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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.

Michael Hardy
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Shreyas Shridharan
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  • Thank you in advance! – Shreyas Shridharan Jan 22 '18 at 02:51
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    Since $\sin$ is a periodic function, it is pretty much impossible to take the integral until infinity. – John Lou Jan 22 '18 at 02:53
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    I'm assuming you're taking the integral in terms of $x$. But then you also have $x$ as a lower bound. This is very ambiguous and typically frowned upon. – Badr B Jan 22 '18 at 02:56
  • I see. But If I were to take it to an arbitrarily large number, what would it equal? I don't see the logic in it (yet). (Reply to John Lou) – Shreyas Shridharan Jan 22 '18 at 02:56
  • Sorry, I've changed it. That must have slipped my mind... (Reply to Badr B) – Shreyas Shridharan Jan 22 '18 at 02:58
  • Let's take $I = \int_0^x \sin(x) dx$. Now, $x = 2\pi$. Clearly the area under the curve is $0$. Now let's go to $x = 3 \pi$. Now the area is $1$. Extending this logic, we have that if $x = 2\pi k$, the area is $0$. Since $\infty$ is a "number" we can't define, it's obvious that we can't tell what the area must be. – John Lou Jan 22 '18 at 03:14
  • I understand, but I find that it converges and ends up being sort of like $1 + (-1) + 1 + (-1)...$, which eventually, at $\lim_{x \to \infty}$, is 1/2. That certainly does not equal the areas of these curves. – Shreyas Shridharan Jan 22 '18 at 03:27
  • The sum you mention in your last comment is a well-known series and is actually divergent with Cauchy PV (or alternatively Caesaro sum) $1/2$ – Prasun Biswas Jan 22 '18 at 03:39
  • By changing the bounds to $0$ and $x$, there’s no reason to mention $\infty$ at all. Perhaps, you’re not thinking of the integral as a signed area, but instead thinking of $\int |\sin x|\ dx $ – pjs36 Jan 22 '18 at 04:34
  • You seem to have completely misunderstood the answer that you're quoting. It shows how to compute $\int_0^x$ using a Riemann sum (splitting the interval in many small parts, and then letting the number of parts tend to infinity while at the same time letting the parts themselves become smaller and smaller). That's something completely different from computing $\int_0^\infty$ or $\int_{-\infty}^\infty$. – Hans Lundmark Jan 22 '18 at 06:19
  • I want to close this question, as it is unclear, but I can't seem to do it. Can someone please vote to close this question? Thanks! – Shreyas Shridharan Jan 23 '18 at 00:58

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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.

Michael Hardy
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Nairit
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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 $x.$

Michael Hardy
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