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Math Help - Improper integral

  1. #1
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    Improper integral

    Evaluate \int_0^1\ln x\,dx
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by liyi View Post
    Evaluate \int_0^1\ln x\,dx
    \int_0^1ln(x)~dx = \lim_{a \to 0}\int_a^1ln(x)~dx

    = \lim_{a \to 0}(x~ln(x) - x)|_a^1 = -1 - \lim_{a \to 0}(a~ln(a) - a)

    So what is \lim_{a \to 0}(a~ln(a) - a)? Obviously this translates to the problem of finding
    \lim_{a \to 0}a~ln(a)

    This is a 0 \cdot \infty form, so try the following:
    \lim_{a \to 0}a~ln(a) = \lim_{a \to 0}\frac{ln(a)}{\frac{1}{a}}

    By L'Hopital's rule
    = \lim_{a \to 0} \frac{\frac{1}{a}}{- \frac{1}{a^2}} = \lim_{a \to 0}(-a) = 0

    So finally:
    \int_0^1ln(x)~dx = \lim_{a \to 0}\int_a^1ln(x)~dx = -1 - \lim_{a \to 0}(a~ln(a) - a) = -1 - 0 = -1

    -Dan
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  3. #3
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    \ln x=\int_1^x\frac1u\,du. So,

    \begin{aligned}<br />
\int_0^1 {\ln x\,dx} &= - \int_0^1 {\int_x^1 {\frac{1}<br />
{u}\,du} \,dx}\\<br />
&= - \int_0^1 {\int_0^u {\frac{1}<br />
{u}\,dx} \,du}\\<br />
&= - \int_0^1 {du}\\<br />
&= - 1.<br />
\end{aligned}
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  4. #4
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    Quote Originally Posted by liyi View Post
    Evaluate \int_0^1\ln x\,dx
    \int_0^1 \ln x dx = \int_0^1 \ln [(x-1)+1] dx, \mbox{ let }t=x-1

    \int_{-1}^0 \ln (t+1) dt = \int_{-1}^0  t - \frac{t^2}{2}+\frac{t^3}{3} - ... dt

    \frac{t^2}{1\cdot 2} - \frac{t^3}{2\cdot 3} + \frac{t^4}{3\cdot 4}-... \big|_{-1}^0 = -\left( \frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+... \right)

    = - \sum_{n=1}^{\infty} \frac{1}{n(n+1)} = -1
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  5. #5
    Forum Admin topsquark's Avatar
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    Gee, I'm starting to feel kind of mundane here...

    -Dan
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by topsquark View Post
    Gee, I'm starting to feel kind of mundane here...

    -Dan
    hehe, you and me both. i would have done it exactly the way you did
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