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Math Help - integral

  1. #1
    Newbie AgentSmith's Avatar
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    Exclamation integral

    How do I find integrals like this?

    \int_{0}^{1}\frac{\log(x)}{x-1}dx
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Re: integral

    If you mean what is the definition of this kind of improper integral: f(x)=\frac{\log x}{x-1} is continuous in (0,1] , as a consequence integrable in [\epsilon,1]\;(0<\epsilon\leq 1) ) . Besides, \lim_{x\to 0^+}f(x)=+\infty so \int_0^1f(x)=\lim_{\epsilon\to 0^+}\int_{\epsilon}^1f(x)\;dx (by definition).
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  3. #3
    Member sbhatnagar's Avatar
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    Re: integral

    Quote Originally Posted by AgentSmith View Post
    How do I find integrals like this?

    \int_{0}^{1}\frac{\log(x)}{x-1}dx
    Are you asked to solve this integral?
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  4. #4
    Newbie AgentSmith's Avatar
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    Re: integral

    Quote Originally Posted by sbhatnagar View Post
    Are you asked to solve this integral?
    yes
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  5. #5
    MHF Contributor chisigma's Avatar
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    Re: integral

    Some time ago I 'discovered' the general formula...

    \int_{0}^{1} x^{m}\ \ln ^{n} x\ dx = (-1)^{n}\ \frac{n!}{(m+1)^{n+1}} (1)

    ... and if we take into account that...

    \frac{1}{1-x}= \sum_{m=0}^{\infty} x^{m} (2)

    ... we obtain...

    \int_{0}^{1} \frac{\ln x}{x-1}\ dx = \sum_{m=0}^{\infty} \frac{1}{(m+1)^{2}}= \frac{\pi^{2}}{6}

    Kind regards

    \chi \sigma
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  6. #6
    Member sbhatnagar's Avatar
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    Re: integral

    Here's what I did...

    1.
    \begin{align*} I=\int_{0}^{1}\frac{\ln(x)}{(x-1)}dx &= \int_{0}^{1} \frac{(x-1)-\frac{1}{2}(x-1)^2+\frac{1}{3}(x-1)^3-\cdots}{x-1}dx \\ &= \int_{0}^{1} \left[ (1-\frac{1}{2}(x-1)+\frac{1}{3}(x-1)^2-\cdots\right]dx\end{align*}

    2. Using the property \int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx, I got

    \begin{align*} I &= \int_{0}^{1}\left[ 1-\frac{1}{2}(-x)+\frac{1}{3}(-x)^2-\frac{1}{4}(-x)^3+\cdots\right] dx \\ &= \int_{0}^{1}\left[ 1+\frac{1}{2}(x)+\frac{1}{3}(x)^2+\frac{1}{4}(x)^3  +\cdots\right] dx \\ &= (1-0)+\frac{1}{2}\Big( \frac{1}{2} -0\Big) + \frac{1}{3}\Big( \frac{1}{3}-0\Big) + \frac{1}{4}\Big( \frac{1}{4}-0\Big)+\cdots \\ &= \sum_{k=1}^{\infty}\frac{1}{k^2} \\ &= \frac{\pi^2}{6}\end{align*}
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  7. #7
    MHF Contributor chisigma's Avatar
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    Re: integral

    Quote Originally Posted by chisigma View Post
    Some time ago I 'discovered' the general formula...

    \int_{0}^{1} x^{m}\ \ln ^{n} x\ dx = (-1)^{n}\ \frac{n!}{(m+1)^{n+1}} (1)

    ... and if we take into account that...

    \frac{1}{1-x}= \sum_{m=0}^{\infty} x^{m} (2)

    ... we obtain...

    \int_{0}^{1} \frac{\ln x}{x-1}\ dx = \sum_{m=0}^{\infty} \frac{1}{(m+1)^{2}}= \frac{\pi^{2}}{6}
    ... of course it is not a difficult task to extend the concept arriving, starting from (1), to the more general formula...

    \int_{0}^{1} \frac{\ln^{n} x}{1-x}\ dx= (-1)^{n}\ n!\ \zeta(n+1)

    ... where \zeta(*) is the 'Riemann Zeta Function'...

    Kind regards

    \chi \sigma
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