Why is the function $f(x) = \sin(2\pi x)$ not Lebesgue Integrable over $E$ such that $E = [1, \infty)$

Question

I am curious as to why the function $f(x) = \sin(2\pi x)$ is not Lebesgue integrable over $E$. My professor claimed it wasn't but didn't really tell us why.

Answer 1

$$\int_1^{\infty} |\sin{2\pi x}|dx \geq \frac{1}{2\pi}\int_{2\pi}^{\infty}|\sin{t}|dt= \frac{1}{2\pi}\sum_{n=1}^{\infty}\int_{2k\pi }^{(2k+1)\pi}|\sin{t}|dt $$$$=\frac{1}{2\pi}\sum_{n=1}^{\infty}\int_{2k\pi }^{(2k+1)\pi}\sin{t}dt=\frac{1}{2\pi}\sum_{k=1}^{\infty}2=+\infty$$

Answer 2

if $f$ was Lebesgue integrable, then as $f$ is continuous, the improper Riemann integral would exist and be equal to the Lebesgue integral. Now, the improper Riemann integral does not exist, hence a contradiction

Answer 3

Well, such examples play on technicalities in some of the definitions of "integrable"... as opposed to genuine mathematical phenomena.

The usual definition of "Lebesgue integral" requires that we be able to separately integrate the positive part and negative part (and positive/negative parts of imaginary part if complex-valued...), so this version of the official definition cannot cope with cancellation. A more relevant example to illustrate the failure is $\int_0^\infty {\sin x\over x}\;dx$, which gives "undefined" as Lebesgue integral, while gives an interesting value (which I unfortunately forget at this moment) as Riemann integral, due to cancellation.

With $\int_1^\infty \sin x\;dx$ or similar, no definition of "integral" returns a value. The words used to describe "does not return a value" vary, but the functionality is the same: there is no value that can be attached to this compatible with certain constraints.

On another hand, which seems significant as well as provocative in an immediate way, such an integral could be construed as trying to compute the Fourier transform of the characteristic function of $[1,\infty)$. This is, in fact, a sensible thing, and interacts well with other parts of mathematics...