By using Taylor series, I managed to see that the $t \rightarrow 0$, $\frac{2\cos(t)\sin(t)}{t} = 1 -\frac{4t^2}{6}$ whose integral converges. As for the case when $t \rightarrow \infty$, I don't know that to do. It equals to $\int^{\infty}_{0}\frac{\sin(2t)}{t}dt$. What can I do?
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Not that it helps, but you can simplify slightly to $\int_0^{\infty}\frac{\sin u}{u};du$ by taking $u=2t$. Do you know about convergence of this integral? – MPW Dec 29 '16 at 18:41
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@MPW I have a feeling it doesn't converge, but can't show it. – user401855 Dec 29 '16 at 18:43
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Consider the tail of the integral from R to infinity and perform a partial integration where you differentiate the 1/t factor. – Count Iblis Dec 29 '16 at 18:46
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what is $$\sint(t)$$? or is this a typo? – Dr. Sonnhard Graubner Dec 29 '16 at 18:46
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http://math.stackexchange.com/questions/5248/solving-the-integral-int-0-infty-frac-sinxx-dx-frac-pi2 – Behrouz Maleki Dec 29 '16 at 18:47
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the result should be $$\frac{\pi}{2}$$ – Dr. Sonnhard Graubner Dec 29 '16 at 18:48
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Yes, it does converge. Try googling "sinc" – MPW Dec 29 '16 at 19:39
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$$\eqalign{\int_0^{\infty} \frac{\sin(2t)}{t}\; dt &= \sum_{n=0}^\infty \int_{n\pi/2}^{(n+1)\pi/2} \frac{\sin(2t)}{t}\; dt \cr &= \sum_{n=0}^\infty \int_{0}^{\pi/2} \sin(2t) \left(\frac{1}{t+n\pi/2} - \frac{1}{t+(n+1)\pi/2}\right)\cr &= \sum_{n=0}^\infty \int_0^{\pi/2} \frac{2\pi\;\sin(2t)}{n^2 \pi^2 + \ldots}}$$ Now estimate...
Robert Israel
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$\int^{\infty}_{0}\frac{sin(2t)}{t}dt$. exists if $\int^{\infty}_{0}\frac{sin(2t)} {t}dt$ exists
exists if $(1)\lim_{a\to0}\int^{1}_{a}\frac{sin(t)}{t}dt$ exists and $(2)\lim_{a\to\infty}\int^{b}_{1}\frac{sin(t)}{t}dt$ exists
$(1)$ on $[0;1], \sin(x)\leq x, \sin(x)/x\leq1$
$\int^{1}_{a}\frac{sin(t)}{t}dt \leq \int^{1}_{a}dt$, so $\lim_{a\to0}\int^{1}_{a}\frac{sin(t)}{t}dt$ exists
$(2)$ let $u_n = \int^{\pi(n+2)}_{\pi(n+1)}\frac{sin(2t)}{t}dt $
$\lim_{a\to\infty}\int^{b}_{1}\frac{sin(t)}{t}dt$ exists if $\sum u_n$ converges. $\sum u_n$ is an alternating series, so $\lim_{a\to\infty}\int^{b}_{1}\frac{sin(t)}{t}dt$ exists.
Baconaro
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