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Math Help - Prove the improper integral below

  1. #1
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    Prove the improper integral below

    Prove that

    <br />
\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)}}


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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

    <br />
\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)}}


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    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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    <br />
I = \int_0^\infty  {\tfrac{{\ln \left( x \right)}}<br />
{{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} <br />
<br />
 = \int_0^\infty  {e^{ - y}  \cdot \left( {\int_0^\infty  {\ln \left( x \right)e^{ - y \cdot x^a } dx} } \right)dy} <br />

    Note that: <br />
\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} }}<br />
{{\partial s}}dx}  = \int_0^\infty  {\ln \left( x \right) \cdot x^{s - 1}  \cdot e^{ - x} dx} <br />

    Now: <br />
\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} <br />
and: <br />
\int_0^\infty  {\ln \left( x \right)e^{ - x^a } dx} \underbrace  = _{x^a  = u}\tfrac{1}<br />
{{a^2 }} \cdot \int_0^\infty  {\ln \left( u \right) \cdot u^{\tfrac{1}<br />
{a} - 1}  \cdot e^{ - u} d} u = \tfrac{1}<br />
{{a{}^2}} \cdot \Gamma '\left( {\tfrac{1}<br />
{a}} \right)<br />

    Thus we have, by properties of the loagrithm: <br />
\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}  = <br />
<br />
\tfrac{{\sqrt[a]{y} \cdot \ln \left( y \right)}}<br />
{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} <br />

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

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

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

    By Euler's Reflection Formula: <br />
\Gamma \left( x \right) \cdot \Gamma \left( {1 - x} \right) = \frac{\pi }<br />
{{\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)}}<br />
{{\sin ^2 \left( {\pi  \cdot x} \right)}}<br />

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