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Math Help - sin(1/x)

  1. #1
    MHF Contributor chiph588@'s Avatar
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    sin(1/x)

    Today I approximated  \lim_{a \to 0^{+}}\int_{a}^1 \sin(\frac{1}{x}) dx as  .5042 using a Riemann sum applet. So i was wondering if  \lim_{a \to 0^{+}}\int_{a}^1 \sin(\frac{1}{x}) dx = \frac{1}{2} or if this is just coincidence.

    Web Mathematica says \int \sin(\frac{1}{x}) dx = x\sin(\frac{1}{x})-Ci(\frac{1}{x}), where  Ci(r) = -\int_{r}^\infty \frac{\cos(t)}{t} dt .

     \lim_{a \to 0^{+}}\int_{a}^1 \sin(\frac{1}{x}) dx = \sin(1)-Ci(1)-\lim_{a \to 0^{+}}x\sin(\frac{1}{x})+\lim_{t \to \infty}Ci(t) = \sin(1)-Ci(1)

    So my guess is that  \sin(1)-Ci(1) \approx .5042 , otherwise it would imply  Ci(1) = \sin(1)-\frac{1}{2} which I doubt is true.
    Last edited by chiph588@; January 2nd 2009 at 05:53 PM.
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  2. #2
    MHF Contributor
    Jester's Avatar
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    Quote Originally Posted by chiph588@ View Post
    Today I approximated  \lim_{a \to 0^{+}}\int_{a}^1 \sin(\frac{1}{x}) dx as  .5042 using a Riemann sum applet. So i was wondering if  \lim_{a \to 0^{+}}\int_{a}^1 \sin(\frac{1}{x}) dx = \frac{1}{2} or if this is just coincidence.

    Web Mathematica says \int \sin(\frac{1}{x}) dx = x\sin(\frac{1}{x})-Ci(\frac{1}{x}), where  Ci(r) = -\int_{r}^\infty \frac{\cos(t)}{t} dt .

     \lim_{a \to 0^{+}}\int_{a}^1 \sin(\frac{1}{x}) dx = \sin(1)-Ci(1)-\lim_{a \to 0^{+}}x\sin(\frac{1}{x})+\lim_{t \to \infty}Ci(t) = \sin(1)-Ci(1)

    So my guess is that  \sin(1)-Ci(1) \approx .5042 , otherwise it would imply  Ci(1) = \sin(1)-\frac{1}{2} which I doubt is true.
    Not sure where you got

     \lim_{a \to 0^{+}}\int_{a}^1 \sin(\frac{1}{x}) dx = \frac{1}{2} ?

    As you said mathematica gave you

     \lim_{a \to 0^{+}}\int_{a}^1 \sin(\frac{1}{x}) dx = .5042 actually .5040670619 from Maple.
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  3. #3
    MHF Contributor chiph588@'s Avatar
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    Ok, then its just coincidence the answer is close to  \frac{1}{2} .
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by chiph588@ View Post
    Today I approximated  \lim_{a \to 0^{+}}\int_{a}^1 \sin(\frac{1}{x}) dx as  .5042 using a Riemann sum applet. So i was wondering if  \lim_{a \to 0^{+}}\int_{a}^1 \sin(\frac{1}{x}) dx = \frac{1}{2} or if this is just coincidence.

    Web Mathematica says \int \sin(\frac{1}{x}) dx = x\sin(\frac{1}{x})-Ci(\frac{1}{x}), where  Ci(r) = -\int_{r}^\infty \frac{\cos(t)}{t} dt .

     \lim_{a \to 0^{+}}\int_{a}^1 \sin(\frac{1}{x}) dx = \sin(1)-Ci(1)-\lim_{a \to 0^{+}}x\sin(\frac{1}{x})+\lim_{t \to \infty}Ci(t) = \sin(1)-Ci(1)

    So my guess is that  \sin(1)-Ci(1) \approx .5042 , otherwise it would imply  Ci(1) = \sin(1)-\frac{1}{2} which I doubt is true.
    Why not just use a little algebra to see if its feasible?

    \text{Ci}(1)=\sin(1)-\frac{1}{2}\implies \text{Ci}(1)=\int_0^1\left\{\cos(x)-x\right\}dx=

    Check to see if it can be transformed into \text{Ci}(1)...I dont think so though.
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