Integral of sin(x)

Question

I often hear that $$\int_0^x \sin(x)\,dx$$ is equal to $\cos(0)−\cos(x)$:

Now if we take the limit as $n \to \infty $ we see $ \frac{x}{2n} \to 0$ and $$\int_0^x \sin t \, dt = \lim_{n \to \infty}\frac{x}{n}\sum_{k=1}^n\sin \left(\frac{kx}{n} \right) = 2\sin^2 \left(\frac{x}{2}\right) = 1 - \cos x = \cos 0 - \cos x.$$

(Sourced from here), but I still can't figure out why this number seems so small for an integral that bounds from $-\infty$ to $\infty$. Can someone please explain?

Note: I have also heard that it equals $0$, but that seems even more unreasonable due to the definition of an integral - the area between the curve and the x axis.

Answer 1

The Fundamental Theorem of Calculus says that if $F(x)$ is the indefinite integral of $f(x)$ then $\displaystyle\int_a^b f(x) = F(b)-F(a)$. It is well known that $\displaystyle\int \sin(x) = -\cos(x)+C$, so $$\displaystyle\int_{-\infty}^{\infty} \sin(x)=-\cos(\infty)-(-\cos(-\infty)) = \cos(-\infty)-\cos(\infty)$$ To compute this, we need to compute $\displaystyle\lim_{x\to\infty}\sin(x)$. However, we can't compute $\displaystyle\lim_{x\to\infty}\sin(x)$ as it turns out to be divergent. (This follows from the Limit Divergence Criterion Test).

This means that the quantity $\displaystyle\int_{-\infty}^\infty \sin(x)\,dx$ is undefined.

Answer 2

It is not from $-\infty$ to $+\infty;$ it is from $0$ to $x.$

Notice that you wrote $\displaystyle\int_0^x \sin t\,dt = \lim_{n\to\infty} \cdots\cdots.$

The thing to the right of the "equals" sign must be equal to the thing to its left, and that is an integral from $0$ to $x.$