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**chiph588@** 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.