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I'm trying to prove that $\displaystyle\int_{0}^{\infty}\frac{\sin^{2}x}{x^{1+\alpha}}dx$ is convergent when $\alpha\in(0,2).$

Split the integral in $\displaystyle\int_{0}^{1}\frac{\sin^{2}x}{x^{1+\alpha}}dx+\displaystyle\int_{1}^{\infty}\frac{\sin^{2}x}{x^{1+\alpha}}dx$ and check the cases when $\alpha\in(0,1)$ and $\alpha\in[1,2)$ are the cases that I've proved the convergence of the integral except the next:

I'm stuck proving the case $\alpha\in[1,2)$ for $\displaystyle\int_{0}^{1}\frac{\sin^{2}x}{x^{1+\alpha}}dx.$ I can't find a way to bound that integral with another convergent.

Any kind of help is thanked in advanced.

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The sine function satisfies $|\sin x| \le |x|$ for all $x$; as such,

$$\int_0^1 \frac{\sin^2 x}{x^{1 + \alpha}} \, dx \le \int_0^1 x^{2 - 1 - \alpha} \, dx$$

converges whenever $1 - \alpha > -1$. That is, whenever $\alpha < 2$.

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  • $\begingroup$ Thanks @user296602. I forgot these fact of sine function. $\endgroup$ – Squird37 Mar 2 '18 at 18:23
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$$0\leq \int_{0}^{+\infty}\frac{\sin^2(x)}{x^{1+\alpha}}\,dx = \int_{0}^{1}\frac{\sin^2(x)}{x^{1+\alpha}}\,dx +\int_{1}^{+\infty}\frac{\sin^2(x)}{x^{1+\alpha}}\,dx\\ \leq \int_{0}^{1}\frac{x^2}{x^{1+\alpha}}\,dx +\int_{1}^{+\infty}\frac{1}{x^{1+\alpha}}\,dx = \frac{2}{a(2-a)}.$$ The actual value of the integral is $\frac{\pi\cdot 2^{a-2}}{\Gamma(a+1)\sin\frac{\pi a}{2}}.$

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  • $\begingroup$ Thanks for help me @Jack D'Aurizio. I voted for the other answer only for the time to answer. $\endgroup$ – Squird37 Mar 2 '18 at 18:24

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