1. ## 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.

2. Originally Posted by chiph588@
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.

3. Ok, then its just coincidence the answer is close to $\frac{1}{2}$.

4. Originally Posted by chiph588@
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.