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  1. #1
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    Improper integral

    Evaluate \int_0^1\sin(\ln x)\,dx
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  2. #2
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    Quote Originally Posted by liyi View Post
    Evaluate \int_0^1\sin(\ln x)\,dx
    You can write, \int_0^1 \sin (\ln x) dx = \int_0^1 x\sin (\ln x) \cdot (\ln x)' dx

    Now let t=\ln x and we get, \int_{-\infty}^0 e^t \sin t ~dt

    This integral is doable.
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  3. #3
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    You could try it this way:

    \int_{0}^{1}sin(ln(x))dx

    \lim_{L\rightarrow{1}}\int_{0}^{L}sin(ln(x))dx

    One way is you can use Integration by Parts for this:

    Let u=sin(ln(x)), \;\ dv=dx, \;\ du=\frac{cos(ln(x))}{x}, \;\ v=x

    Then you get xsin(ln(x))-\int{cos(ln(x))}dx

    Now, use parts again:

    u=cos(ln(x)), \;\ dv=dx, \;\ du=\frac{-sin(ln(x))}{x}, \;\ v=x

    Putting it together, we end up with:

    \lim_{L\rightarrow{1}}\left[\frac{Lsin(ln(L))}{2}-\frac{Lcos(ln(L))}{2}\right]

    Now, as you can see, as L approaches 1, we have:

    0-\frac{1}{2}=\frac{-1}{2}
    Last edited by galactus; January 13th 2008 at 12:23 PM.
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  4. #4
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    Quote Originally Posted by liyi View Post
    Evaluate \int_0^1\sin(\ln x)\,dx
    Double integration leads TPH's integral too.

    \sin (\ln x) = \int_0^{\ln x} {\cos u\,du} . So

    \int_0^1 {\sin (\ln x)\,dx}  =  - \int_0^1 {\int_{\ln x}^0 {\cos u\,du} \,dx}  =  - \int_{ - \infty }^0 {\underbrace {\int_0^{e^u } {dx} }_{e^u }\cos u\,du} .

    Hence \int_0^1 {\sin (\ln x)\,dx}  =  - \int_{ - \infty }^0 {e^u \cos u\,du} .

    Now this is routine, substitute u=-v,

    - \int_0^\infty  {e^{ - v} \cos v\,dv}  =  - \text{Re} \int_0^\infty  {e^{ - (1 - i)v} \,dv}  =  - \text{Re}\, \frac{1}<br />
{{1 - i}} =  - \frac{1}<br />
{2}.

    The rest follows.
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