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Thread: Prove the improper integral below

  1. #1
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    Prove the improper integral below

    Prove that

    $\displaystyle
    \int_0^{\infty}\frac{\ln(x)}{1+x^s}dx=-\left(\frac{\pi}{s}\right)^2\cdot{\cot\left(\frac{ \pi}{s}\right)\cdot{\text{cosec}\left(\frac{\pi}{s }\right)}}$


    $\displaystyle s > 1 $
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by mathwizard View Post
    Prove that

    $\displaystyle
    \int_0^{\infty}\frac{\ln(x)}{1+x^s}dx=-\left(\frac{\pi}{s}\right)^2\cdot{\cot\left(\frac{ \pi}{s}\right)\cdot{\text{cosec}\left(\frac{\pi}{s }\right)}}$


    $\displaystyle s > 1 $
    haha, that's a question for PaulRS. it's his signature
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  3. #3
    Super Member PaulRS's Avatar
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    $\displaystyle
    I = \int_0^\infty {\tfrac{{\ln \left( x \right)}}
    {{1 + x^a }}dx} = \int_0^\infty {\ln \left( x \right)\left( {\int_0^\infty {e^{ - y \cdot \left( {1 + x^a } \right)} dy} } \right)dx}
    $$\displaystyle
    = \int_0^\infty {e^{ - y} \cdot \left( {\int_0^\infty {\ln \left( x \right)e^{ - y \cdot x^a } dx} } \right)dy}
    $

    Note that: $\displaystyle
    \Gamma \left( s \right) = \int_0^\infty {x^{s - 1} \cdot e^{ - x} dx} \Rightarrow \Gamma '\left( s \right) = \int_0^\infty {e^{ - x} \cdot \tfrac{{\partial x^{s - 1} }}
    {{\partial s}}dx} = \int_0^\infty {\ln \left( x \right) \cdot x^{s - 1} \cdot e^{ - x} dx}
    $

    Now: $\displaystyle
    \sqrt[a]{y} \cdot \int_0^\infty {\ln \left( {\sqrt[a]{y} \cdot x} \right)e^{ - \left( {\sqrt[a]{y} \cdot x} \right)^a } dx} = \int_0^\infty {\ln \left( x \right)e^{ - x^a } dx}
    $ and: $\displaystyle
    \int_0^\infty {\ln \left( x \right)e^{ - x^a } dx} \underbrace = _{x^a = u}\tfrac{1}
    {{a^2 }} \cdot \int_0^\infty {\ln \left( u \right) \cdot u^{\tfrac{1}
    {a} - 1} \cdot e^{ - u} d} u = \tfrac{1}
    {{a{}^2}} \cdot \Gamma '\left( {\tfrac{1}
    {a}} \right)
    $

    Thus we have, by properties of the loagrithm: $\displaystyle
    \sqrt[a]{y} \cdot \int_0^\infty {\ln \left( {\sqrt[a]{y} \cdot x} \right)e^{ - \left( {\sqrt[a]{y} \cdot x} \right)^a } dx} =
    $$\displaystyle
    \tfrac{{\sqrt[a]{y} \cdot \ln \left( y \right)}}
    {a} \cdot \int_0^\infty {e^{ - \left( {\sqrt[a]{y} \cdot x} \right)^a } dx} + \sqrt[a]{y} \cdot \int_0^\infty {\ln \left( x \right)e^{ - \left( {\sqrt[a]{y} \cdot x} \right)^a } dx}
    $

    So that: $\displaystyle
    \sqrt[a]{y} \cdot \int_0^\infty {\ln \left( {\sqrt[a]{y} \cdot x} \right)e^{ - \left( {\sqrt[a]{y} \cdot x} \right)^a } dx} =
    $$\displaystyle
    \tfrac{{\ln \left( y \right)}}
    {{a^2 }} \cdot \Gamma \left( {\tfrac{1}
    {a}} \right) + \sqrt[a]{y} \cdot \int_0^\infty {\ln \left( x \right)e^{ - \left( {\sqrt[a]{y} \cdot x} \right)^a } dx}
    $

    and: $\displaystyle
    \int_0^\infty {\ln \left( x \right)e^{ - \left( {\sqrt[a]{y} \cdot x} \right)^a } dx} = \frac{{\tfrac{1}
    {{a{}^2}} \cdot \Gamma '\left( {\tfrac{1}
    {a}} \right) - \tfrac{{\ln \left( y \right)}}
    {{a^2 }} \cdot \Gamma \left( {\tfrac{1}
    {a}} \right)}}
    {{\sqrt[a]{y}}}
    $

    Integrating: $\displaystyle
    I = \int_0^\infty {e^{ - y} \cdot \left( {\frac{{\tfrac{1}
    {{a{}^2}} \cdot \Gamma '\left( {\tfrac{1}
    {a}} \right) - \tfrac{{\ln \left( y \right)}}
    {{a^2 }} \cdot \Gamma \left( {\tfrac{1}
    {a}} \right)}}
    {{\sqrt[a]{y}}}} \right)dy} = \frac{{\Gamma '\left( {\tfrac{1}
    {a}} \right)\Gamma \left( {1 - \tfrac{1}
    {a}} \right) - \Gamma \left( {\tfrac{1}
    {a}} \right)\Gamma '\left( {1 - \tfrac{1}
    {a}} \right)}}
    {{a^2 }}
    $

    By Euler's Reflection Formula: $\displaystyle
    \Gamma \left( x \right) \cdot \Gamma \left( {1 - x} \right) = \frac{\pi }
    {{\sin \left( {\pi \cdot x} \right)}} \Rightarrow \Gamma '\left( x \right) \cdot \Gamma \left( {1 - x} \right) - \Gamma \left( x \right) \cdot \Gamma '\left( {1 - x} \right) = - \frac{{\pi ^2 \cdot \cos \left( {\pi \cdot x} \right)}}
    {{\sin ^2 \left( {\pi \cdot x} \right)}}
    $

    Thus: $\displaystyle
    {I = - \left( {\tfrac{\pi }
    {a}} \right)^2 \cdot \frac{{\cos \left( {\tfrac{\pi }
    {a}} \right)}}
    {{\sin ^2 \left( {\tfrac{\pi }
    {a}} \right)}} = - \left( {\tfrac{\pi }
    {a}} \right)^2 \cdot \cot \left( {\tfrac{\pi }
    {a}} \right) \cdot \text{cosec} \left( {\tfrac{\pi }
    {a}} \right)}
    $
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