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Math Help - Today's calculation of integral #18

  1. #1
    Math Engineering Student
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    Today's calculation of integral #18

    Find \int_{0}^{\pi /2}{\frac{\tan ^{\alpha }x-1}{\ln (\tan x)}\,dx},\,0<\alpha <1.
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  2. #2
    Super Member wingless's Avatar
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    Quote Originally Posted by Krizalid View Post
    Find \int_{0}^{\pi /2}{\frac{\tan ^{\alpha }x-1}{\ln (\tan x)}\,dx},\,0<\alpha <1.
    \int_{0}^{\pi /2}{\frac{\tan^{n}x-1}{\ln (\tan x)}\,dx}

    \frac{\tan ^{n}x-1}{\ln (\tan x)} = \int_0^n \tan^y x~dy

    It becomes,

    \int_{0}^{\pi /2} \int_0^n \tan^y x~dy ~dx

    \int_0^n \int_{0}^{\pi /2}  \tan^y x~dx ~dy

    I'm not in a condition to go further. It's been too late I better go sleep some ..
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  3. #3
    Math Engineering Student
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    It's okay. That was exactly the idea.
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Krizalid View Post
    It's okay. That was exactly the idea.
    Haha...I cant remember the last integral you put up that wasn't double integration
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