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Thread: sin(1/x)

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

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

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

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

    So my guess is that $\displaystyle \sin(1)-Ci(1) \approx .5042 $, otherwise it would imply $\displaystyle Ci(1) = \sin(1)-\frac{1}{2} $ which I doubt is true.
    Last edited by chiph588@; Jan 2nd 2009 at 04:53 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 $\displaystyle \lim_{a \to 0^{+}}\int_{a}^1 \sin(\frac{1}{x}) dx $ as $\displaystyle .5042 $ using a Riemann sum applet. So i was wondering if $\displaystyle \lim_{a \to 0^{+}}\int_{a}^1 \sin(\frac{1}{x}) dx = \frac{1}{2}$ or if this is just coincidence.

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

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

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

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

    As you said mathematica gave you

    $\displaystyle \lim_{a \to 0^{+}}\int_{a}^1 \sin(\frac{1}{x}) dx = .5042 $ actually $\displaystyle .5040670619$ 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 $\displaystyle \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 $\displaystyle \lim_{a \to 0^{+}}\int_{a}^1 \sin(\frac{1}{x}) dx $ as $\displaystyle .5042 $ using a Riemann sum applet. So i was wondering if $\displaystyle \lim_{a \to 0^{+}}\int_{a}^1 \sin(\frac{1}{x}) dx = \frac{1}{2}$ or if this is just coincidence.

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

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

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

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

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