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Thread: How to solve improper integral ln(x) / (x^2 + 1)

  1. #1
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    Exclamation How to solve improper integral ln(x) / (x^2 + 1)

    Hello, guys, I just want you to help me of finding the integral of $\displaystyle \int_{0}^{1}\dfrac{\ln(x)}{x^2 + 1}\mathrm{d}x$. I just know that is improper.
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    Re: How to solve improper integral ln(x) / (x^2 + 1)



    Use the method of integration by parts. As above, let u = ln(x), dv = 1/(x^2+1). Let me know if you need more help.
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    Re: How to solve improper integral ln(x) / (x^2 + 1)

    Quote Originally Posted by lebdim View Post
    Hello, guys, I just want you to help me of finding the integral of $\displaystyle \int_{0}^{1}\dfrac{\ln(x)}{x^2 + 1}\mathrm{d}x$. I just know that is improper.
    Is your difficulty with the anti-derivative $\displaystyle \displaystyle\int {\frac{{\log (x)}}{{{x^2} + 1}}dx}~? $
    There is no elementary solution. The solution involves the Polylogarithm function: SEE HERE
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    Re: How to solve improper integral ln(x) / (x^2 + 1)

    integration by parts we get

    $\displaystyle -\int_0^1 \frac{\tan ^{-1}x}{x} \, dx$

    $\displaystyle -\int_0^1 \left(\sum _{k=0}^{\infty } \frac{(-1)^kx^{2k}}{(2k+1)}\right) \, dx$

    $\displaystyle -\sum _{k=0}^{\infty } \frac{(-1)^k}{(2k+1)}\int_0^1 x^{2k} \, dx$

    $\displaystyle -\sum _{k=0}^{\infty } \frac{(-1)^k}{(2k+1)^2}$

    $\displaystyle -Catalan$ constant

    $-0.915966$
    Thanks from romsek
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    Re: How to solve improper integral ln(x) / (x^2 + 1)

    Oh, thank you!
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