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I was thinking of proving $\int_{0}^{\infty} \sin(x) dx$ is unbounded?

Graphically the areas get added below the curve, but it seems to be adding equal positive and negative areas, just like $+A -A +A -A +A -A +....$ where $A$ represents the area between the curve and the x coordinates $0$ to $\pi$ but why the definite integral is unbounded?. Is it is bounded though?

Next I thought of proving it by contradiction, like let the definite integral be bounded then I was thinking how do we get a contradiction?

user
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BAYMAX
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    $\lim_{x \to \infty} \sin(x)$ doesn't exist. – Matthew Cassell May 14 '18 at 12:21
  • If you can't rearrange the terms, the infinite sum $A - A + A - A + \ldots$ is bounded above by $A$ and below by $0$. – Bilbottom May 14 '18 at 12:23
  • @Mattos did you mean to write $\lim_{x \to \infty} \cos(x)$? Otherwise I don't see how the remark is relevant. For if $f = \chi_{\mathbb{N}}$ is the indicator function of the natural numbers, then $\lim_{x \to \infty} f(x)$ does not exist, yet $\int_0^\infty f$ converges. – Qeeko May 14 '18 at 12:45

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It's not unbounded. However, the limit $\displaystyle\lim_{x\to\infty} \int_0^x \sin t \,dt$ does not exist as you point out.

Ayman Hourieh
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