infty}\frac{(-1)^n}{(n+1)^a}-\sum_{n=1}^{\infty}\frac{(-1)^n}{(n+1)^{a+1}}\right\}\\ &= \Gamma(a+1)\left(\left(1-2^{-a}\right)\zeta(a+1)-\left(1-2^{-(a+1)}\right)\zeta(a)\right)\end{aligned}

    Last edited by Drexel28; January 13th 2011 at 07:17 PM.
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  2. #62
    Super Member Random Variable's Avatar
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    My solution only involves  \zeta(a) and  \Gamma(a+1). Maybe your solution can be simplified.

    EDIT: I found an error in my solution. Your solution is most likely correct.
    Last edited by Random Variable; January 13th 2011 at 06:03 PM.
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  3. #63
    Super Member Random Variable's Avatar
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    I would never have thought to break up the summation like that.
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  4. #64
    Super Member PaulRS's Avatar
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    Here's a nice series:

    Compute: \displaystyle\sum_{k=1}^{\infty}{\frac{(-1)^{k-1}}{k}\cdot \lfloor\log_2(k) \rfloor}
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  5. #65
    Super Member Random Variable's Avatar
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    Quote Originally Posted by PaulRS View Post
    Here's a nice series:

    Compute: \displaystyle\sum_{k=1}^{\infty}{\frac{(-1)^{k-1}}{k}\cdot \lfloor\log_2(k) \rfloor}
    Perhaps a hint?
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  6. #66
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by PaulRS View Post
    Here's a nice series:

    Compute: \displaystyle\sum_{k=1}^{\infty}{\frac{(-1)^{k-1}}{k}\cdot \lfloor\log_2(k) \rfloor}
    This proof is invalid because the series is conditionally convergent, but perhaps it can be extended by introducing an error term of some sort.

     \displaystyle\sum_{k=1}^{\infty}{\frac{(-1)^{k-1}}{k}\cdot \lfloor\log_2(k) \rfloor} = 1 - \log2 +\sum_{k=1}^{\infty}{\frac{(-1)^{k-1}}{k}\cdot (\lfloor\log_2(k) \rfloor-1)}

     \displaystyle = 1+\tfrac12 - \log2^2 +\sum_{k=1}^{\infty}{\frac{(-1)^{k-1}}{k}\cdot (\lfloor\log_2(k) \rfloor-2)} = 1+\tfrac12+\tfrac13+\tfrac14 - \log2^3 +\sum_{k=1}^{\infty}{\frac{(-1)^{k-1}}{k}\cdot (\lfloor\log_2(k) \rfloor-3)} = \ldots

     \displaystyle = H_{2^n} - \log2^n + \sum_{k=1}^{\infty}{\frac{(-1)^{k-1}}{k}\cdot (\lfloor\log_2(k) \rfloor-n)} = \lim_{n\to\infty}\left(H_{2^n-\log2^n}\right) = \gamma

    Edit: Wait... I'm not confident in this solution all of the sudden..
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  7. #67
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by chiph588@ View Post
    This proof is invalid because the series is conditionally convergent, but perhaps it can be extended by introducing an error term of some sort.

     \displaystyle\sum_{k=1}^{\infty}{\frac{(-1)^{k-1}}{k}\cdot \lfloor\log_2(k) \rfloor} = 1 - \log2 +\sum_{k=1}^{\infty}{\frac{(-1)^{k-1}}{k}\cdot (\lfloor\log_2(k) \rfloor-1)}

     \displaystyle = 1+\tfrac12 - \log2^2 +\sum_{k=1}^{\infty}{\frac{(-1)^{k-1}}{k}\cdot (\lfloor\log_2(k) \rfloor-2)} = 1+\tfrac12+\tfrac13+\tfrac14 - \log2^3 +\sum_{k=1}^{\infty}{\frac{(-1)^{k-1}}{k}\cdot (\lfloor\log_2(k) \rfloor-3)} = \ldots

     \displaystyle = H_{2^n} - \log2^n + \sum_{k=1}^{\infty}{\frac{(-1)^{k-1}}{k}\cdot (\lfloor\log_2(k) \rfloor-n)} = \lim_{n\to\infty}\left(H_{2^n-\log2^n}\right) = \gamma

    Edit: Wait... I'm not confident in this solution all of the sudden..
    I'm not quite sure how the last step (technically second to last, where you conclude that the sum but with \lfloor \lg(k)\rfloor is replaced with \lfloor\lg(k)\rfloor-n tends to zero) is achieved in your proof? Your answer is correct though. I'll leave it for someone else to do, unless no one else wants to?
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  8. #68
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by Drexel28 View Post
    I'm not quite sure how the last step (technically second to last, where you conclude that the sum but with \lfloor \lg(k)\rfloor is replaced with \lfloor\lg(k)\rfloor-n tends to zero) is achieved in your proof? Your answer is correct though. I'll leave it for someone else to do, unless no one else wants to?
    Yeah, that 'proof' is complete garbage...
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  9. #69
    Super Member PaulRS's Avatar
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    Quote Originally Posted by chiph588@ View Post
    This proof is invalid because the series is conditionally convergent, but perhaps it can be extended by introducing an error term of some sort.

     \displaystyle\sum_{k=1}^{\infty}{\frac{(-1)^{k-1}}{k}\cdot \lfloor\log_2(k) \rfloor} = 1 - \log2 +\sum_{k=1}^{\infty}{\frac{(-1)^{k-1}}{k}\cdot (\lfloor\log_2(k) \rfloor-1)}

     \displaystyle = 1+\tfrac12 - \log2^2 +\sum_{k=1}^{\infty}{\frac{(-1)^{k-1}}{k}\cdot (\lfloor\log_2(k) \rfloor-2)} = 1+\tfrac12+\tfrac13+\tfrac14 - \log2^3 +\sum_{k=1}^{\infty}{\frac{(-1)^{k-1}}{k}\cdot (\lfloor\log_2(k) \rfloor-3)} = \ldots

     \displaystyle = H_{2^n} - \log2^n + \sum_{k=1}^{\infty}{\frac{(-1)^{k-1}}{k}\cdot (\lfloor\log_2(k) \rfloor-n)} = \lim_{n\to\infty}\left(H_{2^n-\log2^n}\right) = \gamma

    Edit: Wait... I'm not confident in this solution all of the sudden..
    If you read carefully, the first line implies: \log(2) = \frac{1}{2} ... problem with the signs and the indices of the sums.

    Hint:
    Spoiler:
    try to exploit that \lfloor \log_2(n) \rfloor is constant in certain intervals.
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  10. #70
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by PaulRS View Post
    try to exploit that \lfloor \log_2(n) \rfloor is constant in certain intervals.
    That's what I tried to do above... I just did a very bad job at it haha. Let me try again.

    Define  \displaystyle s_n = \sum_{k=1}^{2^n}\frac{(-1)^{k-1}}k and  r_n = \log2-s_n .

    Observe  \displaystyle \lim_{n\to\infty} nr_n = 0 \implies \gamma = \lim_{n\to\infty} (H_{2^n}-n\log2) = \lim_{n\to\infty} (H_{2^n}-ns_n) .

    Now note  \displaystyle H_{2^n}-ns_n - H_2-s_1 = \sum_{k=1}^{n-1} (H_{2^{k+1}}-(k+1)s_{k+1}-H_{2^k}+ks_k) implies  \displaystyle \gamma = 1-\sum_{n=1}^\infty n(s_{n+1}-s_n) = 1-\sum_{n=1}^\infty \sum_{k=2^{n-1}+1}^{2^n} \frac n{2k(2k+1)}

    ------

    Going along the hint we get  \displaystyle \sum_{k=1}^{\infty}{\frac{(-1)^k}{k}\cdot \lfloor\log_2(k)\rfloor = \sum_{n=1}^\infty n\sum_{k=2^n}^{2^{n+1}-1} \frac{(-1)^k}{k} = \sum_{n=1}^\infty n\sum_{k=2^{n-1}}^{2^n-1} \frac1{2k(2k+1)}

     \displaystyle = 1-\sum_{n=1}^\infty \sum_{k=2^{n-1}+1}^{2^n} \frac n{2k(2k+1)} = \gamma
    Last edited by chiph588@; January 16th 2011 at 03:37 PM.
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  11. #71
    MHF Contributor chiph588@'s Avatar
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    Compute  \displaystyle \int_0^1 \frac{1-e^{-t}}t dt -\int_1^\infty \frac{e^{-t}}t dt
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  12. #72
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by chiph588@ View Post
    Compute  \displaystyle \int_0^1 \frac{1-e^{-t}}t dt -\int_1^\infty \frac{e^{-t}}t dt
    Solution:

    Spoiler:


    Recall the common fact that \Gamma'(1)=-\gamma. Note though that



    \displaystyle \Gamma'(x)=\frac{d}{dx}\int_0^{\infty}t^{x-1}e^{-t}\text{ }dt=\int_0^{\infty}\frac{\partial }{\partial x}t^{x-1}e^{-t}\text{ }dt=\int_0^{\infty}t^{x-1}\log(t)e^{-t}\text{ }dt


    and thus



    \displaystyle -\gamma=\int_0^{\infty}\log(t)e^{-t}\text{ }dt=\int_0^1 \log(t)e^{-t}\text{ }dt+\int_1^{\infty}\log(t)e^{-t}\text{ }dt


    integrate by parts in each to get that


    \displaystyle -\gamma=\left((-e^{-t}+1)\log(t)\right)\bigg\mid_0^1-\int_0^1\frac{e^{-t}+1}{t}\text{ }dt+\left(-e^{-t}\log(t)\right)\bigg\mid_1^{\infty}-\int_1^{\infty}\frac{e^{-t}}{t}\text{ }dt


    Note though that both the non-integral evaluations are zero and so we may conclude that


    \displaystyle -\gamma=\int_0^1\frac{e^{-t}-1}{t}\text{ }dt+\int_1^{\infty}\frac{e^{-t}}{t}\text{ }dt


    Multiplying both sides by -1 gives the correct result.
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  13. #73
    MHF Contributor Drexel28's Avatar
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    What about \displaystyle \int_0^{\infty}e^{-x^2}\log(x)\text{ }dx?
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  14. #74
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by Drexel28 View Post
    What about \displaystyle \int_0^{\infty}e^{-x^2}\log(x)\text{ }dx?
    Spoiler:


    Consider  \displaystyle I(a) = \int_0^\infty x^ae^{-x^2} dx , we're after  \displaystyle I'(0) .

    Let  \displaystyle x=\sqrt{t} \implies dx = \frac{dt}{2\sqrt{t}} , so  \displaystyle I(a) = \frac12\int_0^\infty t^{(a-1)/2}e^{-t} dt = \frac12\Gamma\left(\frac{a+1}2\right) .

    Thus  \displaystyle I'(a) = \frac14\Gamma'\left(\frac{a+1}2\right) .

    So  \displaystyle I'(0) = \frac14\Gamma'\left(\frac12\right) =-\frac14\sqrt{\pi}(\gamma+2\log2) . (the last equality is not trivial )

    Last edited by chiph588@; January 16th 2011 at 11:46 PM.
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  15. #75
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by chiph588@ View Post
    [spoiler]

    (the last equality is not trivial )
    So then I suppose it will be impressive when you prove it, right!
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