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Math Help - Prove some identities! (Round two)

  1. #46
    Super Member Random Variable's Avatar
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     \displaystyle \int^{\infty}_{0} \frac{x^{a-1} \ln x}{1+x} \ dx \ \text{for} \ 0<a<1
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  2. #47
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Random Variable View Post
     \displaystyle \int^{\infty}_{0} \frac{x^{a-1} \ln x}{1+x} \ dx \ \text{for} \ 0<a<1
    Spoiler:

    Let

    \displaystyle I(a)=\int_0^{\infty}\frac{x^{a-1}}{1+x}\text{ }dx\quad a\in(0,1)

    Let x^{a-1}=u then x=u^{\frac{1}{a-1}} and so \displaystyle dx=\frac{1}{(a-1)}u^{\frac{1}{a-1}-1} du and thus


    \displaystyle I(a)=\int_0^{\infty}\frac{u}{1+u^{\frac{1}{a-1}}} \frac{1}{(a-1)}u^{1-\frac{1}{1-a}}\text{ }du=\frac{1}{a-1}\int_0^{\infty}\frac{du}{1+u^{\frac{1}{1-a}}}


    but by an earlier post this is equal to \displaystyle I(a)==\frac{\pi}{\sin(\pi a)}. Thus,



    \displaystyle -\pi^2\cot(\pi a)\csc(\pi a)=I'(a)=\int_0^{\infty}\frac{x^{a-1}\log(x)}{1+x}\text{ }dx


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  3. #48
    Super Member Random Variable's Avatar
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    I used contour integration and the Pac-Man contour to find I(a), unaware that by substitution I could make it look like that integral I asked about.

    You and simplependulum are too good. I give up.
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  4. #49
    Super Member Random Variable's Avatar
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    Doesn't  \displaystyle  \frac{1}{a-1} \int^{\infty}_{0} \frac{1}{1+ u^{\frac{1}{1-a}}} \ du = \frac{1}{a-1}  \frac{\pi (1-a)}{\sin \pi(1-a)} = -\frac{\pi}{\sin \pi a} ?
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  5. #50
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Random Variable View Post
    Doesn't  \displaystyle  \frac{1}{a-1} \int^{\infty}_{0} \frac{1}{1+ u^{\frac{1}{1-a}}} \ du = \frac{1}{a-1}  \frac{\pi (1-a)}{\sin \pi(1-a)} = -\frac{\pi}{\sin \pi a} ?
    See! we all make stupid mistakes sometimes (especially me ). Find mine! .Meanwhile, here is a correct proof:


    Spoiler:



    Let


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


    Let x^a=u and thus x^{a-1}=u^{1-\frac{1}{a}} and \displaystyle dx=\frac{1}{a}u^{\frac{1}{a}-1}. Thus,



    \displaystyle I(a)=\frac{1}{a}\int_0^{\infty}\frac{u^{1-\frac{1}{a}}}{1+u^{\frac{1}{a}}}u^{\frac{1}{a}-1}\text{ }dx=\frac{1}{a}\int_0^{\infty}\frac{du}{1+u^{\frac  {1}{a}}}


    So, by the same previous problem \displaystyle I(a)=\frac{\pi}{\sin(\pi a)} and thus


    \displaystyle -\pi^2\cot(\pi a)\csc(\pi a)=I'(a)=\int_0^{\infty}\frac{x^{a-1}\log(x)}{1+x}\text{ }dx


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  6. #51
    Super Member Random Variable's Avatar
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    The only other thing I see wrong in your first solution is what you substituted back in for dx. But even if you did that correctly, you would still get the negative of what you should get. So what else is wrong?
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  7. #52
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Random Variable View Post
    The only other thing I see wrong in your first solution is what you substituted back in for dx. But even if you did that correctly, you would still get the negative of what you should get. So what else is wrong?
    I let u=x^{a-1} and since a<1 we have that x^{a-1}\to 0 as x\to\infty, not \infty!
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  8. #53
    Super Member Random Variable's Avatar
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    So that's where the negative was hiding. Sneaky bastard.

    EDIT: I was referring to the negative; not you.
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  9. #54
    Super Member Random Variable's Avatar
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    How about  \displaystyle \int^{\infty}_{0} \frac{x^{a-1} \ln x}{1+x^{2}} \ dx \ , \ \text{for} \ 0<a<2

    This one I worked out without using contour integration.
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  10. #55
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Random Variable View Post
    How about  \displaystyle \int^{\infty}_{0} \frac{x^{a-1} \ln x}{1+x^{2}} \ dx \ , \ \text{for} \ 0<a<2

    This one I worked out without using contour integration.
    How about more generally:


    Spoiler:



    Let


    \displaystyle I(a,m)=\int_0^{\infty}\frac{x^{a-1}}{1+x^m},\quad m>1,\;\; 0<a<m



    Then, let x^m=u to get that



    \displaystyle I(a,m)=\int_0^{\infty}\frac{u^{\frac{a-1}{m}}}{1+u}\frac{du}{m u^{1-\frac{1}{m}}}=\frac{1}{m}\int_0^{\infty}\frac{u^{\  frac{a}{m}-1}}{1+u}\text{ }du



    Note though that since \displaystyle 0<\frac{a}{m}<1 we may apply the methods in the previous problem to conclude that


    \displaystyle I(a,m)=\frac{1}{m}\frac{\pi}{\sin\left(\frac{\pi a}{m}}\right)}


    Thus,


    \displaystyle \int_0^{\infty}\frac{x^{a-1}\log(x)}{1+x^m}\text{ }dx=\frac{\partial}{\partial a}I(a,m)=-\left(\frac{\pi}{m}\right)^2\cot\left(\frac{\pi a}{m}\right)\csc\left(\frac{\pi a}{m}\right)


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  11. #56
    Super Member Random Variable's Avatar
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    I can generalize my solution, which doesn't rely on the previous problem.


     \displaystyle \int^{\infty}_{0} \frac{x^{a-1}}{1+x^{m}} \ dx = \int^{\infty}_{0} \int^{\infty}_{0} x^{a-1} e^{-(1+x^m)t} \ dt \ dx

     \displaystyle = \int^{\infty}_{0} \int^{\infty}_{0} e^{-t} x^{a-1} e^{-tx^{m}} \ dt \ dx

    reverse the order the integration and let  \displaystyle  x = \Big(\frac{u}{t}\Big)^{1/m}

     \displaystyle  = \frac{1}{m}\int^{\infty}_{0} \int^{\infty}_{0} e^{-t} \Big(\frac{u}{t}\Big)^{\frac{1}{m}(a-1)} e^{-u} \Big(\frac{u}{t}\Big)^{\frac{1}{m} -1} \ \frac{1}{t} \ dt \ du

     \displaystyle =  \frac{1}{m} \int^{\infty}_{0} t^{-\frac{a}{m}} \ e^{-t} \ dt \int^{\infty}_{0} u^{\frac{a}{m} - 1} \ e^{-u} \ du

     = \frac{1}{m} \Gamma(1- \frac{a}{m}) \Gamma(\frac{a}{m}) =  \frac{1}{m} \pi \csc (\frac{\pi a}{m})
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  12. #57
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Random Variable View Post
    I can generalize my solution, which doesn't rely on the previous problem.


     \displaystyle \int^{\infty}_{0} \frac{x^{a-1}}{1+x^{m}} \ dx = \int^{\infty}_{0} \int^{\infty}_{0} x^{a-1} e^{-(1+x^m)t} \ dt \ dx

     \displaystyle = \int^{\infty}_{0} \int^{\infty}_{0} e^{-t} x^{a-1} e^{-tx^{m}} \ dt \ dx

    reverse the order the integration and let  \displaystyle  x = \Big(\frac{u}{t}\Big)^{1/m}

     \displaystyle  = \frac{1}{m}\int^{\infty}_{0} \int^{\infty}_{0} e^{-t} \Big(\frac{u}{t}\Big)^{\frac{1}{m}(a-1)} e^{-u} \Big(\frac{u}{t}\Big)^{\frac{1}{m} -1} \ \frac{1}{t} \ dt \ du

     \displaystyle =  \frac{1}{m} \int^{\infty}_{0} t^{-\frac{a}{m}} \ e^{-t} \ dt \int^{\infty}_{0} u^{\frac{a}{m} - 1} \ e^{-u} \ du

     = \frac{1}{m} \Gamma(1- \frac{a}{m}) \Gamma(\frac{a}{m}) =  \frac{1}{m} \pi \csc (\frac{\pi a}{m})
    Nice!
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  13. #58
    Super Member Random Variable's Avatar
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    EDIT: On second thought, it's really not that interesting. Nevermind.
    Last edited by Random Variable; January 12th 2011 at 12:55 PM.
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  14. #59
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Random Variable View Post
    EDIT: On second thought, it's really not that interesting. Nevermind.
    Way to get me excited only to yank it away!
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  15. #60
    Super Member Random Variable's Avatar
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    Quote Originally Posted by Drexel28 View Post
    Way to get me excited only to yank it away!
    Sorry. I have another one.

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