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Math Help - Integral

  1. #1
    Super Member dhiab's Avatar
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    Integral

    Calculate : \int\limits_0^\infty {\left( {\frac{{\ln \left( x \right)}}{{x^2 + 1}}} \right)} dx<br />
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  2. #2
    MHF Contributor chisigma's Avatar
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    With the substitution \ln x=t \rightarrow x=e^{t} \rightarrow dx= t\cdot e^{t} the integral becomes...

    \int _{0}^{\infty} \frac{\ln x}{1+x^{2}}\cdot dx = \int_{-\infty}^{+\infty}\frac{t\cdot e^{t}}{1+e^{2t}}\cdot dt = \int_{-\infty}^{+\infty}\frac{t}{2\cdot \cosh t}\cdot dt

    But t is an odd function and \cosh t an even function, so that their ratio is an odd function, so that is...

    \int _{0}^{\infty} \frac{\ln x}{1+x^{2}}\cdot dx = 0

    Kind regards

    \chi \sigma
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  3. #3
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    Quote Originally Posted by dhiab View Post
    Calculate : \int\limits_0^\infty {\left( {\frac{{\ln \left( x \right)}}{{x^2 + 1}}} \right)} dx<br />
    If you let t=\frac{1}{x} then you get:
    \int_{\infty}^0 \frac{\log \left( \frac{1}{t}\right)}{t^2+1} dt = - \int_0^{\infty} \frac{\log t}{t^2+1}dt

    Thus, \int_0^{\infty} \frac{\log t}{t^2+1}dt =- \int_0^{\infty} \frac{\log t}{t^2+1}dt \implies \int_0^{\infty} \frac{\log t}{t^2+1}dt = 0
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