I already tried to solve it myself and to type it into the calculator, but it gives out some weird stuff with an endless long solution.
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1What if you switch the integrals? – Michael Dec 12 '16 at 15:16
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Then my calculator tells me that there is no limit. – Julian Dec 12 '16 at 15:17
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2Try doing it without your calculator. – Michael Dec 12 '16 at 15:20
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Yes, I did, but I don't have an approach for it. – Julian Dec 12 '16 at 15:22
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1What if you switch the integrals? – Michael Dec 12 '16 at 15:22
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In order to calculate it by hand? Is that allowed? – Julian Dec 12 '16 at 15:23
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2Yes, calculating by hand is allowed. – Michael Dec 12 '16 at 15:23
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No, to switch the integrals. – Julian Dec 12 '16 at 15:24
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1My guess is that this is a "Fubini theorem" type problem, where part of the problem is justifying the switch. But before that, you should at least do a calculation to show that a switch is advantageous. – Michael Dec 12 '16 at 15:26
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1Ah, I got it. Switching it gives me a cosinus-like result, and putting in the values gives me $0$. – Julian Dec 12 '16 at 15:27
2 Answers
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Since $\left|\sin\right|\leq 1$, Fubini's theorem applies, hence: $$ I=\int_{0}^{2\pi}\int_{1}^{+\infty}\frac{\sin(s-t)}{t^3}\,dt\,ds = \int_{1}^{+\infty}\frac{1}{t^3}\int_{0}^{2\pi}\sin(s-t)\,ds\,dt = 0. $$ But you don't even need Fubini's theorem to state the same: $$ \int_{1}^{+\infty}\frac{\sin(s-t)}{t^3}\,dt = \sin(s)\int_{1}^{+\infty}\frac{\cos(t)}{t^3}\,dt - \cos(s)\int_{1}^{+\infty}\frac{\sin(t)}{t^3}\,dt $$ where both $\int_{1}^{+\infty}\frac{\cos(t)}{t^3}\,dt$ and $\int_{1}^{+\infty}\frac{\sin(t)}{t^3}\,dt$ are converging integrals, bounded by $\int_{1}^{+\infty}\frac{dt}{t^3}=\frac{1}{2}$ in absolute value. It follows that: $$ I = \int_{0}^{2\pi}\left(A \sin(s)+B\cos(s)\right)\,ds = 0.$$
Jack D'Aurizio
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Zero is the correct result.
Switching the integrals you have
$$\int_1^{+\infty}\frac{\text{d}t}{t^3}\int_0^{+2\pi} \sin(s-t)\ \text{d}s = \int_1^{+\infty}\frac{\text{d}t}{t^3}\left(-\cos(s-t)\bigg|_{0}^{+2\pi}\right)$$
But
$$\left(-\cos(s-t)\bigg|_{0}^{+2\pi}\right) = 0$$
Hence
$$\int_1^{+\infty} \frac{\text{d}t}{t^3}\cdot 0 = 0$$
Enrico M.
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@user159517 I don't have to give an entire lesson about Fubini's theorem and so on. If one studies, then he can understand by his own why it's legit to switch. – Enrico M. Dec 12 '16 at 16:17
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2What a strange reasoning. Why give the half you gave then? The same reasoning would apply, no? – Did Dec 12 '16 at 16:38