i saw this $\int_{0}^{∞} sin(x)dx$ but i also saw the graph, plot here
i tried doing
$\lim_{a \to\ ∞} \int_{0}^{a} sin(x)dx$
but i dont know i get a $cos(a)-1$ and that seems to have no limit
i did this then, like grabbing chunks or intervals of 2π
$\lim_{n\to\infty} $ $\sum_{i=1}^{n}\int_{2π (n-1)}^{2πn} sin(x)dx=0$ but i don't know if that if rigorous or valid,
thanks
On
The limit of interest does not exist in the context of classical analysis.
However, in the theory of Generalized Functions (i.e., distribution theory), the limit as $\lim_{n\to\infty} e^{inx}=0$.
To see this, we appeal to the Riemann Lebesgue Lemma, which guarantees that if $\phi(x)$ is $L^1$ integrable, then
$$\lim_{n\to \infty}\int_{-\infty}^\infty \phi(x)e^{inx}\,dx=0 \tag 1$$
Therefore, in the context of the regularization of a distributiton, we can write
$$\lim_{n\to \infty}e^{inx} \sim 0 \tag 2$$
where $(2)$ is interpreted to mean that $(1)$ holds.
NOTE:
Formal (non-rigorous) application of distribution theory is used pervasively by physicists and engineers to obtain results (often correct ones) very quickly without the need to enforce rigor. An example of such application is the formal use of "Dirac Delta" to define Fourier integrals that do not converge classically.
Since the antiderivative of $\sin(x)$ is $-\cos(x)$, we have that
$$\int_a^b \sin x\ dx = -\cos(b)+\cos(a)$$
So,
$$\lim_{a\to\infty} \int_0^a\sin x\ dx = \lim_{a\to\infty} 1-\cos(a)$$
(All of this you had.) Since $\cos(a)$ oscillates between $-1$ and $1$, this limit is undefined, and so is the integral.