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Rewriting I get

$$\lim_{n\rightarrow \infty}\int_{1}^{n}\frac{\sin{x^2}}{x^{\alpha}}dx.$$

Substituting $t=x^2,\quad dt=2xdx$ I get

$$\frac{1}{2}\lim_{n\rightarrow \infty}\int_{1}^{n}\frac{\sin{t}}{t\cdot x^{\alpha-1}}dt.$$

I don't really see where this leads me.

Jack D'Aurizio
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Parseval
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  • A formalist reader would ask what $x^{\alpha-1}$ is in your second integral. –  Oct 18 '17 at 07:33
  • Well, it's actually $$\frac{1}{2}\lim_{n\rightarrow \infty}\int_{1}^{n}\frac{\sin{t}}{x\cdot x^{\alpha}}dt.$$ Now I factor out an $x$ from $x^\alpha$ to get a $t$, but then I have to subtract $1$ from the exponent. – Parseval Oct 18 '17 at 07:37
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    You are missing the point: either you make a substitution (and thus you substitute everything that needs to be substituted) or you don't. There is no in-between. –  Oct 18 '17 at 07:44
  • You suggest I replace $x^{\alpha-1}$ with $\sqrt{t}^{\alpha-1}?$ How will this help me? – Parseval Oct 18 '17 at 08:10
  • I don't know, on top of my head. But at least you know what you are working with. –  Oct 18 '17 at 08:12
  • We can show that all $\alpha\geq 1$ works. Denoting $f(x)=\sin(x^2)$ and $g(x)=x^{-\alpha}$, note that $g(x)$ is bounded and monotonic on $[1,\infty)$ and $g(x)\to 0$ as $x\to\infty$. Also, note that $\int\limits_1^X f(x),\mathrm dx$ is bounded for all $X\geq 1$ (see Fresnel integral), so by Dirichlet's test, we see that $\int\limits_1^\infty f(x)g(x),\mathrm dx$ converges for all $\alpha\geq 1$ – Prasun Biswas Oct 18 '17 at 08:17
  • @PrasunBiswas - This is single variable calculus, we are not yet introduced to Fresnel integral. – Parseval Oct 18 '17 at 08:49
  • @G.Sassatelli $e^{(\alpha-1)\log(x)}$ – M. Van Oct 18 '17 at 10:28

1 Answers1

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Let us assume $\alpha> -1$, then perform the substitution $x\mapsto\sqrt{z}$ and a step of integration by parts: $$\int_{1}^{n}\frac{\sin(x^2)}{x^\alpha}\,dx = \frac{1}{2}\int_{1}^{n^2}\frac{\sin(z)}{z^{\frac{\alpha+1}{2}}}\,dz\stackrel{\text{IBP}}{=}\frac{1}{2}\left[\frac{1-\cos z}{z^{\frac{\alpha+1}{2}}}\right]_{1}^{n^2}+\frac{\alpha+1}{4}\int_{1}^{n^2}\frac{1-\cos z}{z^{\frac{\alpha+3}{2}}}\,dz$$ We get that $\lim_{n\to +\infty}\text{LHS}$ is finite iff $\lim_{n\to +\infty}\int_{1}^{n^2}\frac{1-\cos z}{z^{\frac{\alpha+3}{2}}}\,dz$ is finite. $1-\cos z$ is a periodic non-negative function with mean value $1$, hence (by integration by parts again) the wanted limit is finite iff $\int_{1}^{+\infty}\frac{dz}{z^{\frac{\alpha+3}{2}}}$ is finite, and any value of $\alpha > -1$ does the job.

The cases $\alpha=-1$ and $\alpha <-1$ can be studied through the same approach.

Jack D'Aurizio
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