Is sin(1/x) measurable?

**Amanda1990** · February 10th 2009, 02:05 AM

I'm asked to find the Lebesgue integral of sin (1/x) on the interval (0,1], if it exists. But is sin(1/x) even a measurable function? It seems to me that the answer is no, but I cannot prove this.

**Laurent** · February 10th 2009, 02:29 AM

Originally Posted by Amanda1990

I'm asked to find the Lebesgue integral of sin (1/x) on the interval (0,1], if it exists. But is sin(1/x) even a measurable function? It seems to me that the answer is no, but I cannot prove this.

It is continuous on $(0,1]$ , hence it is of course measurable. In addition, it is bounded, so that it is Lebesgue-integrable on this bounded interval (Indeed, $\left|\sin\frac{1}{x}\right|\leq 1$ and $x\mapsto 1$ is integrable on $(0,1]$ , so that $x\mapsto\sin\frac{1}{x}$ is integrable on $(0,1]$ as well).

**Amanda1990** · February 10th 2009, 02:47 AM

If it is integrable, then what is the integral over (0,1]??

**Laurent** · February 10th 2009, 02:56 AM

Originally Posted by Amanda1990

If it is integrable, then what is the integral over (0,1]??

This is not an easy integral to compute...

By letting $u=\frac{1}{x}$ , you get $\int_0^1\sin\frac{1}{x}dx=\int_1^\infty \frac{\sin u}{u^2} du$

I don't think there is a simple formula for this one; does anyone know?

**CaptainBlack** · February 10th 2009, 05:24 AM

Originally Posted by Laurent

This is not an easy integral to compute...

By letting $u=\frac{1}{x}$ , you get $\int_0^1\sin\frac{1}{x}dx=\int_1^\infty \frac{\sin u}{u^2} du$

I don't think there is a simple formula for this one; does anyone know?

Integration by parts gives:

$\int \sin(1/x)\ dx = x \sin\left(\frac{1}{x}\right)-\text{Ci}\left(\frac{1}{x}\right) + C$

So:

$\int_0^1 \sin(1/x)\ dx = \sin(1)-\text{Ci}(1)-\lim_{a \to 0}\left[ a\sin\left(\frac{1}{a}\right)-\text{Ci}\left(\frac{1}{a}\right)\right]$ $=\sin(1)-\text{Ci}(1)\approx 0.504067$

CB

**Amanda1990** · February 10th 2009, 06:55 AM

Ok thanks very much for this. I see now that I was being a bit silly when I suggested that sin(1/x) was probably not measurable.

Perhaps this is a better question to highlight the problems I'm having with Lebesgue integration:

Is f(x) tan x integrable over the interval (0,pi/2)?

Now we can show that f is measurable on the interval, and hence continuous. However, it is unbounded and hence not necessarily integrable. I know we can say " $\int tan x = \log \sec x$ " and probably this to show the integral doesn't exist, but this isn't really the point. I'm after something that show from first principals or standard results (eg Monotone convergence theorem etc) why this isn't integrable over the stated interval. Thanks.

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