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$\int_{0}^{\infty} \cos(x) , dx$

It's like the grandi series I integrated it separately From 0 to $\frac{\pi}{2}$ Then from $\frac{\pi}{2}$ to $\pi$ Then from $\pi$ to $\frac{3\pi}{2}$ And lastly from $\frac{3\pi}{2}$ to $2\pi$

This will repeat You'll get the grandi series That is, 1-1+1-1+1-1+1-1.......... And we know it's $\frac{1}{2}$

I am asking it seriously

Martin.s
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    Grandi is divergent. – David G. Stork Sep 25 '23 at 21:09
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    "That is, $1-1+1-1+1-1\dots$ and we know it's $\frac{1}{2}$" Next you'll try to convince people that $1+2+3+\dots$ is equal to $-\frac{1}{12}$. Spoiler alert... it's not. You seem to have misread and misunderstood certain factoids and assumed that they are in their pop-sci layman descriptions actually accurate identities when they are not. They come with heavy caveats and require nonstandard interpretations and approaches which fundamentally change the questions from how they are otherwise interpreted in the usual contexts. – JMoravitz Sep 25 '23 at 21:15
  • https://math.stackexchange.com/questions/1889404/how-can-we-know-the-answer-to-1-11-11 and https://math.stackexchange.com/questions/635324/checking-my-understanding-1-1-1-1-1-frac12 – Al.G. Sep 25 '23 at 21:23

1 Answers1

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This improper integral diverges.

$$ \int_0^T\cos(x)\;dx = \sin T, \\ \text{the limit} \quad \lim_{T \to +\infty} \sin T \quad \text{does not exist}. $$

However, the "mean value" does exist in the sense $$ \lim_{T \to +\infty}\frac{1}{T}\int_0^T\cos x\;dx = 0 . $$

Some physicists will tell you $$ \delta(t) = \frac{1}{2\pi}\int_{-\infty}^{+\infty} e^{i x t}\,dx $$ so that, if $t \ne 0$ is real, $$ 0 = \delta(t) = \frac{1}{\pi} \int_{0}^{+\infty} \cos(x t)\,dx . $$ So $t=1$ gives us the OP. If you know about Schwartz distributions, you can make sense of this. (But if you don't, you can't.)

GEdgar
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