$$f(x)=\int_x^{x+1}\sin(t^2)dt$$ I have to prove that $|f(x)| < \frac{1}{x}$ for $x>0$. then I have to show that $2xf(x) = \cos(x^2) - \cos(x+1)^2 + r(x)$ where $|r(x)| < \frac{c}{x}$ for some constant $c$. I used the fundamental theorem of calculus and calculated its derivative, but I don't know how to continue. Any hint?
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5$2t\sin(t^2)\cdot{1\over 2 t}$ and integration by parts. – Ryszard Szwarc Jun 09 '23 at 20:17
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1You can prove the first statement by using an upper and lower Reimann sum. If it find one that’s less than $\frac{1}{x}$ then the integral will be less than that as well – moboDawn_φ Jun 09 '23 at 20:20
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1Compare https://math.stackexchange.com/q/259313/42969 – Martin R Jun 09 '23 at 20:21
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1Also: https://math.stackexchange.com/q/3244801/42969, https://math.stackexchange.com/q/2243337/42969 – Martin R Jun 09 '23 at 20:21
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@OliverDíaz I doubt it: in that post, these two questions are supposed to be solved and the problem is about another question. – Anne Bauval Jun 11 '23 at 06:56
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In fact, by IBP, one has $$f(x)=\int_x^{x+1}\sin(t^2)dt=-\int_x^{x+1}\frac{1}{2t}d\cos(t^2)=-\frac{1}{2t}\cos(t^2)\bigg|_x^{x+1}+\frac12\int_x^{x+1}\frac{\cos(t^2)}{t^2}dt$$ which implies $$ |f(x)|\le \bigg|\frac{1}{2t}\cos(t^2)\bigg|_x^{x+1}\bigg|+\frac12\int_x^{x+1}\frac{1}{t^2}dt\le\frac12(\frac1x+\frac1{x+1})+\frac12(\frac{1}{x}-\frac1{x+1})=\frac1x.$$
xpaul
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